| INTERPOLATE function - Question [message #92833] |
Wed, 09 March 2016 04:34  |
dmfl0590
Messages: 17
Registered: December 2015
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Junior Member |
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Hi all
I wrote the following code because I'm interested to understand how the INTERPOLATE (bilinear) function works.
Big = randomu(2,136,136)
nint = size(Big, /dimensions)
Small = fltarr(4,4)
Small[0,0]=0.1
n = size(Small, /dimensions)
n = n[1:*]
X = (n[0]-1)*findgen(nint[0])/(nint[0]-1E)
Y = (n[0]-1)*findgen(nint[0])/(nint[0]-1E)
Small_int = fltarr(nint[0],nint[1])
Small_int = INTERPOLATE(reform(Small[*,*]), X, Y, /GRID)
The Small array which is the array I want to interpolate has only one non-zero entry. When I interpolated from [4,4] to [136,136] I noticed that Small_int[0:44,0:44] its the non-zero part of the matrix (2025 non-zero pixels), i.e. that part of matrix affected by interpolation.
I did the same test but this time the Small=[8,8]. I interpolated to [136,136] and I got 400 non-zero pixels (i.e. Small_int[0:19,0:19]).
Does anyone knows if it's possible to know how many pixels will be affected when we interpolate our matrices?
Thanks in advance
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| Re: INTERPOLATE function - Question [message #92838 is a reply to message #92833] |
Wed, 09 March 2016 11:52  |
wlandsman
Messages: 743
Registered: June 2000
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Senior Member |
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On Wednesday, March 9, 2016 at 7:34:04 AM UTC-5, dmfl...@gmail.com wrote:
> Hi all
>
> I wrote the following code because I'm interested to understand how the INTERPOLATE (bilinear) function works.
>
> Big = randomu(2,136,136)
> nint = size(Big, /dimensions)
>
> Small = fltarr(4,4)
> Small[0,0]=0.1
> n = size(Small, /dimensions)
> n = n[1:*]
>
> X = (n[0]-1)*findgen(nint[0])/(nint[0]-1E)
> Y = (n[0]-1)*findgen(nint[0])/(nint[0]-1E)
>
> Small_int = fltarr(nint[0],nint[1])
> Small_int = INTERPOLATE(reform(Small[*,*]), X, Y, /GRID)
>
> The Small array which is the array I want to interpolate has only one non-zero entry. When I interpolated from [4,4] to [136,136] I noticed that Small_int[0:44,0:44] its the non-zero part of the matrix (2025 non-zero pixels), i.e. that part of matrix affected by interpolation.
>
Your formulae for X and Y are incorrect. The output values currently go from 0 to 3 whereas you want them to go from 0 to 4. (In one dimension "0" refers to the left edge of the first pixel, and "4" refers to the right edge of the last pixel.)
Y = n[0]*findgen(nint[0])/nint[0]
if you do this you will the find same percentage of non-zero pixels in the small array as the big one. In your case, 1 out of 4 pixels should be nonzero so 136/4. =34 pixels in each dimension, small_int[0:33,0:33]
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| Re: INTERPOLATE function - Question [message #92845 is a reply to message #92838] |
Thu, 10 March 2016 11:44 |
dmfl0590
Messages: 17
Registered: December 2015
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Junior Member |
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Hi
Thank you for pointing out this. I changed the code but I still have something that I don't understand.
When Small[0,0]=0.1
I got 34x34=1156 nonzero pixels i.e. small_int[0:33,0:33] nonzero
When I changed the Small[1,0]=0.1 I didn't got 1156 nonzero pixels as I was expecting (i.e. small_int[34:67,34:67] nonzero). Please see the code below:
Big = randomu(2,136,136)
nint = size(Big, /dimensions)
Small = fltarr(4,4)
Small[1,0]=0.1
n = size(Small, /dimensions)
n = n[0:*]
X = n[0]*findgen(nint[0])/nint[0]
Y = n[0]*findgen(nint[0])/nint[0]
Small_int = fltarr(nint[0],nint[1])
Small_int = INTERPOLATE(reform(Small[*,*]), X, Y, /GRID)
index = WHERE(small_int gt 0, count)
print, count
> IDL 2278
Also for Small[3,3]=0.1 I got 4489 nonzero pixels (the other entries in the Small matrix are zero). I was expecting small_int[101:135,101:135] to be the nonzero part (34 pixels in each direction). How can we explain that? Maybe I did something wrong again...
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| Re: INTERPOLATE function - Question [message #92846 is a reply to message #92838] |
Thu, 10 March 2016 13:43 |
wlandsman
Messages: 743
Registered: June 2000
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Senior Member |
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On Wednesday, March 9, 2016 at 2:52:45 PM UTC-5, wlandsman wrote:
>
> if you do this you will the find same percentage of non-zero pixels in the small array as the big one. In your case, 1 out of 4 pixels should be nonzero so 136/4. =34 pixels in each dimension, small_int[0:33,0:33]
OK, I will somewhat modify my previous answer. In general, there is no reason to expect the same fraction of non-zero pixels when you are *interpolating*. If you want to have the same fraction then use
small_int = rebin(small,nint[0],nint[1], /sample)
or
small_int = congrid(small,nint[0],nint[1] )
But any fractional pixel position less than 1 pixel away will be interpolated and thus non-zero. So in dimension if small[1] is non-zero, then any pixel position between 0 and 2 will be interpolated.
The corner pixel [0,0] is a special case, because all interpolated positions will have a X,Y value less than one. (Remember that [0,0] specifies the lower left hand corner of a pixel, with [0.5, 0.5] in the middle.)
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