| Interpolation [message #2184] |
Thu, 02 June 1994 10:44  |
jun
Messages: 9
Registered: June 1994
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Junior Member |
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I am using IDL to process and display 3D data. At some stage, I have to
interpolate a volume p=p(x,y,z) or a surface z=z(x,y) defined on an irregular
grid onto a regular grid. Does any one know any IDL program that has this
capability?
Thanks in advance.
jun@eureka.columbia.edu
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| Re: Interpolation [message #2192 is a reply to message #2184] |
Fri, 03 June 1994 01:14  |
stl
Messages: 70
Registered: February 1994
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Member |
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In article <2sl5qo$h1q@apakabar.cc.columbia.edu> jun@eureka.phys.columbia.edu (Jun Xu) writes:
> I am using IDL to process and display 3D data. At some stage, I have to
> interpolate a volume p=p(x,y,z) or a surface z=z(x,y) defined on an irregular
> grid onto a regular grid. Does any one know any IDL program that has this
> capability?
hi,
Well, it depends on how mathimaticly correct you need this stuff. I
believe you could do some very very complicated stuff here, but there is
a relatively simple solution.
Use the trigrid and triangulate procedures in IDL. In the Reference
manual, there is a good example in the Trigrid section of how this
works.
-hope this help (sorry, but I am no mathimatician. Cannot spell
fro squate either!)
-stephen
--
Stephen C Strebel / SKI TO DIE
stl@maz.sma.ch / and
Swiss Meteorological Institute, Zuerich / LIVE TO TELL ABOUT IT
01 256 93 85 / (and pray for snow)
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| Re: Interpolation [message #59871 is a reply to message #2184] |
Mon, 14 April 2008 13:13 |
tarequeaziz
Messages: 20
Registered: November 2007
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Junior Member |
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On Apr 14, 11:34 am, Spon <christoph.b...@gmail.com> wrote:
> On Apr 13, 1:56 am, tarequea...@gmail.com wrote:
>
>> Step one: Real data in a XY frame
>> Step two: 'Design a new xy frame, say X'Y' frame, whose values are
>> generated from a chosen r_vec and theta_vec.
>> Step 3: Now interpolate from XY to X'Y'.
>
>> Tareque
>
> Hi Tareque,
>
> I'm guessing you know whereabouts your small frame is, within your big
> frame, right?
> So, if you take your big normal (x'y') frame, your small frame can be
> defined by two points,
> bottom-left and top-right - let's call them (b,l) & (t,r) - in terms
> of x'y' grid co-oords.
>
> -----
> ; Once you've worked out where these two points are, you can use
> CONGRID on your xy dataset:
> tempx = r - l ; How many data points of the x'y' grid does the xy grid
> span
> tempy = t - b ; in each dimension?
>
> ; interpolate to new sub-grid
> newdata = congrid(data, tempx, tempy)
>
> ; Your x'y' frame co-ordinates for this data are
> newx = l + lindgen(tempx)
> newy = b + lindgen(tempy)
>
> ; (this bit is just array juggling to avoid for loops)
> newx = rebin(newx,tempx,tempy)
> newy = rebin(reform(newy,1,tempy),tempx,tempy)
> newx = reform(newx,n_elements(newx))
> newy = reform(newy,n_elements(newy))
>
> ; x'y' co-ordinates for ever datapoint in 'newdata'
> xycoords = transpose([[newx],[newy]])
>
> ; so your new data should be at r/theta co-ordinates defined by:
> polarcoords = cv_coord(from_rect = xycoords, /to_polar)
> -----
>
> I've assumed that your big circle is centered on the origin.
> I've also assumed your small circle is in the upper-right quadrant of
> your large circle here,
> so I don't have to wrap my mind around minus-signs and the like...
>
> I hope this helps and that I've understood your question
> correctly. :-)
>
> Regards,
> Chris
Hi Chris,
Thank you so much for getting back at this.
Without your permission I sent a picture of my set up. Hope that will
be able to shed some light on it.
Once again, much appreciated!
Best,
Tareque
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| Re: Interpolation [message #59877 is a reply to message #2184] |
Mon, 14 April 2008 08:34 |
Spon
Messages: 178
Registered: September 2007
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Senior Member |
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On Apr 13, 1:56 am, tarequea...@gmail.com wrote:
> Step one: Real data in a XY frame
> Step two: 'Design a new xy frame, say X'Y' frame, whose values are
> generated from a chosen r_vec and theta_vec.
> Step 3: Now interpolate from XY to X'Y'.
>
> Tareque
Hi Tareque,
I'm guessing you know whereabouts your small frame is, within your big
frame, right?
So, if you take your big normal (x'y') frame, your small frame can be
defined by two points,
bottom-left and top-right - let's call them (b,l) & (t,r) - in terms
of x'y' grid co-oords.
-----
; Once you've worked out where these two points are, you can use
CONGRID on your xy dataset:
tempx = r - l ; How many data points of the x'y' grid does the xy grid
span
tempy = t - b ; in each dimension?
; interpolate to new sub-grid
newdata = congrid(data, tempx, tempy)
; Your x'y' frame co-ordinates for this data are
newx = l + lindgen(tempx)
newy = b + lindgen(tempy)
; (this bit is just array juggling to avoid for loops)
newx = rebin(newx,tempx,tempy)
newy = rebin(reform(newy,1,tempy),tempx,tempy)
newx = reform(newx,n_elements(newx))
newy = reform(newy,n_elements(newy))
; x'y' co-ordinates for ever datapoint in 'newdata'
xycoords = transpose([[newx],[newy]])
; so your new data should be at r/theta co-ordinates defined by:
polarcoords = cv_coord(from_rect = xycoords, /to_polar)
-----
I've assumed that your big circle is centered on the origin.
I've also assumed your small circle is in the upper-right quadrant of
your large circle here,
so I don't have to wrap my mind around minus-signs and the like...
I hope this helps and that I've understood your question
correctly. :-)
Regards,
Chris
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| Re: Interpolation [message #59883 is a reply to message #2184] |
Sat, 12 April 2008 17:56 |
tarequeaziz
Messages: 20
Registered: November 2007
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Junior Member |
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On Apr 12, 6:26 pm, "ben.bighair" <ben.bigh...@gmail.com> wrote:
> On Apr 12, 4:24 pm, tarequea...@gmail.com wrote:
>
>
>
>> On Apr 12, 11:57 am, "ben.bighair" <ben.bigh...@gmail.com> wrote:
>
>>> On Apr 11, 11:05 pm, tarequea...@gmail.com wrote:
>
>>>> Hello All (IDL Gods),
>
>>>> I am back with yet another problem.
>>>> I know...I know...its friday night. I apologize for that. But I am
>>>> really stuck here for a while.
>
>>>> The problem:
>
>>>> In a certain part of my code I need to do interpolation. The data that
>>>> I am dealing with are from a XY grid. I need to convert them to polar
>>>> coordinate. So, what I do is following:
>
>>>> a. I generate a radius vector r_vec and a theta array containing
>>>> values from zero to 2 pi.
>>>> b. Now use the simple polar-to-rectangular coordinate transform i.e. x
>>>> = r cos(theta) etc.
>>>> c. Using these values I have a xy grid generated through a known r-
>>>> theta values.
>>>> d. Now think of superimposing the new xy grid(lets call it x'y' ) on
>>>> to the old xy grid which contains the real data.
>>>> e. This is the part where I need interpolation. I do interpolations to
>>>> get the x'y' values from the xy points.
>
>>>> So here's the question:
>
>>>> ----- Is there any elegant way of doing this coordinate
>>>> transformation ? (And in case you are thinking, "well you already have
>>>> the xy data, so why not just convert to r-theta?", I have to say that
>>>> the interpolation method actually gives me a way nicer dataset ).
>
>>>> My 2nd trouble is, and this is probably the biggest and dumbest
>>>> problem for me.
>
>>>> ----- i was playing around with several interpolation routines from
>>>> IDl. My boss's suggestion was to use 'bilinear'.
>>>> but I thought to give others' a shot too. Problem is, when I am
>>>> done with interpolation, result is nothing like what I was
>>>> expecting. A run down version of the code is shown below:
>
>>>> =======================start of code================================
>
>>>> Nth= 10.
>>>> dth= 1/Nth
>>>> r_vec= findgen(Nth)/Nth
>>>> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>
>>>> for i=0L,Nth - 1 do begin
>
>>>> x[i]= r_vec[i]* cos(theta_vec)
>
>>>> endfor
>
>>>> for j=0L,Nth - 1 do begin
>
>>>> y[j]= r_vec[j]* sin(theta_vec)
>
>>>> endfor
>
>>>> ;print,y
>
>>>> ;plot,x,y
>
>>>> ------------------------------------------------------------ -----------
>>>> Now I create the 'main' dataset on which I am going to use
>>>> interpolation scheme.
>
>>>> rr = findgen(20.)/30.
>>>> tht = findgen(20.)/30. *2*!Pi
>
>>>> m = fltarr(20,20)
>
>>>> for j=0,19 do begin
>>>> for i=0,19 do begin
>
>>>> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
>>>> ;print,i
>>>> endfor
>>>> endfor
>
>>>> m_p=bilinear(m,x,y)
>
>>>> End
>>>> =======================End of Code=================================
>
>>>> The problem is, as I mentioned above, when I plot m and the
>>>> interpolated m_p, they do not look like similar at all.
>
>>>> Any help will be greatly appreciated.
>
>>>> Thanks in advance.
>
>>>> ~tareque
>
>>> Hi,
>
>>> Coordinate transforms are very easy with CV_COORD().
>
>>> I don't understand where you want to go with the interpolation. The
>>> input coordinates to BILINEAR are rectangular and formed as indexed
>>> based (as in subscript indices in X and Y.) As near as I can
>>> reconstruct, X ranges from -0.500000 to 0.728115 and Y ranges
>>> from -0.760845 to 0.285317. My reconstruction of your code is
>>> below.
>
>>> Cheers,
>>> Ben
>
>>> PRO tareque
>
>>> Nth= 10L
>>> dth= 1.0/Nth
>>> r_vec= findgen(Nth)/Nth
>>> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>>> x = fltarr(nth)
>>> y = fltarr(nth)
>
>>> for i=0L,Nth - 1 do begin
>
>>> x[i]= r_vec[i]* cos(theta_vec[i])
>
>>> endfor
>
>>> for j=0L,Nth - 1 do begin
>
>>> y[j]= r_vec[j]* sin(theta_vec[j])
>
>>> endfor
>
>>> ;print,y
>
>>> ;plot,x,y
>
>>> ;----------------------------------------------------------- ------------
>>> ;Now I create the 'main' dataset on which I am going to use
>>> ;interpolation scheme.
>
>>> rr = findgen(20.)/30.
>>> tht = findgen(20.)/30. *2*!Pi
>
>>> m = fltarr(20,20)
>
>>> for j=0,19 do begin
>>> for i=0,19 do begin
>
>>> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
>>> ;print,i
>>> endfor
>>> endfor
>
>>> m_p=bilinear(m,x,y)
>
>>> STOP
>>> End
>
>> hi Ben,
>
>> Thanks for getting back to me on this.
>
>> I guess I could not clearly state my problem before. So, I am going to
>> give it another try:
>
>> The data I deal with comes from a ccd camera and this goes some
>> tinkering and tweaking (to account for background noise and stuff).
>> Then this 'processed' data is in need for interpolation. why? well the
>> reason being this, that these processed data are actually in XY
>> frame.
>> We want them to be on a polar coordinate. The reason I cannot use
>> 'cv_coord' is because in that case I will get the data in a r-theta
>> plane which is
>> directly correlated to the experimental XY grid. If we use
>> interpolation, then we can be on a XY grid (yes, it is XY grid) which
>> is carefully 'designed' according to
>> our 'own' r-theta values. So in this way we have more leverage over
>> data manipulation.
>
>> Was it any clearer than before?
>
>> And one more thing, I could not find much changes in your version
>> except for an inclusion of 'STOP'. Did i miss something here??
>
>> Thanks again for your time.
>> Much appreciated !
>
>> ~ Tareque
>
> Hi Again,
>
> The only thing I did other than STOP (which I forgot to remove) was
> redefine X and Y, which in your example were missing. I thought it
> would be helpful for others if the whole thing worked. I should have
> been clearer.
>
> Speaking of clearer... I think I am more confused now. You might
> need the intervention from the IDL gods that you originally
> petitioned. Regardless of the parts I don't understand, the thing I
> do get is that BILINEAR is looking for you to provide interpolation
> coords that fit within the dimensions of the input array to be
> sampled.
>
> Take a peek at the second example in the online help for BILINEAR...
> (slightly modified here...)
> P = FINDGEN(3,3)
> IX = [[0.5, 1.9], [1.1, 2.2]] ;Define the X subscripts.
> JY = [[0.1, 0.9], [1.2, 1.8]] ;Define the Y subscripts.
> Z = BILINEAR(P, IX, JY, MISSING = !VALUES.F_NAN) ;Interpolate.
> PRINT, P
> PRINT,Z
> P prints as...
> 0.00000 1.00000 2.00000
> 3.00000 4.00000 5.00000
> 6.00000 7.00000 8.00000
> Z prints as ...
> 0.800000 4.60000
> 4.70000 NaN
>
> Note that IX and JY are mostly falling within the 0-2 range of indices
> for a 3x3 array. BILINEAR can extrapolate - just remove the MISSING
> keyword - but it can only extrapolate a little ways. In your example
> IX are sampled at
>
> 0.00000 0.0809017 0.0618034 -0.0927051 -0.323607
> -0.500000 -0.485410 -0.216312 0.247214 0.728115
>
> and JY are sampled at ...
>
> 0.00000
> 0.0587785
> 0.190211
> 0.285317
> 0.235114
> -4.37114e-08
> -0.352671
> -0.665740
> -0.760845
> -0.529007
>
> Which to my way of thinking your are interpolating all around either
> side of the [0,0] location. Is that what you intended? It is
> possible that in my redo of your example I corrupted the X and Y
> values you intended. In which case I would be not helping very much
> at all!
>
> Cheers,
> Ben
Hi Ben,
Thanks once again for your reply and also taking time to look into
this problem.
I am trying to attach a picture of my problem. well, looks like I cant
attach a picture here.
Anyways, I am going to give another shot at this. And,once again Boss,
thanks so much for your help.
Alright, lets start:
Think of a big circle. Close to the periphery, there is another little
circle inside of that big circle. Now, this little guy's positions are
the
bones of contentions. The coordinate system used are just rectangular
ones. We want them to be in polar.
At the same time I want the these positions in such a way that they
corresponds to a r-theta grid 'designed' by me.
That's why we are trying to interpolate.We want to get the
interpolated points based on our desired r-theta values.
In other words
Step one: Real data in a XY frame
Step two: 'Design a new xy frame, say X'Y' frame, whose values are
generated from a chosen r_vec and theta_vec.
Step 3: Now interpolate from XY to X'Y'.
Now how does it sound?
Once again, Thank you so much for your help.
Best,
Tareque
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| Re: Interpolation [message #59885 is a reply to message #2184] |
Sat, 12 April 2008 15:26 |
ben.bighair
Messages: 221
Registered: April 2007
|
Senior Member |
|
|
On Apr 12, 4:24 pm, tarequea...@gmail.com wrote:
> On Apr 12, 11:57 am, "ben.bighair" <ben.bigh...@gmail.com> wrote:
>
>
>
>> On Apr 11, 11:05 pm, tarequea...@gmail.com wrote:
>
>>> Hello All (IDL Gods),
>
>>> I am back with yet another problem.
>>> I know...I know...its friday night. I apologize for that. But I am
>>> really stuck here for a while.
>
>>> The problem:
>
>>> In a certain part of my code I need to do interpolation. The data that
>>> I am dealing with are from a XY grid. I need to convert them to polar
>>> coordinate. So, what I do is following:
>
>>> a. I generate a radius vector r_vec and a theta array containing
>>> values from zero to 2 pi.
>>> b. Now use the simple polar-to-rectangular coordinate transform i.e. x
>>> = r cos(theta) etc.
>>> c. Using these values I have a xy grid generated through a known r-
>>> theta values.
>>> d. Now think of superimposing the new xy grid(lets call it x'y' ) on
>>> to the old xy grid which contains the real data.
>>> e. This is the part where I need interpolation. I do interpolations to
>>> get the x'y' values from the xy points.
>
>>> So here's the question:
>
>>> ----- Is there any elegant way of doing this coordinate
>>> transformation ? (And in case you are thinking, "well you already have
>>> the xy data, so why not just convert to r-theta?", I have to say that
>>> the interpolation method actually gives me a way nicer dataset ).
>
>>> My 2nd trouble is, and this is probably the biggest and dumbest
>>> problem for me.
>
>>> ----- i was playing around with several interpolation routines from
>>> IDl. My boss's suggestion was to use 'bilinear'.
>>> but I thought to give others' a shot too. Problem is, when I am
>>> done with interpolation, result is nothing like what I was
>>> expecting. A run down version of the code is shown below:
>
>>> =======================start of code================================
>
>>> Nth= 10.
>>> dth= 1/Nth
>>> r_vec= findgen(Nth)/Nth
>>> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>
>>> for i=0L,Nth - 1 do begin
>
>>> x[i]= r_vec[i]* cos(theta_vec)
>
>>> endfor
>
>>> for j=0L,Nth - 1 do begin
>
>>> y[j]= r_vec[j]* sin(theta_vec)
>
>>> endfor
>
>>> ;print,y
>
>>> ;plot,x,y
>
>>> ------------------------------------------------------------ -----------
>>> Now I create the 'main' dataset on which I am going to use
>>> interpolation scheme.
>
>>> rr = findgen(20.)/30.
>>> tht = findgen(20.)/30. *2*!Pi
>
>>> m = fltarr(20,20)
>
>>> for j=0,19 do begin
>>> for i=0,19 do begin
>
>>> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
>>> ;print,i
>>> endfor
>>> endfor
>
>>> m_p=bilinear(m,x,y)
>
>>> End
>>> =======================End of Code=================================
>
>>> The problem is, as I mentioned above, when I plot m and the
>>> interpolated m_p, they do not look like similar at all.
>
>>> Any help will be greatly appreciated.
>
>>> Thanks in advance.
>
>>> ~tareque
>
>> Hi,
>
>> Coordinate transforms are very easy with CV_COORD().
>
>> I don't understand where you want to go with the interpolation. The
>> input coordinates to BILINEAR are rectangular and formed as indexed
>> based (as in subscript indices in X and Y.) As near as I can
>> reconstruct, X ranges from -0.500000 to 0.728115 and Y ranges
>> from -0.760845 to 0.285317. My reconstruction of your code is
>> below.
>
>> Cheers,
>> Ben
>
>> PRO tareque
>
>> Nth= 10L
>> dth= 1.0/Nth
>> r_vec= findgen(Nth)/Nth
>> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>> x = fltarr(nth)
>> y = fltarr(nth)
>
>> for i=0L,Nth - 1 do begin
>
>> x[i]= r_vec[i]* cos(theta_vec[i])
>
>> endfor
>
>> for j=0L,Nth - 1 do begin
>
>> y[j]= r_vec[j]* sin(theta_vec[j])
>
>> endfor
>
>> ;print,y
>
>> ;plot,x,y
>
>> ;----------------------------------------------------------- ------------
>> ;Now I create the 'main' dataset on which I am going to use
>> ;interpolation scheme.
>
>> rr = findgen(20.)/30.
>> tht = findgen(20.)/30. *2*!Pi
>
>> m = fltarr(20,20)
>
>> for j=0,19 do begin
>> for i=0,19 do begin
>
>> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
>> ;print,i
>> endfor
>> endfor
>
>> m_p=bilinear(m,x,y)
>
>> STOP
>> End
>
> hi Ben,
>
> Thanks for getting back to me on this.
>
> I guess I could not clearly state my problem before. So, I am going to
> give it another try:
>
> The data I deal with comes from a ccd camera and this goes some
> tinkering and tweaking (to account for background noise and stuff).
> Then this 'processed' data is in need for interpolation. why? well the
> reason being this, that these processed data are actually in XY
> frame.
> We want them to be on a polar coordinate. The reason I cannot use
> 'cv_coord' is because in that case I will get the data in a r-theta
> plane which is
> directly correlated to the experimental XY grid. If we use
> interpolation, then we can be on a XY grid (yes, it is XY grid) which
> is carefully 'designed' according to
> our 'own' r-theta values. So in this way we have more leverage over
> data manipulation.
>
> Was it any clearer than before?
>
> And one more thing, I could not find much changes in your version
> except for an inclusion of 'STOP'. Did i miss something here??
>
> Thanks again for your time.
> Much appreciated !
>
> ~ Tareque
Hi Again,
The only thing I did other than STOP (which I forgot to remove) was
redefine X and Y, which in your example were missing. I thought it
would be helpful for others if the whole thing worked. I should have
been clearer.
Speaking of clearer... I think I am more confused now. You might
need the intervention from the IDL gods that you originally
petitioned. Regardless of the parts I don't understand, the thing I
do get is that BILINEAR is looking for you to provide interpolation
coords that fit within the dimensions of the input array to be
sampled.
Take a peek at the second example in the online help for BILINEAR...
(slightly modified here...)
P = FINDGEN(3,3)
IX = [[0.5, 1.9], [1.1, 2.2]] ;Define the X subscripts.
JY = [[0.1, 0.9], [1.2, 1.8]] ;Define the Y subscripts.
Z = BILINEAR(P, IX, JY, MISSING = !VALUES.F_NAN) ;Interpolate.
PRINT, P
PRINT,Z
P prints as...
0.00000 1.00000 2.00000
3.00000 4.00000 5.00000
6.00000 7.00000 8.00000
Z prints as ...
0.800000 4.60000
4.70000 NaN
Note that IX and JY are mostly falling within the 0-2 range of indices
for a 3x3 array. BILINEAR can extrapolate - just remove the MISSING
keyword - but it can only extrapolate a little ways. In your example
IX are sampled at
0.00000 0.0809017 0.0618034 -0.0927051 -0.323607
-0.500000 -0.485410 -0.216312 0.247214 0.728115
and JY are sampled at ...
0.00000
0.0587785
0.190211
0.285317
0.235114
-4.37114e-08
-0.352671
-0.665740
-0.760845
-0.529007
Which to my way of thinking your are interpolating all around either
side of the [0,0] location. Is that what you intended? It is
possible that in my redo of your example I corrupted the X and Y
values you intended. In which case I would be not helping very much
at all!
Cheers,
Ben
|
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| Re: Interpolation [message #59887 is a reply to message #2184] |
Sat, 12 April 2008 13:24 |
tarequeaziz
Messages: 20
Registered: November 2007
|
Junior Member |
|
|
On Apr 12, 11:57 am, "ben.bighair" <ben.bigh...@gmail.com> wrote:
> On Apr 11, 11:05 pm, tarequea...@gmail.com wrote:
>
>
>
>> Hello All (IDL Gods),
>
>> I am back with yet another problem.
>> I know...I know...its friday night. I apologize for that. But I am
>> really stuck here for a while.
>
>> The problem:
>
>> In a certain part of my code I need to do interpolation. The data that
>> I am dealing with are from a XY grid. I need to convert them to polar
>> coordinate. So, what I do is following:
>
>> a. I generate a radius vector r_vec and a theta array containing
>> values from zero to 2 pi.
>> b. Now use the simple polar-to-rectangular coordinate transform i.e. x
>> = r cos(theta) etc.
>> c. Using these values I have a xy grid generated through a known r-
>> theta values.
>> d. Now think of superimposing the new xy grid(lets call it x'y' ) on
>> to the old xy grid which contains the real data.
>> e. This is the part where I need interpolation. I do interpolations to
>> get the x'y' values from the xy points.
>
>> So here's the question:
>
>> ----- Is there any elegant way of doing this coordinate
>> transformation ? (And in case you are thinking, "well you already have
>> the xy data, so why not just convert to r-theta?", I have to say that
>> the interpolation method actually gives me a way nicer dataset ).
>
>> My 2nd trouble is, and this is probably the biggest and dumbest
>> problem for me.
>
>> ----- i was playing around with several interpolation routines from
>> IDl. My boss's suggestion was to use 'bilinear'.
>> but I thought to give others' a shot too. Problem is, when I am
>> done with interpolation, result is nothing like what I was
>> expecting. A run down version of the code is shown below:
>
>> =======================start of code================================
>
>> Nth= 10.
>> dth= 1/Nth
>> r_vec= findgen(Nth)/Nth
>> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>
>> for i=0L,Nth - 1 do begin
>
>> x[i]= r_vec[i]* cos(theta_vec)
>
>> endfor
>
>> for j=0L,Nth - 1 do begin
>
>> y[j]= r_vec[j]* sin(theta_vec)
>
>> endfor
>
>> ;print,y
>
>> ;plot,x,y
>
>> ------------------------------------------------------------ -----------
>> Now I create the 'main' dataset on which I am going to use
>> interpolation scheme.
>
>> rr = findgen(20.)/30.
>> tht = findgen(20.)/30. *2*!Pi
>
>> m = fltarr(20,20)
>
>> for j=0,19 do begin
>> for i=0,19 do begin
>
>> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
>> ;print,i
>> endfor
>> endfor
>
>> m_p=bilinear(m,x,y)
>
>> End
>> =======================End of Code=================================
>
>> The problem is, as I mentioned above, when I plot m and the
>> interpolated m_p, they do not look like similar at all.
>
>> Any help will be greatly appreciated.
>
>> Thanks in advance.
>
>> ~tareque
>
> Hi,
>
> Coordinate transforms are very easy with CV_COORD().
>
> I don't understand where you want to go with the interpolation. The
> input coordinates to BILINEAR are rectangular and formed as indexed
> based (as in subscript indices in X and Y.) As near as I can
> reconstruct, X ranges from -0.500000 to 0.728115 and Y ranges
> from -0.760845 to 0.285317. My reconstruction of your code is
> below.
>
> Cheers,
> Ben
>
> PRO tareque
>
> Nth= 10L
> dth= 1.0/Nth
> r_vec= findgen(Nth)/Nth
> theta_vec = findgen(Nth)/Nth * 2.*!Pi
> x = fltarr(nth)
> y = fltarr(nth)
>
> for i=0L,Nth - 1 do begin
>
> x[i]= r_vec[i]* cos(theta_vec[i])
>
> endfor
>
> for j=0L,Nth - 1 do begin
>
> y[j]= r_vec[j]* sin(theta_vec[j])
>
> endfor
>
> ;print,y
>
> ;plot,x,y
>
> ;----------------------------------------------------------- ------------
> ;Now I create the 'main' dataset on which I am going to use
> ;interpolation scheme.
>
> rr = findgen(20.)/30.
> tht = findgen(20.)/30. *2*!Pi
>
> m = fltarr(20,20)
>
> for j=0,19 do begin
> for i=0,19 do begin
>
> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
> ;print,i
> endfor
> endfor
>
> m_p=bilinear(m,x,y)
>
> STOP
> End
hi Ben,
Thanks for getting back to me on this.
I guess I could not clearly state my problem before. So, I am going to
give it another try:
The data I deal with comes from a ccd camera and this goes some
tinkering and tweaking (to account for background noise and stuff).
Then this 'processed' data is in need for interpolation. why? well the
reason being this, that these processed data are actually in XY
frame.
We want them to be on a polar coordinate. The reason I cannot use
'cv_coord' is because in that case I will get the data in a r-theta
plane which is
directly correlated to the experimental XY grid. If we use
interpolation, then we can be on a XY grid (yes, it is XY grid) which
is carefully 'designed' according to
our 'own' r-theta values. So in this way we have more leverage over
data manipulation.
Was it any clearer than before?
And one more thing, I could not find much changes in your version
except for an inclusion of 'STOP'. Did i miss something here??
Thanks again for your time.
Much appreciated !
~ Tareque
|
|
|
|
| Re: Interpolation [message #59889 is a reply to message #2184] |
Sat, 12 April 2008 08:57 |
ben.bighair
Messages: 221
Registered: April 2007
|
Senior Member |
|
|
On Apr 11, 11:05 pm, tarequea...@gmail.com wrote:
> Hello All (IDL Gods),
>
> I am back with yet another problem.
> I know...I know...its friday night. I apologize for that. But I am
> really stuck here for a while.
>
> The problem:
>
> In a certain part of my code I need to do interpolation. The data that
> I am dealing with are from a XY grid. I need to convert them to polar
> coordinate. So, what I do is following:
>
> a. I generate a radius vector r_vec and a theta array containing
> values from zero to 2 pi.
> b. Now use the simple polar-to-rectangular coordinate transform i.e. x
> = r cos(theta) etc.
> c. Using these values I have a xy grid generated through a known r-
> theta values.
> d. Now think of superimposing the new xy grid(lets call it x'y' ) on
> to the old xy grid which contains the real data.
> e. This is the part where I need interpolation. I do interpolations to
> get the x'y' values from the xy points.
>
> So here's the question:
>
> ----- Is there any elegant way of doing this coordinate
> transformation ? (And in case you are thinking, "well you already have
> the xy data, so why not just convert to r-theta?", I have to say that
> the interpolation method actually gives me a way nicer dataset ).
>
> My 2nd trouble is, and this is probably the biggest and dumbest
> problem for me.
>
> ----- i was playing around with several interpolation routines from
> IDl. My boss's suggestion was to use 'bilinear'.
> but I thought to give others' a shot too. Problem is, when I am
> done with interpolation, result is nothing like what I was
> expecting. A run down version of the code is shown below:
>
> =======================start of code================================
>
> Nth= 10.
> dth= 1/Nth
> r_vec= findgen(Nth)/Nth
> theta_vec = findgen(Nth)/Nth * 2.*!Pi
>
> for i=0L,Nth - 1 do begin
>
> x[i]= r_vec[i]* cos(theta_vec)
>
> endfor
>
> for j=0L,Nth - 1 do begin
>
> y[j]= r_vec[j]* sin(theta_vec)
>
> endfor
>
> ;print,y
>
> ;plot,x,y
>
> ------------------------------------------------------------ -----------
> Now I create the 'main' dataset on which I am going to use
> interpolation scheme.
>
> rr = findgen(20.)/30.
> tht = findgen(20.)/30. *2*!Pi
>
> m = fltarr(20,20)
>
> for j=0,19 do begin
> for i=0,19 do begin
>
> m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
>
> ;print,i
> endfor
> endfor
>
> m_p=bilinear(m,x,y)
>
> End
> =======================End of Code=================================
>
> The problem is, as I mentioned above, when I plot m and the
> interpolated m_p, they do not look like similar at all.
>
> Any help will be greatly appreciated.
>
> Thanks in advance.
>
> ~tareque
Hi,
Coordinate transforms are very easy with CV_COORD().
I don't understand where you want to go with the interpolation. The
input coordinates to BILINEAR are rectangular and formed as indexed
based (as in subscript indices in X and Y.) As near as I can
reconstruct, X ranges from -0.500000 to 0.728115 and Y ranges
from -0.760845 to 0.285317. My reconstruction of your code is
below.
Cheers,
Ben
PRO tareque
Nth= 10L
dth= 1.0/Nth
r_vec= findgen(Nth)/Nth
theta_vec = findgen(Nth)/Nth * 2.*!Pi
x = fltarr(nth)
y = fltarr(nth)
for i=0L,Nth - 1 do begin
x[i]= r_vec[i]* cos(theta_vec[i])
endfor
for j=0L,Nth - 1 do begin
y[j]= r_vec[j]* sin(theta_vec[j])
endfor
;print,y
;plot,x,y
;----------------------------------------------------------- ------------
;Now I create the 'main' dataset on which I am going to use
;interpolation scheme.
rr = findgen(20.)/30.
tht = findgen(20.)/30. *2*!Pi
m = fltarr(20,20)
for j=0,19 do begin
for i=0,19 do begin
m[i,j] = rr[i]*cos(tht[j]) + 5.*rr[i]*sin(tht[j])
;print,i
endfor
endfor
m_p=bilinear(m,x,y)
STOP
End
|
|
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|
| Re: interpolation [message #78571 is a reply to message #2184] |
Thu, 08 December 2011 13:24 |
Yngvar Larsen
Messages: 134
Registered: January 2010
|
Senior Member |
|
|
On Dec 7, 6:50 pm, anil <akpinar.a...@gmail.com> wrote:
> On Dec 7, 6:42 pm, Russell <rryan....@gmail.com> wrote:
>
>> I'm not exactly sure what the issue is, but interpol seems to work
>> fine?
>
>> readcol,'data.dat',depth,temp,salinity,pot_temp,density
>> thisdepth=findgen(300)*5 ;the values where you want the temperature
>> -- whose derivative changes sign
>> thistemp=interpol(temp,depth,thisdepth)
>
>> You might add /quad or /lsq for a better fit, or even use a better
>> spline routine...
>
>> Russell
>
> This is not working for sure! When I do this I get negative values and
> numbers as -3000,-4165 etc.. whereas my temperature range is between 4
> and 26.
Maybe missing numbers represented as -9999, or something similar, in
some of the files?
--
Yngvar
|
|
|
|
| Re: interpolation [message #78572 is a reply to message #2184] |
Thu, 08 December 2011 12:57 |
Brian Wolven
Messages: 94
Registered: May 2011
|
Member |
|
|
|
I would add only that it seems difficult to justify using any type of higher order scheme for data of this type, but that's more of a philosophical and aesthetic issue to debate. Getting values like -4000 when the data range from 4 to 26 is way more than just an overshoot problem!
|
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| Re: interpolation [message #78576 is a reply to message #2184] |
Thu, 08 December 2011 04:15 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 10:13 pm, David Fanning <n...@dfanning.com> wrote:
> Jeremy Bailin writes:
>> Russell's code certainly works perfectly well on the data you provided.
>
> You may have to consider the unthinkable: a problem with
> your own code! ;-)
>
> Cheers,
>
> David
>
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming:http://www.idlcoyote.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
Ok. I've found the problem.It is because of the loops etc.. i have
used previously. I've fixed it and now Russell's code works fine for
my case. The only problem I have now is that:
I have certain number of desired depth values which I assign as :
depth2=indgen(310)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
and temp2,sal2,potemp2,den2 are the values corresponding to these
depths. I should have a depth range of 5 to 1550(or 1540)m but in some
data ,I only have depths up to 100m or 300 in some other. So i have
done this instead:
step=(fix(max(p))-fix(min(p)))/5
depth2=indgen(step)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
because of fix I lose 1 or 2 data points. But it is not so important
for my case. Thank you all for helping out.
Regards,
ANIL
|
|
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| Re: interpolation [message #78577 is a reply to message #2184] |
Thu, 08 December 2011 04:15 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 10:13 pm, David Fanning <n...@dfanning.com> wrote:
> Jeremy Bailin writes:
>> Russell's code certainly works perfectly well on the data you provided.
>
> You may have to consider the unthinkable: a problem with
> your own code! ;-)
>
> Cheers,
>
> David
>
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming:http://www.idlcoyote.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
Ok. I've found the problem.It is because of the loops etc.. i have
used previously. I've fixed it and now Russell's code works fine for
my case. The only problem I have now is that:
I have certain number of desired depth values which I assign as :
depth2=indgen(310)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
and temp2,sal2,potemp2,den2 are the values corresponding to these
depths. I should have a depth range of 5 to 1550(or 1540)m but in some
data ,I only have depths up to 100m or 300 in some other. So i have
done this instead:
step=(fix(max(p))-fix(min(p)))/5
depth2=indgen(step)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
because of fix I lose 1 or 2 data points. But it is not so important
for my case. Thank you all for helping out.
Regards,
ANIL
|
|
|
|
| Re: interpolation [message #78578 is a reply to message #2184] |
Thu, 08 December 2011 04:15 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 10:13 pm, David Fanning <n...@dfanning.com> wrote:
> Jeremy Bailin writes:
>> Russell's code certainly works perfectly well on the data you provided.
>
> You may have to consider the unthinkable: a problem with
> your own code! ;-)
>
> Cheers,
>
> David
>
> --
> David Fanning, Ph.D.
> Fanning Software Consulting, Inc.
> Coyote's Guide to IDL Programming:http://www.idlcoyote.com/
> Sepore ma de ni thui. ("Perhaps thou speakest truth.")
Ok. I've found the problem.It is because of the loops etc.. i have
used previously. I've fixed it and now Russell's code works fine for
my case. The only problem I have now is that:
I have certain number of desired depth values which I assign as :
depth2=indgen(310)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
and temp2,sal2,potemp2,den2 are the values corresponding to these
depths. I should have a depth range of 5 to 1550(or 1540)m but in some
data ,I only have depths up to 100m or 300 in some other. So i have
done this instead:
step=(fix(max(p))-fix(min(p)))/5
depth2=indgen(step)*5+5
temp2=interpol(t,p,depth2)
sal2=interpol(s,p,depth2)
potemp2=interpol(q,p,depth2)
den2=interpol(d,p,depth2)
because of fix I lose 1 or 2 data points. But it is not so important
for my case. Thank you all for helping out.
Regards,
ANIL
|
|
|
|
| Re: interpolation [message #78584 is a reply to message #2184] |
Wed, 07 December 2011 14:25 |
Kenneth P. Bowman
Messages: 585
Registered: May 2000
|
Senior Member |
|
|
In article < 12296280.447.1323289672656.JavaMail.geo-discussion-forums@yq q16 >,
Brian Wolven <brian.wolven@gmail.com> wrote:
> It does seem unlikely that one could *interpolate* and get values that were
> outside the bounds of the original data. ;)
With linear interpolation that is certainly true, but it can occur
with higher order schemes. Interpolation usually guarantees only
that the interpolating function passes through the known data points.
If you take the spline interpolation example in my book
http://csrp.tamu.edu/pdf/idl/sample_chapter.pdf
and compute the result at high resolution, you will see that
the spline function overshoots the tabulated points.
Cheers, Ken Bowman
|
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|
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| Re: interpolation [message #78592 is a reply to message #2184] |
Wed, 07 December 2011 12:13 |
David Fanning
Messages: 11724
Registered: August 2001
|
Senior Member |
|
|
Jeremy Bailin writes:
> Russell's code certainly works perfectly well on the data you provided.
You may have to consider the unthinkable: a problem with
your own code! ;-)
Cheers,
David
--
David Fanning, Ph.D.
Fanning Software Consulting, Inc.
Coyote's Guide to IDL Programming: http://www.idlcoyote.com/
Sepore ma de ni thui. ("Perhaps thou speakest truth.")
|
|
|
|
| Re: interpolation [message #78593 is a reply to message #2184] |
Wed, 07 December 2011 12:07 |
Jeremy Bailin
Messages: 618
Registered: April 2008
|
Senior Member |
|
|
On 12/7/11 5:50 PM, anil wrote:
> On Dec 7, 6:42 pm, Russell<rryan....@gmail.com> wrote:
>> I'm not exactly sure what the issue is, but interpol seems to work
>> fine?
>>
>> readcol,'data.dat',depth,temp,salinity,pot_temp,density
>> thisdepth=findgen(300)*5 ;the values where you want the temperature
>> -- whose derivative changes sign
>> thistemp=interpol(temp,depth,thisdepth)
>>
>> You might add /quad or /lsq for a better fit, or even use a better
>> spline routine...
>>
>> Russell
>>
>>
>>
>
> This is not working for sure! When I do this I get negative values and
> numbers as -3000,-4165 etc.. whereas my temperature range is between 4
> and 26.
Russell's code certainly works perfectly well on the data you provided.
-Jeremy.
|
|
|
|
| Re: interpolation [message #78594 is a reply to message #2184] |
Wed, 07 December 2011 09:50 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 6:42 pm, Russell <rryan....@gmail.com> wrote:
> I'm not exactly sure what the issue is, but interpol seems to work
> fine?
>
> readcol,'data.dat',depth,temp,salinity,pot_temp,density
> thisdepth=findgen(300)*5 ;the values where you want the temperature
> -- whose derivative changes sign
> thistemp=interpol(temp,depth,thisdepth)
>
> You might add /quad or /lsq for a better fit, or even use a better
> spline routine...
>
> Russell
>
>
>
This is not working for sure! When I do this I get negative values and
numbers as -3000,-4165 etc.. whereas my temperature range is between 4
and 26.
|
|
|
|
| Re: interpolation [message #78596 is a reply to message #2184] |
Wed, 07 December 2011 08:42 |
Russell[1]
Messages: 101
Registered: August 2011
|
Senior Member |
|
|
I'm not exactly sure what the issue is, but interpol seems to work
fine?
readcol,'data.dat',depth,temp,salinity,pot_temp,density
thisdepth=findgen(300)*5 ;the values where you want the temperature
-- whose derivative changes sign
thistemp=interpol(temp,depth,thisdepth)
You might add /quad or /lsq for a better fit, or even use a better
spline routine...
Russell
On Dec 7, 8:51 am, anil <akpinar.a...@gmail.com> wrote:
> On Dec 7, 3:46 pm, anil <akpinar.a...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
>> Hi,
>> I have many ascii files which contain a header , 5 columns of data .
>> Number of rows for each data is different. Here is an example of the
>> data:
>
>> 7.5 14.552 17.698 14.551 12.7623
>> 9.6 14.547 17.700 14.546 12.7648
>> 14.6 14.462 17.713 14.460 12.7903
>> 19.4 14.341 17.737 14.339 12.8307
>> 24.4 14.161 17.780 14.158 12.8962
>> 29.3 13.704 17.921 13.701 13.0851
>> 34.6 12.313 17.986 12.309 13.3670
>> 39.2 9.622 18.062 9.619 13.8134
>> 44.5 8.659 18.165 8.655 14.0121
>> 49.7 8.451 18.194 8.447 14.0590
>> 54.3 8.221 18.257 8.217 14.1343
>> 59.4 7.938 18.324 7.934 14.2180
>> 64.8 7.808 18.441 7.803 14.3237
>> 69.5 7.880 18.617 7.875 14.4536
>> 74.5 7.923 18.924 7.917 14.6890
>> 79.4 8.040 19.190 8.034 14.8840
>> 84.3 8.100 19.372 8.093 15.0196
>> 89.8 8.050 19.613 8.043 15.2137
>> 94.6 8.069 19.781 8.062 15.3429
>> 99.3 8.076 19.922 8.068 15.4524
>> 109.7 8.235 20.314 8.226 15.7404
>> 169.3 8.564 21.193 8.550 16.3880
>> 179.3 8.602 21.269 8.587 16.4428
>> 189.3 8.632 21.327 8.616 16.4844
>> 199.1 8.659 21.378 8.642 16.5210
>> 209.6 8.680 21.424 8.662 16.5544
>> 219.3 8.708 21.482 8.689 16.5963
>> 229.2 8.733 21.527 8.713 16.6283
>> 239.4 8.755 21.570 8.734 16.6592
>> 249.4 8.768 21.617 8.746 16.6944
>> 259.3 8.777 21.649 8.754 16.7183
>> 268.9 8.788 21.678 8.764 16.7397
>> 278.9 8.796 21.700 8.771 16.7559
>> 289.1 8.806 21.733 8.781 16.7805
>> 298.9 8.834 21.764 8.808 16.8013
>> 324.2 8.834 21.813 8.805 16.8398
>> 349.1 8.854 21.864 8.823 16.8773
>> 374.4 8.861 21.896 8.828 16.9017
>> 399.5 8.861 21.916 8.825 16.9176
>> 424.1 8.866 21.935 8.828 16.9321
>> 449.1 8.871 21.958 8.831 16.9497
>> 474.3 8.878 21.982 8.835 16.9678
>> 499.5 8.881 22.005 8.836 16.9857
>> 524.2 8.885 22.028 8.837 17.0034
>> 549.1 8.889 22.048 8.839 17.0188
>> 574.0 8.892 22.072 8.840 17.0375
>> 599.0 8.895 22.100 8.840 17.0592
>> 649.4 8.900 22.137 8.840 17.0881
>> 699.5 8.907 22.169 8.842 17.1128
>> 749.4 8.913 22.195 8.843 17.1329
>> 799.3 8.921 22.215 8.847 17.1481
>> 849.3 8.929 22.230 8.850 17.1595
>> 899.5 8.938 22.246 8.853 17.1714
>> 949.5 8.947 22.258 8.857 17.1803
>> 999.3 8.954 22.266 8.859 17.1863
>> 1049.6 8.965 22.276 8.865 17.1933
>> 1099.1 8.971 22.282 8.866 17.1979
>> 1149.0 8.980 22.288 8.870 17.2021
>> 1198.9 8.989 22.294 8.873 17.2063
>> 1249.3 8.996 22.297 8.875 17.2084
>> 1299.4 9.003 22.301 8.877 17.2113
>> 1349.4 9.011 22.304 8.879 17.2133
>> 1399.0 9.018 22.305 8.881 17.2139
>> 1449.1 9.025 22.308 8.882 17.2160
>> 1499.2 9.033 22.309 8.885 17.2165
>> 1549.2 9.040 22.311 8.886 17.2179
>> 1551.8 9.041 22.311 8.887 17.2178
>
>> The columns are depth,temperature,salinity,potential temperature and
>> density respectively. The depth values are changing in each data set
>> (7.5 is 6.7 and 9.6 is 10 in the next data and so on...). In some data
>> files, depth values end at 300 or 400 etc... . What I want to do is to
>> interpolate these depth values and their corresponding
>> temperature,salinity, etc.. to certain levels i assign
>> [5,10,15,20,25.........200,210,220.... 300,325,350,.....] or to 1m
>> depth levels starting from 5m, such as (5,6,7,......50,....1550). The
>> problem here is with the 2nd and 4th column where the data first
>> descends to a certain level and then starts ascending again. Therefore
>> i can not use function interpol. I have done such an interpolation
>> earlier:
>
>> pval=100; depth value
>> ind=0
>> if (p[0] le pval) then begin
>> while (ind le (nrows-2) and p[ind] le 100) do ind=ind+1
>> tempx(i)=t[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(t[ind]-t[in d-1]))
>> salx(i)=s[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(s[ind]-s[ind -1]))
>> potempx(i)=q[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(q[ind]-q[ ind-1]))
>> denx(i)=d[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(d[ind]-d[ind -1]))
>> endif
>
>> ; i is the data index( 1st file,2nd file etc..) and nrows is the
>> number of rows, which are assigned earlier in the code.
>
>> with this, I interpolate the values and obtain temperature,salinity
>> etc.., for this case at 100m.
>
>> I think i can find all the values i want with this way, but there
>> should be a much easier way I guess. I need help on this.
>> Regards
>
> My aim here is to obtain a regular data set. And use it to contour
> such a figure:http://argo.jcommops.org/FTPRoot/Argo/Doc/Floats/2008 -03/podens200803...
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| Re: interpolation [message #78598 is a reply to message #2184] |
Wed, 07 December 2011 05:51 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 3:46 pm, anil <akpinar.a...@gmail.com> wrote:
> Hi,
> I have many ascii files which contain a header , 5 columns of data .
> Number of rows for each data is different. Here is an example of the
> data:
>
> 7.5 14.552 17.698 14.551 12.7623
> 9.6 14.547 17.700 14.546 12.7648
> 14.6 14.462 17.713 14.460 12.7903
> 19.4 14.341 17.737 14.339 12.8307
> 24.4 14.161 17.780 14.158 12.8962
> 29.3 13.704 17.921 13.701 13.0851
> 34.6 12.313 17.986 12.309 13.3670
> 39.2 9.622 18.062 9.619 13.8134
> 44.5 8.659 18.165 8.655 14.0121
> 49.7 8.451 18.194 8.447 14.0590
> 54.3 8.221 18.257 8.217 14.1343
> 59.4 7.938 18.324 7.934 14.2180
> 64.8 7.808 18.441 7.803 14.3237
> 69.5 7.880 18.617 7.875 14.4536
> 74.5 7.923 18.924 7.917 14.6890
> 79.4 8.040 19.190 8.034 14.8840
> 84.3 8.100 19.372 8.093 15.0196
> 89.8 8.050 19.613 8.043 15.2137
> 94.6 8.069 19.781 8.062 15.3429
> 99.3 8.076 19.922 8.068 15.4524
> 109.7 8.235 20.314 8.226 15.7404
> 169.3 8.564 21.193 8.550 16.3880
> 179.3 8.602 21.269 8.587 16.4428
> 189.3 8.632 21.327 8.616 16.4844
> 199.1 8.659 21.378 8.642 16.5210
> 209.6 8.680 21.424 8.662 16.5544
> 219.3 8.708 21.482 8.689 16.5963
> 229.2 8.733 21.527 8.713 16.6283
> 239.4 8.755 21.570 8.734 16.6592
> 249.4 8.768 21.617 8.746 16.6944
> 259.3 8.777 21.649 8.754 16.7183
> 268.9 8.788 21.678 8.764 16.7397
> 278.9 8.796 21.700 8.771 16.7559
> 289.1 8.806 21.733 8.781 16.7805
> 298.9 8.834 21.764 8.808 16.8013
> 324.2 8.834 21.813 8.805 16.8398
> 349.1 8.854 21.864 8.823 16.8773
> 374.4 8.861 21.896 8.828 16.9017
> 399.5 8.861 21.916 8.825 16.9176
> 424.1 8.866 21.935 8.828 16.9321
> 449.1 8.871 21.958 8.831 16.9497
> 474.3 8.878 21.982 8.835 16.9678
> 499.5 8.881 22.005 8.836 16.9857
> 524.2 8.885 22.028 8.837 17.0034
> 549.1 8.889 22.048 8.839 17.0188
> 574.0 8.892 22.072 8.840 17.0375
> 599.0 8.895 22.100 8.840 17.0592
> 649.4 8.900 22.137 8.840 17.0881
> 699.5 8.907 22.169 8.842 17.1128
> 749.4 8.913 22.195 8.843 17.1329
> 799.3 8.921 22.215 8.847 17.1481
> 849.3 8.929 22.230 8.850 17.1595
> 899.5 8.938 22.246 8.853 17.1714
> 949.5 8.947 22.258 8.857 17.1803
> 999.3 8.954 22.266 8.859 17.1863
> 1049.6 8.965 22.276 8.865 17.1933
> 1099.1 8.971 22.282 8.866 17.1979
> 1149.0 8.980 22.288 8.870 17.2021
> 1198.9 8.989 22.294 8.873 17.2063
> 1249.3 8.996 22.297 8.875 17.2084
> 1299.4 9.003 22.301 8.877 17.2113
> 1349.4 9.011 22.304 8.879 17.2133
> 1399.0 9.018 22.305 8.881 17.2139
> 1449.1 9.025 22.308 8.882 17.2160
> 1499.2 9.033 22.309 8.885 17.2165
> 1549.2 9.040 22.311 8.886 17.2179
> 1551.8 9.041 22.311 8.887 17.2178
>
> The columns are depth,temperature,salinity,potential temperature and
> density respectively. The depth values are changing in each data set
> (7.5 is 6.7 and 9.6 is 10 in the next data and so on...). In some data
> files, depth values end at 300 or 400 etc... . What I want to do is to
> interpolate these depth values and their corresponding
> temperature,salinity, etc.. to certain levels i assign
> [5,10,15,20,25.........200,210,220.... 300,325,350,.....] or to 1m
> depth levels starting from 5m, such as (5,6,7,......50,....1550). The
> problem here is with the 2nd and 4th column where the data first
> descends to a certain level and then starts ascending again. Therefore
> i can not use function interpol. I have done such an interpolation
> earlier:
>
> pval=100; depth value
> ind=0
> if (p[0] le pval) then begin
> while (ind le (nrows-2) and p[ind] le 100) do ind=ind+1
> tempx(i)=t[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(t[ind]-t[in d-1]))
> salx(i)=s[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(s[ind]-s[ind -1]))
> potempx(i)=q[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(q[ind]-q[ ind-1]))
> denx(i)=d[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(d[ind]-d[ind -1]))
> endif
>
> ; i is the data index( 1st file,2nd file etc..) and nrows is the
> number of rows, which are assigned earlier in the code.
>
> with this, I interpolate the values and obtain temperature,salinity
> etc.., for this case at 100m.
>
> I think i can find all the values i want with this way, but there
> should be a much easier way I guess. I need help on this.
> Regards
My aim here is to obtain a regular data set. And use it to contour
such a figure:
http://argo.jcommops.org/FTPRoot/Argo/Doc/Floats/2008-03/pod ens200803.jpg
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| Re: interpolation [message #78600 is a reply to message #2184] |
Wed, 07 December 2011 05:48 |
anil
Messages: 34
Registered: August 2009
|
Member |
|
|
On Dec 7, 3:46 pm, anil <akpinar.a...@gmail.com> wrote:
> Hi,
> I have many ascii files which contain a header , 5 columns of data .
> Number of rows for each data is different. Here is an example of the
> data:
>
> 7.5 14.552 17.698 14.551 12.7623
> 9.6 14.547 17.700 14.546 12.7648
> 14.6 14.462 17.713 14.460 12.7903
> 19.4 14.341 17.737 14.339 12.8307
> 24.4 14.161 17.780 14.158 12.8962
> 29.3 13.704 17.921 13.701 13.0851
> 34.6 12.313 17.986 12.309 13.3670
> 39.2 9.622 18.062 9.619 13.8134
> 44.5 8.659 18.165 8.655 14.0121
> 49.7 8.451 18.194 8.447 14.0590
> 54.3 8.221 18.257 8.217 14.1343
> 59.4 7.938 18.324 7.934 14.2180
> 64.8 7.808 18.441 7.803 14.3237
> 69.5 7.880 18.617 7.875 14.4536
> 74.5 7.923 18.924 7.917 14.6890
> 79.4 8.040 19.190 8.034 14.8840
> 84.3 8.100 19.372 8.093 15.0196
> 89.8 8.050 19.613 8.043 15.2137
> 94.6 8.069 19.781 8.062 15.3429
> 99.3 8.076 19.922 8.068 15.4524
> 109.7 8.235 20.314 8.226 15.7404
> 169.3 8.564 21.193 8.550 16.3880
> 179.3 8.602 21.269 8.587 16.4428
> 189.3 8.632 21.327 8.616 16.4844
> 199.1 8.659 21.378 8.642 16.5210
> 209.6 8.680 21.424 8.662 16.5544
> 219.3 8.708 21.482 8.689 16.5963
> 229.2 8.733 21.527 8.713 16.6283
> 239.4 8.755 21.570 8.734 16.6592
> 249.4 8.768 21.617 8.746 16.6944
> 259.3 8.777 21.649 8.754 16.7183
> 268.9 8.788 21.678 8.764 16.7397
> 278.9 8.796 21.700 8.771 16.7559
> 289.1 8.806 21.733 8.781 16.7805
> 298.9 8.834 21.764 8.808 16.8013
> 324.2 8.834 21.813 8.805 16.8398
> 349.1 8.854 21.864 8.823 16.8773
> 374.4 8.861 21.896 8.828 16.9017
> 399.5 8.861 21.916 8.825 16.9176
> 424.1 8.866 21.935 8.828 16.9321
> 449.1 8.871 21.958 8.831 16.9497
> 474.3 8.878 21.982 8.835 16.9678
> 499.5 8.881 22.005 8.836 16.9857
> 524.2 8.885 22.028 8.837 17.0034
> 549.1 8.889 22.048 8.839 17.0188
> 574.0 8.892 22.072 8.840 17.0375
> 599.0 8.895 22.100 8.840 17.0592
> 649.4 8.900 22.137 8.840 17.0881
> 699.5 8.907 22.169 8.842 17.1128
> 749.4 8.913 22.195 8.843 17.1329
> 799.3 8.921 22.215 8.847 17.1481
> 849.3 8.929 22.230 8.850 17.1595
> 899.5 8.938 22.246 8.853 17.1714
> 949.5 8.947 22.258 8.857 17.1803
> 999.3 8.954 22.266 8.859 17.1863
> 1049.6 8.965 22.276 8.865 17.1933
> 1099.1 8.971 22.282 8.866 17.1979
> 1149.0 8.980 22.288 8.870 17.2021
> 1198.9 8.989 22.294 8.873 17.2063
> 1249.3 8.996 22.297 8.875 17.2084
> 1299.4 9.003 22.301 8.877 17.2113
> 1349.4 9.011 22.304 8.879 17.2133
> 1399.0 9.018 22.305 8.881 17.2139
> 1449.1 9.025 22.308 8.882 17.2160
> 1499.2 9.033 22.309 8.885 17.2165
> 1549.2 9.040 22.311 8.886 17.2179
> 1551.8 9.041 22.311 8.887 17.2178
>
> The columns are depth,temperature,salinity,potential temperature and
> density respectively. The depth values are changing in each data set
> (7.5 is 6.7 and 9.6 is 10 in the next data and so on...). In some data
> files, depth values end at 300 or 400 etc... . What I want to do is to
> interpolate these depth values and their corresponding
> temperature,salinity, etc.. to certain levels i assign
> [5,10,15,20,25.........200,210,220.... 300,325,350,.....] or to 1m
> depth levels starting from 5m, such as (5,6,7,......50,....1550). The
> problem here is with the 2nd and 4th column where the data first
> descends to a certain level and then starts ascending again. Therefore
> i can not use function interpol. I have done such an interpolation
> earlier:
>
> pval=100; depth value
> ind=0
> if (p[0] le pval) then begin
> while (ind le (nrows-2) and p[ind] le 100) do ind=ind+1
> tempx(i)=t[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(t[ind]-t[in d-1]))
> salx(i)=s[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(s[ind]-s[ind -1]))
> potempx(i)=q[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(q[ind]-q[ ind-1]))
> denx(i)=d[ind]-((p[ind]-100)/(p[ind]-p[ind-1])*(d[ind]-d[ind -1]))
> endif
>
> ; i is the data index( 1st file,2nd file etc..) and nrows is the
> number of rows, which are assigned earlier in the code.
>
> with this, I interpolate the values and obtain temperature,salinity
> etc.., for this case at 100m.
>
> I think i can find all the values i want with this way, but there
> should be a much easier way I guess. I need help on this.
> Regards
My aim here is to obtain a regular data set and then use it to contour
such a figure:
http://argo.jcommops.org/FTPRoot/Argo/Doc/Floats/2008-03/pod ens200803.jpg
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|
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