| Gradient idl [message #90553] |
Mon, 09 March 2015 22:12  |
siumtesfai
Messages: 62
Registered: April 2013
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Member |
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Hello all,
How can calculate gradient a 2-Dimensional scalar field with missing data.
Best regards
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| Re: Gradient idl [message #90558 is a reply to message #90553] |
Tue, 10 March 2015 09:00  |
siumtesfai
Messages: 62
Registered: April 2013
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Member |
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On Tuesday, March 10, 2015 at 1:12:31 AM UTC-4, siumt...@gmail.com wrote:
> Hello all,
>
> How can calculate gradient a 2-Dimensional scalar field with missing data.
>
>
>
> Best regards
do you mean to calculate gradient of a scalar function (2ddata) can be solved this way
kernel = [ [-1,0,1],[-1,0,1]]
result = CONVOL(2Ddata, kernel,/NAN)
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| Re: Gradient idl [message #90559 is a reply to message #90558] |
Tue, 10 March 2015 09:13 |
Jeremy Bailin
Messages: 618
Registered: April 2008
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Senior Member |
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On Tuesday, March 10, 2015 at 11:00:30 AM UTC-5, siumt...@gmail.com wrote:
> On Tuesday, March 10, 2015 at 1:12:31 AM UTC-4, siumt...@gmail.com wrote:
>> Hello all,
>>
>> How can calculate gradient a 2-Dimensional scalar field with missing data.
>>
>>
>>
>> Best regards
>
> do you mean to calculate gradient of a scalar function (2ddata) can be solved this way
>
> kernel = [ [-1,0,1],[-1,0,1]]
> result = CONVOL(2Ddata, kernel,/NAN)
No, you'll need two separate convolutions, one for the x-component of the gradient and one for the y-component:
result_x = CONVOL(Data, [-1,0,1], /NAN)
result_y = CONVOL(Data, TRANSPOSE([-1,0,1]), /NAN)
See the CONVOL help page for details about edge options, etc.
-Jeremy.
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| Re: Gradient idl [message #90561 is a reply to message #90559] |
Tue, 10 March 2015 09:39 |
siumtesfai
Messages: 62
Registered: April 2013
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Member |
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On Tuesday, March 10, 2015 at 12:13:31 PM UTC-4, Jeremy Bailin wrote:
> On Tuesday, March 10, 2015 at 11:00:30 AM UTC-5, siumt...@gmail.com wrote:
>> On Tuesday, March 10, 2015 at 1:12:31 AM UTC-4, siumt...@gmail.com wrote:
>>> Hello all,
>>>
>>> How can calculate gradient a 2-Dimensional scalar field with missing data.
>>>
>>>
>>>
>>> Best regards
>>
>> do you mean to calculate gradient of a scalar function (2ddata) can be solved this way
>>
>> kernel = [ [-1,0,1],[-1,0,1]]
>> result = CONVOL(2Ddata, kernel,/NAN)
>
> No, you'll need two separate convolutions, one for the x-component of the gradient and one for the y-component:
>
> result_x = CONVOL(Data, [-1,0,1], /NAN)
> result_y = CONVOL(Data, TRANSPOSE([-1,0,1]), /NAN)
>
> See the CONVOL help page for details about edge options, etc.
>
> -Jeremy.
Hello
I am interested in the magnitude of the gradient
Result_xy=sqrt(result_x^2 +result_y^2)
right ?
I looked at the following function but it does not handle missing data
function gradient, image, vector=vector, norm=norm, $
forward=forward,central=central,$
scale=scale
;+
; NAME:
; gradient
; PURPOSE:
; Compute the gradient vector at every pixel of an image using
; neighbor pixels. Default is to return the Euclidean norm of gradient
; (2-D array), or specify keywords for other options.
; CALLING:
; gradim = gradient( image )
; INPUTS:
; image = 2-D array.
; KEYWORDS:
; scale=scale,or [xscale,yscale]; scale means [scale,scale] for a single value, scale=[xscale,yscale]
; /vector then 2 images are returned (3-D array) with d/dx and d/dy.
; /forward then x & y-partial derivatives computed as forward difference.
; /central, or default cenral difference
; /norm then magnitude of gradient is computed as norm.
; OUTPUTS:
; Function returns gradient norm (2-D array) or gradient vector (3-D).
; HISTORY:
; 10-Feb-2011, by Ding Yuan( Ding.Yuan@warwick.ac.uk)
;
;-
if ~exist(image) then on_error
sz = size( image )
if (sz[0] ne 2) then message,"Please input an 2D image (array)"
nx=sz[1] & ny=sz[2]
option=0
if exist(forward) then option=1
case option of
0: begin
didx = ( shift( image, -1, 0 ) - shift( image, 1, 0 ) ) * 0.5
didx[0,*] = image(1,*) - image(0,*)
didy = ( shift( image, 0, -1 ) - shift( image, 0, 1 ) ) * 0.5
didy[*,0] = image(*,1) - image(*,0)
end
1: begin
didx = shift( image, -1, 0 ) - image
didy = shift( image, 0, -1 ) - image
end
else: on_error,2
endcase
didx[nx-1,*] = image[nx-1,*] - image[nx-2,*]
didy[*,ny-1] = image[*,ny-1] - image[*,ny-2]
xscale=1 & yscale=1
case n_elements(scale) of
0: begin
xscale=1 & yscale=1
end
1:begin
xscale=scale & yscale=scale
end
2:begin
xscale=scale[0] & yscale=scale[2]
end
else: message,'scale should be a scalar or 2 element vector'
endcase
didx = didx * xscale
didy = didy * yscale
if keyword_set(vector ) then return, [ [[didx]], [[didy]] ] else return , sqrt( didx*didx + didy*didy )
end
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| Re: Gradient idl [message #91785 is a reply to message #90577] |
Wed, 26 August 2015 09:02 |
siumtesfai
Messages: 62
Registered: April 2013
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Member |
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Hello
I used a function called Gradient by Ding Yuan( Ding.Yuan@warwick.ac.uk)
to calculate gradient of geopotential height.
It uses shift and centered finite difference. My question would be what would be the unit of the gradient. Unitless
But I input to the gradient function with climate model output data (i.e. geopotential height ( Scalar), which is interpolated to 2.5 by 2.5 degree latitude and longitude). Do I need to convert the units from degree to meters ?
result=gradient(data)
This would be in units of meter/degree. I would think i need to convert my result to unitless by multiplying the result with [2.5* 111000 degree/meter].
1 degree = 111km
I have data with a resolution of 2.5 by 2.5 degree .
What do you all think ?
Thanks
Best regards
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