| Interpolation question [message #25895] |
Wed, 25 July 2001 11:18  |
rkj
Messages: 66
Registered: February 1996
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Member |
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I have an image containg some bad values. I would like
to replace these points with the average value of their
neighbors. Is there an easy way to do this without loops?
For instance, a 3x3 array as follows:
1 1 1
1 0 1
1 1 1
would become
1 1 1
1 1 1
1 1 1.
I can't get the boundary conditions to work correctly when
I use CONVOL. It either zeros the edges or does not process
them.
For instance,
1 1 1
0 1 1
1 1 0
should be
1 1 1
1 1 1
1 1 1
as well.
Kyle J.
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| Re: interpolation question [message #48393 is a reply to message #25895] |
Thu, 20 April 2006 07:50  |
chen123.dian
Messages: 4
Registered: April 2006
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Junior Member |
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Thanks, Peter,
Your code works well for 1-D data. It will be nice to extend it to 2-D
data.
Regards,
Chen
Peter Albert wrote:
> Hi Chen,
>
> I haven't tested this extensively, but I would try
>
> nn = y[round(interpol(indgen(n_elements(x)),x, u))]
>
> where "x" and "y" are the, well, x and y values and "u" is the vector
> with x-values for which the nearest neighbours are to be found.
>
> What goes on?
>
> The interpol command interpolates the indices of the x and y vectors to
> the new locations, round actually finds the nearest neighbour in terms
> of index, and then we just assign y with those indices.
>
> Regards,
>
> Peter
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| Re: interpolation question [message #48394 is a reply to message #25895] |
Thu, 20 April 2006 01:20 |
peter.albert@gmx.de
Messages: 108
Registered: July 2005
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Senior Member |
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Hi Chen,
I haven't tested this extensively, but I would try
nn = y[round(interpol(indgen(n_elements(x)),x, u))]
where "x" and "y" are the, well, x and y values and "u" is the vector
with x-values for which the nearest neighbours are to be found.
What goes on?
The interpol command interpolates the indices of the x and y vectors to
the new locations, round actually finds the nearest neighbour in terms
of index, and then we just assign y with those indices.
Regards,
Peter
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| Re: interpolation question [message #48454 is a reply to message #25895] |
Mon, 24 April 2006 12:47 |
news.verizon.net
Messages: 47
Registered: August 2003
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Member |
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> I am so wonder that why IDL has no simple function like MATLAB's
> 'interp2'.
I agree with you that IDL appears deficient to MATLAB in providing an
easy and consistent set of interpolation routines. First of all,
while IDL has interpol.pro for 1d interpolation, there is no equivalent
function for 2-d interpolation. (bilinear.pro and interpolate
require you to supply indicies). And while interpol.pro provides
several interpolation methods, it doesn't include the simplest (nearest
neighbor), though this would be easy to add (as in
http://idlastro.gsfc.nasa.gov/ftp/pro/math/linterp.pro). It makes
much more sense to have functions interp1d and interp2d, each with a
variety of interpolation methods available.
> Another problem is for value_locate. Some suggestions
> mentioned to use value_locate. Here is a example to show my problem.
>
> IDL> vec = [2.0, 5.0, 8.0, 10.0]
> IDL> print, vec
> 2.00000 5.00000 8.00000 10.0000
> IDL> loc = VALUE_LOCATE(vec, [0.0, 4.5, 5.0, 6.0, 12. ])
DL> print, loc
-1 0 1 1 3
VALUE_LOCATE is doing what it says it does -- returning a value j such
that
vec[j] < x < vec[j+1]. I don't think anyone suggested that
VALUE_LOCATE can give you the answer by itself, but both JD Smith and
the archive posting from David Fanning (http://tinyurl.com/r9t5s)
showed how you could use VALUE_LOCATE to get
the index of the nearest value. In this case
loc = round(loc+(x-vec[loc]) / (vec[loc+1]-vec[loc]))
(Actually one should first check that 0 < loc < N_Elements(vec)-1, as
in
http://idlastro.gsfc.nasa.gov/ftp/pro/math/tabinv.pro )
--Wayne
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| Re: interpolation question [message #48457 is a reply to message #25895] |
Mon, 24 April 2006 12:09 |
Paul Van Delst[1]
Messages: 1157
Registered: April 2002
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Senior Member |
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chen123.dian@gmail.com wrote:
> Hi,
>
> Thanks so much for all suggestions here.
>
> I am so wonder that why IDL has no simple function like MATLAB's
> 'interp2'. Another problem is for value_locate. Some suggestions
> mentioned to use value_locate. Here is a example to show my problem.
>
> IDL> vec = [2.0, 5.0, 8.0, 10.0]
> IDL> print, vec
> 2.00000 5.00000 8.00000 10.0000
> IDL> loc = VALUE_LOCATE(vec, [0.0, 4.5, 5.0, 6.0, 12.0])
> IDL> print, loc
> -1 0 1 1 3
>
> We can see the value of "4.5" corresponds to index location of "0".
> Actually, the value of "4.5" should correspond to index location of "1"
> because the value of "4.5" is closer to the value of "5.0" having index
> location of "1".
From the documentation:
<quote>
Result = VALUE_LOCATE ( Vector, Value [, /L64 ] )
Return Value
Each return value, Result [i], is an index, j, into Vector, corresponding to the interval
into which the given Value [i] falls. The returned values are in the range -1 £ j £ N-1,
where N is the number of elements in the input vector
</quote>
I presume in IDL-documentation-land "£" means "≤".
So the results you've got seem to be correct.
paulv
--
Paul van Delst Ride lots.
CIMSS @ NOAA/NCEP/EMC Eddy Merckx
Ph: (301)763-8000 x7748
Fax:(301)763-8545
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| Re: interpolation question [message #48458 is a reply to message #25895] |
Mon, 24 April 2006 12:04 |
Jo Klein
Messages: 54
Registered: January 2006
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Member |
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Hi Chen,
VALUE_LOCATE works differently: It returns the location as the interval
in which your value is to be found, vec[location]<=value<vec[location+1]
It's not designed to be a nearest-neighbour-style routine. Granted, the
function name is a bit misleading ...
Cheers,
Jo
chen123.dian@gmail.com wrote:
> Hi,
>
> Thanks so much for all suggestions here.
>
> I am so wonder that why IDL has no simple function like MATLAB's
> 'interp2'. Another problem is for value_locate. Some suggestions
> mentioned to use value_locate. Here is a example to show my problem.
>
> IDL> vec = [2.0, 5.0, 8.0, 10.0]
> IDL> print, vec
> 2.00000 5.00000 8.00000 10.0000
> IDL> loc = VALUE_LOCATE(vec, [0.0, 4.5, 5.0, 6.0, 12.0])
> IDL> print, loc
> -1 0 1 1 3
>
> We can see the value of "4.5" corresponds to index location of "0".
> Actually, the value of "4.5" should correspond to index location of "1"
> because the value of "4.5" is closer to the value of "5.0" having index
> location of "1".
>
> Thanks.
>
> Regards,
>
> Chen
>
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| Re: interpolation question [message #48459 is a reply to message #25895] |
Mon, 24 April 2006 11:35 |
chen123.dian
Messages: 4
Registered: April 2006
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Junior Member |
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Hi,
Thanks so much for all suggestions here.
I am so wonder that why IDL has no simple function like MATLAB's
'interp2'. Another problem is for value_locate. Some suggestions
mentioned to use value_locate. Here is a example to show my problem.
IDL> vec = [2.0, 5.0, 8.0, 10.0]
IDL> print, vec
2.00000 5.00000 8.00000 10.0000
IDL> loc = VALUE_LOCATE(vec, [0.0, 4.5, 5.0, 6.0, 12.0])
IDL> print, loc
-1 0 1 1 3
We can see the value of "4.5" corresponds to index location of "0".
Actually, the value of "4.5" should correspond to index location of "1"
because the value of "4.5" is closer to the value of "5.0" having index
location of "1".
Thanks.
Regards,
Chen
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| Re: interpolation question [message #48490 is a reply to message #48393] |
Thu, 20 April 2006 08:47 |
peter.albert@gmx.de
Messages: 108
Registered: July 2005
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Senior Member |
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Hi Chen,
if your input array is 2D, things are a bit more tricky. By brue force,
you have to calculate all differencences between all grod points and
all "nearest neighbour" points, which, given sufficiently large arrays,
is either very slow or very memory consuing, or both. But luckily,
astronomers seem to have a need for this, therefore it's all done,
after a discussion in this group, David has provided two routines and
the way how to get there at
http://www.dfanning.com/code_tips/slowloops.html
Regards,
Peter
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| Re: Interpolation question [message #49584 is a reply to message #25895] |
Wed, 02 August 2006 09:31 |
JD Smith
Messages: 850
Registered: December 1999
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Senior Member |
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On Wed, 02 Aug 2006 11:21:55 -0500, Mike Wallace wrote:
> I have an array that I need to interpolate and would like to find a
> semi-efficient way to do it. Well, almost anything would be more
> efficient than what I'm currently doing.
>
> I have an array of data and a corresponding array of times when the data
> was taken. I have a third array that represents the times that I want
> to calculate the interpolation. How can I easily (and efficiently)
> calculate the data points corresponding to that array?
>
>
> For example...
>
> data = [12, 6, 1, 4, 8, 8, 10]
> time = [ 0, 1, 4, 7, 8, 11, 14]
>
> Now, I say that I wanted to interpolate the data array but only
> calculate the interpolation for the time values in some new array...
>
> new_time = [1, 4, 5, 8, 9, 12]
new_data=interpol(float(data),time,new_time)
The float is needed or all calcs are done in integer space (probably not
what you want).
JD
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| Re: Interpolation question [message #49585 is a reply to message #25895] |
Wed, 02 August 2006 09:37 |
mchinand
Messages: 66
Registered: September 1996
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Member |
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In article <12d1k3ah497rj6c@corp.supernews.com>,
Mike Wallace <mwallace.no.spam.please@swri.edu.invalid> wrote:
> I have an array that I need to interpolate and would like to find a
> semi-efficient way to do it. Well, almost anything would be more
> efficient than what I'm currently doing.
>
> I have an array of data and a corresponding array of times when the data
> was taken. I have a third array that represents the times that I want
> to calculate the interpolation. How can I easily (and efficiently)
> calculate the data points corresponding to that array?
>
>
> For example...
>
> data = [12, 6, 1, 4, 8, 8, 10]
> time = [ 0, 1, 4, 7, 8, 11, 14]
>
> Now, I say that I wanted to interpolate the data array but only
> calculate the interpolation for the time values in some new array...
>
> new_time = [1, 4, 5, 8, 9, 12]
>
Check out INTERPOL:
IDL> new_data=interpol(data,time,new_time)
IDL> print, new_data
6.00000 1.00000 2.00000 8.00000 8.00000 8.66667
I changed the data array to float, otherwise the resulting new_data array is Integer type which you
may or may not want. INTERPOL has different interpolation types to choose from, linear, quadratic,
or spline.
Hope that helps,
--Mike
--
Michael Chinander
m-chinander@uchicago.edu
Department of Radiology
University of Chicago
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| file_lines and expand path [message #49635 is a reply to message #49556] |
Fri, 04 August 2006 08:23 |
FL
Messages: 17
Registered: April 2006
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Junior Member |
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Hi,
I am implementing file_lines in FL and was wondering what to return when
the input path expands to multiple files. The IDL reference guide says
nothing. I have run IDL and got curious result:
IDL> help, file_lines('*.pro')
<Expression> LONG64 = 15
I have 448 .pro files in the current directory, with a total of 10k lines.
IDL seems to be randomly picking a single file and reporting its length
(probably the first match in readdir()).
What do you expect to get?
Possibilities:
1. never expand (but why do we have NOEXPAND then?)
2. give an error message if the expansion results in more than one file
3. return the total length of files
4. return the total length of files if a new keyword TOTAL was set
5. do as IDL, pick a single file silently
thanks,
lajos
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