| How to represent the spatial distribution of a parameter [message #64172] |
Wed, 10 December 2008 01:16  |
duxiyu@gmail.com
Messages: 88
Registered: March 2007
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Member |
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Dear all,
I have three array X, Y and V.
X and Y represent the position, and they are not distributed uniformly
on the X-Y plane.
V represent the parameter values corresponding to their position.
I want to take some 2D figures which can represent the spatial
distribution of the parameter values.
Is there any recommended representation?
I find some figures in a paper which is shown in the web link below.
http://lh5.ggpht.com/_NSjLmdFf0ac/ST9_aR0A8zI/AAAAAAAAAFA/8H 3m5pkbLO0/1.png
http://lh6.ggpht.com/_NSjLmdFf0ac/ST9_a51bQII/AAAAAAAAAFI/2Z oP59xdyYw/2.png
He divided the X-Y plane into small grids, and used the mean of the
parameter at the points, which are located in each grid, to represent
the parameter value in that grid.
I wander whether there is a simple way to do this in IDL.
For example, I want to use the 0.1*0.1 grid to divide the X-Y plane.
How do I plot the figures like that from X, Y and V?
Best regards,
Du
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| Re: How to represent the spatial distribution of a parameter [message #64247 is a reply to message #64172] |
Thu, 11 December 2008 07:36  |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Dec 11, 10:34 am, Jeremy Bailin <astroco...@gmail.com> wrote:
> On Dec 10, 8:05 am, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
> wrote:
>
>
>
>> Hi,
>
>> I did something similar a while ago... here is part of it:
>
>> newX = Xdata / CellSizeX ;Agregate the data
>> newY = Ydata / CellSizeY
>
>> nbCol = ceil((maxX+1.0) / cellSizeX)
>> nbRow = ceil((maxY+1.0) / cellSizeY)
>> nPoint = n_elements(newX)
>
>> image = lonarr(nbCol, nbRow)
>> nbPointsXY = lonarr(nbCol, nbRow)
>
>> for i = 0L, nPoint-1 do begin
>> image[newX[i], newY[i]] += v[i]
>> nbPointsXY[newX[i], newY[i]] += 1
>> endfor
>
>> image /= nbPointsXY ;do the average
>
>> tvscl, image
>
>> Jean
>
> If there are on average lots of points per grid cell, it would be more
> efficient to use the reverse indices in David's HIST_ND:
>
> histimage = hist_nd(transpose([[X],[Y]], [binX,binY], $
> min=[minX,minY], max=[maxX,maxY], reverse_indices=hiri)
> meanimage = fltarr(size(histimage,/dimen))
> for i=0l,n_elements(histimage)-1 do if histimage[i] gt 0 then $
> meanimage[i] = mean(V[hiri[hiri[i]:hiri[i+1]-1]])
>
> If performance is still a serious issue, you should go readhttp://www.dfanning.com/code_tips/drizzling.html(it's effectively the
> same thing - you just need to divide by histimage at the end). In
> fact, go read it anyway - it's very illuminating. :-)=
>
> -Jeremy.
...although, to answer the original poster's question, I'd agree with
Giorgio - your easiest option is probably to use GRIDDATA.
-Jeremy.
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| Re: How to represent the spatial distribution of a parameter [message #64248 is a reply to message #64172] |
Thu, 11 December 2008 07:34 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Dec 10, 8:05 am, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
wrote:
> Hi,
>
> I did something similar a while ago... here is part of it:
>
> newX = Xdata / CellSizeX ;Agregate the data
> newY = Ydata / CellSizeY
>
> nbCol = ceil((maxX+1.0) / cellSizeX)
> nbRow = ceil((maxY+1.0) / cellSizeY)
> nPoint = n_elements(newX)
>
> image = lonarr(nbCol, nbRow)
> nbPointsXY = lonarr(nbCol, nbRow)
>
> for i = 0L, nPoint-1 do begin
> image[newX[i], newY[i]] += v[i]
> nbPointsXY[newX[i], newY[i]] += 1
> endfor
>
> image /= nbPointsXY ;do the average
>
> tvscl, image
>
> Jean
If there are on average lots of points per grid cell, it would be more
efficient to use the reverse indices in David's HIST_ND:
histimage = hist_nd(transpose([[X],[Y]], [binX,binY], $
min=[minX,minY], max=[maxX,maxY], reverse_indices=hiri)
meanimage = fltarr(size(histimage,/dimen))
for i=0l,n_elements(histimage)-1 do if histimage[i] gt 0 then $
meanimage[i] = mean(V[hiri[hiri[i]:hiri[i+1]-1]])
If performance is still a serious issue, you should go read
http://www.dfanning.com/code_tips/drizzling.html (it's effectively the
same thing - you just need to divide by histimage at the end). In
fact, go read it anyway - it's very illuminating. :-)=
-Jeremy.
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| Re: How to represent the spatial distribution of a parameter [message #64256 is a reply to message #64172] |
Wed, 10 December 2008 13:20 |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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Hi Du,
ok, so you have a set of X and Y coordinates
You want to plot them in a regular grid, so to know in which cell to
plot the corresponding data, you divide the coords by the cell size
ex: X = [1,2,2.5]
cellSizeX = 0.5
NewX = X/cellSizeX
==> 2.00000 4.00000 5.00000
==> on the grid, your point will be in cell 2,4 and 5
grid: 0-0.5-1-1.5-2-2.5-3-3.5-4-4.5-5 etc
Now, you want to know the size of your grid, so you do
nbCol = ceil((maxX+1.0) / cellSizeX)
this assume that the 1st cell of the grid is coord 0;0, do maxX-minX if not.
So basically, you take the highest X value and divide by the cell size.
Up-round this, so that you are sure that the new coord is within your grid.
Create a grid (an array) with the number of columns and rows. Create a
"count" layer, that you will use for doing the average.
For each point that you have, using the new X and Y coordinate, you
1)add the value V to the image,
2) add 1 to the count layer
then, simply divide the image by the count, and you get the average
value in each pixel.
Jean
> Excuse me!
> I do not understand your routine.
> Could you show me an example? or give me some detailed explanations?
>
> Du
>
>
> On Dec 10, 9:05 pm, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
> wrote:
>> Hi,
>>
>> I did something similar a while ago... here is part of it:
>>
>> newX = Xdata / CellSizeX ;Agregate the data
>> newY = Ydata / CellSizeY
>>
>> nbCol = ceil((maxX+1.0) / cellSizeX)
>> nbRow = ceil((maxY+1.0) / cellSizeY)
>> nPoint = n_elements(newX)
>>
>> image = lonarr(nbCol, nbRow)
>> nbPointsXY = lonarr(nbCol, nbRow)
>>
>> for i = 0L, nPoint-1 do begin
>> image[newX[i], newY[i]] += v[i]
>> nbPointsXY[newX[i], newY[i]] += 1
>> endfor
>>
>> image /= nbPointsXY ;do the average
>>
>> tvscl, image
>>
>> Jean
>
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| Re: How to represent the spatial distribution of a parameter [message #64263 is a reply to message #64172] |
Wed, 10 December 2008 09:37 |
duxiyu@gmail.com
Messages: 88
Registered: March 2007
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Member |
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Excuse me!
I do not understand your routine.
Could you show me an example? or give me some detailed explanations?
Du
On Dec 10, 9:05 pm, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
wrote:
> Hi,
>
> I did something similar a while ago... here is part of it:
>
> newX = Xdata / CellSizeX ;Agregate the data
> newY = Ydata / CellSizeY
>
> nbCol = ceil((maxX+1.0) / cellSizeX)
> nbRow = ceil((maxY+1.0) / cellSizeY)
> nPoint = n_elements(newX)
>
> image = lonarr(nbCol, nbRow)
> nbPointsXY = lonarr(nbCol, nbRow)
>
> for i = 0L, nPoint-1 do begin
> image[newX[i], newY[i]] += v[i]
> nbPointsXY[newX[i], newY[i]] += 1
> endfor
>
> image /= nbPointsXY ;do the average
>
> tvscl, image
>
> Jean
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| Re: How to represent the spatial distribution of a parameter [message #64270 is a reply to message #64172] |
Wed, 10 December 2008 05:05 |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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Hi,
I did something similar a while ago... here is part of it:
newX = Xdata / CellSizeX ;Agregate the data
newY = Ydata / CellSizeY
nbCol = ceil((maxX+1.0) / cellSizeX)
nbRow = ceil((maxY+1.0) / cellSizeY)
nPoint = n_elements(newX)
image = lonarr(nbCol, nbRow)
nbPointsXY = lonarr(nbCol, nbRow)
for i = 0L, nPoint-1 do begin
image[newX[i], newY[i]] += v[i]
nbPointsXY[newX[i], newY[i]] += 1
endfor
image /= nbPointsXY ;do the average
tvscl, image
Jean
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