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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93708 is a reply to message #93699] |
Tue, 04 October 2016 03:32   |
Markus Schmassmann
Messages: 129
Registered: April 2016
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Senior Member |
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On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
> Can Someone help me solve this problem in IDL:
> "I have a matrix with duplicate numbers in one of the columns. I
> would
> like to average the rows with duplicate numbers. For example, I have
> duplicate values in a matrix A in column 3:
> A =
> 1 2 1
> 4 4 2
> 5 4 2
> 4 5 2
> 5 5 3
> 10 3 3
>
>
> B =
> 1 2 1
> 4.3333 4.3333 2.0000
> 7.5000 4.0000 3.0000
>
> where each row is the average values of the duplicate rows of column 3.
>
> Can anyone help?"
>
> found here:
> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
if isa(A,/integer) then begin
h=histogram(A[2,*],reverse_indices=ri)
idx=where(h ne 0,n)
B=fltarr(3,n)
for i=0,n-1 do begin
if ri[idx[i]] eq ri[idx[i]+1]-1 then $
B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
endfor
endif else
values=A[2,uniq(A[2,*],sort(A[2,*]))]
; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
n=n_elements(values)
B=fltarr(3,n)
for i=0,n-1 do begin
w=where(A[2,*] eq values[i],cnt)
if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
endfor
endelse
hope that does it, Markus
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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93709 is a reply to message #93708] |
Tue, 04 October 2016 04:17 |
belkaraza
Messages: 6
Registered: June 2016
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Junior Member |
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Am Dienstag, 4. Oktober 2016 12:32:48 UTC+2 schrieb Markus Schmassmann:
> On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
>> Can Someone help me solve this problem in IDL:
>> "I have a matrix with duplicate numbers in one of the columns. I
>> would
>> like to average the rows with duplicate numbers. For example, I have
>> duplicate values in a matrix A in column 3:
>> A =
>> 1 2 1
>> 4 4 2
>> 5 4 2
>> 4 5 2
>> 5 5 3
>> 10 3 3
>>
>>
>> B =
>> 1 2 1
>> 4.3333 4.3333 2.0000
>> 7.5000 4.0000 3.0000
>>
>> where each row is the average values of the duplicate rows of column 3.
>>
>> Can anyone help?"
>>
>> found here:
>> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>
> if isa(A,/integer) then begin
> h=histogram(A[2,*],reverse_indices=ri)
> idx=where(h ne 0,n)
> B=fltarr(3,n)
> for i=0,n-1 do begin
> if ri[idx[i]] eq ri[idx[i]+1]-1 then $
> B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
> B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
> endfor
> endif else
> values=A[2,uniq(A[2,*],sort(A[2,*]))]
> ; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
> n=n_elements(values)
> B=fltarr(3,n)
> for i=0,n-1 do begin
> w=where(A[2,*] eq values[i],cnt)
> if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
> B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
> endfor
> endelse
>
>
> hope that does it, Markus
Hey, thanks for the answer. The last if loop is bugged. if w cnt 1 then B[0,i]
Can't see how to fix that
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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93710 is a reply to message #93709] |
Tue, 04 October 2016 04:23 |
belkaraza
Messages: 6
Registered: June 2016
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Junior Member |
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Am Dienstag, 4. Oktober 2016 13:17:24 UTC+2 schrieb belk...@web.de:
> Am Dienstag, 4. Oktober 2016 12:32:48 UTC+2 schrieb Markus Schmassmann:
>> On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
>>> Can Someone help me solve this problem in IDL:
>>> "I have a matrix with duplicate numbers in one of the columns. I
>>> would
>>> like to average the rows with duplicate numbers. For example, I have
>>> duplicate values in a matrix A in column 3:
>>> A =
>>> 1 2 1
>>> 4 4 2
>>> 5 4 2
>>> 4 5 2
>>> 5 5 3
>>> 10 3 3
>>>
>>>
>>> B =
>>> 1 2 1
>>> 4.3333 4.3333 2.0000
>>> 7.5000 4.0000 3.0000
>>>
>>> where each row is the average values of the duplicate rows of column 3.
>>>
>>> Can anyone help?"
>>>
>>> found here:
>>> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>>
>> if isa(A,/integer) then begin
>> h=histogram(A[2,*],reverse_indices=ri)
>> idx=where(h ne 0,n)
>> B=fltarr(3,n)
>> for i=0,n-1 do begin
>> if ri[idx[i]] eq ri[idx[i]+1]-1 then $
>> B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
>> B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
>> endfor
>> endif else
>> values=A[2,uniq(A[2,*],sort(A[2,*]))]
>> ; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
>> n=n_elements(values)
>> B=fltarr(3,n)
>> for i=0,n-1 do begin
>> w=where(A[2,*] eq values[i],cnt)
>> if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
>> B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
>> endfor
>> endelse
>>
>>
>> hope that does it, Markus
>
>
> Hey, thanks for the answer. The last if loop is bugged. if w cnt 1 then B[0,i]
> Can't see how to fix that
Ok fixed it with "if w[cnt] eq 1 then B[0,i]"
Again thanks alot for your help ;)
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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93711 is a reply to message #93710] |
Tue, 04 October 2016 04:32 |
belkaraza
Messages: 6
Registered: June 2016
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Junior Member |
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Am Dienstag, 4. Oktober 2016 13:23:59 UTC+2 schrieb belk...@web.de:
> Am Dienstag, 4. Oktober 2016 13:17:24 UTC+2 schrieb belk...@web.de:
>> Am Dienstag, 4. Oktober 2016 12:32:48 UTC+2 schrieb Markus Schmassmann:
>>> On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
>>>> Can Someone help me solve this problem in IDL:
>>>> "I have a matrix with duplicate numbers in one of the columns. I
>>>> would
>>>> like to average the rows with duplicate numbers. For example, I have
>>>> duplicate values in a matrix A in column 3:
>>>> A =
>>>> 1 2 1
>>>> 4 4 2
>>>> 5 4 2
>>>> 4 5 2
>>>> 5 5 3
>>>> 10 3 3
>>>>
>>>>
>>>> B =
>>>> 1 2 1
>>>> 4.3333 4.3333 2.0000
>>>> 7.5000 4.0000 3.0000
>>>>
>>>> where each row is the average values of the duplicate rows of column 3.
>>>>
>>>> Can anyone help?"
>>>>
>>>> found here:
>>>> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>>>
>>> if isa(A,/integer) then begin
>>> h=histogram(A[2,*],reverse_indices=ri)
>>> idx=where(h ne 0,n)
>>> B=fltarr(3,n)
>>> for i=0,n-1 do begin
>>> if ri[idx[i]] eq ri[idx[i]+1]-1 then $
>>> B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
>>> B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
>>> endfor
>>> endif else
>>> values=A[2,uniq(A[2,*],sort(A[2,*]))]
>>> ; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
>>> n=n_elements(values)
>>> B=fltarr(3,n)
>>> for i=0,n-1 do begin
>>> w=where(A[2,*] eq values[i],cnt)
>>> if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
>>> B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
>>> endfor
>>> endelse
>>>
>>>
>>> hope that does it, Markus
>>
>>
>> Hey, thanks for the answer. The last if loop is bugged. if w cnt 1 then B[0,i]
>> Can't see how to fix that
> Ok fixed it with "if w[cnt] eq 1 then B[0,i]"
> Again thanks alot for your help ;)
In case someone wants to use it as a function:
FUNCTION tsm,A,columntotal,column
if isa(A,/integer) then begin
h=histogram(A[column,*],reverse_indices=ri)
idx=where(h ne 0,n)
B=fltarr(columntotal,n)
for i=0,n-1 do begin
if ri[idx[i]] eq ri[idx[i]+1]-1 then $
B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
endfor
endif else begin
values=A[column,uniq(A[column,*],sort(A[column,*]))]
; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
n=n_elements(values)
B=fltarr(columntotal,n)
for i=0,n-1 do begin
w=where(A[column,*] eq values[i],cnt)
if w[cnt] eq 1 then B[0,i]=A[*,where(A[column,*] eq values[i])] else $
B[0,i]=mean(A[*,where(A[column,*] eq values[i])],dim=2,/nan)
endfor
endelse
return,B
end
Credits to Mr. Schmassmann
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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93712 is a reply to message #93710] |
Tue, 04 October 2016 04:35 |
Markus Schmassmann
Messages: 129
Registered: April 2016
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Senior Member |
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Am 04.10.2016 um 13:23 schrieb belkaraza@web.de:
> Am Dienstag, 4. Oktober 2016 13:17:24 UTC+2 schrieb belk...@web.de:
>> Am Dienstag, 4. Oktober 2016 12:32:48 UTC+2 schrieb Markus Schmassmann:
>>> On 10/03/2016 11:05 PM, belkaraza@web.de wrote:
>>>> Can Someone help me solve this problem in IDL:
>>>> "I have a matrix with duplicate numbers in one of the columns. I
>>>> would
>>>> like to average the rows with duplicate numbers. For example, I have
>>>> duplicate values in a matrix A in column 3:
>>>> A =
>>>> 1 2 1
>>>> 4 4 2
>>>> 5 4 2
>>>> 4 5 2
>>>> 5 5 3
>>>> 10 3 3
>>>>
>>>>
>>>> B =
>>>> 1 2 1
>>>> 4.3333 4.3333 2.0000
>>>> 7.5000 4.0000 3.0000
>>>>
>>>> where each row is the average values of the duplicate rows of column 3.
>>>>
>>>> Can anyone help?"
>>>>
>>>> found here:
>>>> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>>>
>>> if isa(A,/integer) then begin
>>> h=histogram(A[2,*],reverse_indices=ri)
>>> idx=where(h ne 0,n)
>>> B=fltarr(3,n)
>>> for i=0,n-1 do begin
>>> if ri[idx[i]] eq ri[idx[i]+1]-1 then $
>>> B[0,i]=A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]] else $
>>> B[0,i]=mean(A[*,ri[ri[idx[i]]:ri[idx[i]+1]-1]],dim=2)
>>> endfor
>>> endif else
>>> values=A[2,uniq(A[2,*],sort(A[2,*]))]
>>> ; if A[2,*] is already sorted, A[2,uniq(A[2,*])] is sufficient there
>>> n=n_elements(values)
>>> B=fltarr(3,n)
>>> for i=0,n-1 do begin
>>> w=where(A[2,*] eq values[i],cnt)
>>> if w cnt 1 then B[0,i]=A[*,where(A[2,*] eq values[i])] else $
>>> B[0,i]=mean(A[*,where(A[2,*] eq values[i])],dim=2,/nan)
>>> endfor
>>> endelse
>> Hey, thanks for the answer. The last if loop is bugged. if w cnt 1 then B[0,i]
>> Can't see how to fix that
> Ok fixed it with "if w[cnt] eq 1 then B[0,i]"
> Again thanks alot for your help ;)
if cnt eq 1 then ...
this is to avoid errors of the following type:
IDL> print, mean([0,1,2],dim=2)
% MEAN: Illegal keyword value for DIMENSION.
but if A is an integer type array, that loop does not matter anyway
depending on whether A contains non-integer values, you can choose which
part of the outer if/then/else you want to keep
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| Re: I would like to average the first n columns based on duplicate values of the n+1th column [message #93714 is a reply to message #93699] |
Tue, 04 October 2016 05:34 |
Helder Marchetto
Messages: 520
Registered: November 2011
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Senior Member |
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On Monday, October 3, 2016 at 11:05:31 PM UTC+2, belk...@web.de wrote:
> Hey,
> Can Someone help me solve this problem in IDL:
> "I have a matrix with duplicate numbers in one of the columns. I would like to average the rows with duplicate numbers. For example, I have duplicate values in a matrix A in column 3:
>
> A =
> 1 2 1
> 4 4 2
> 5 4 2
> 4 5 2
> 5 5 3
> 10 3 3
>
>
> B =
> 1 2 1
> 4.3333 4.3333 2.0000
> 7.5000 4.0000 3.0000
>
> where each row is the average values of the duplicate rows of column 3.
>
> Can anyone help?"
>
> found here:
> http://stackoverflow.com/questions/15270019/i-would-like-to- average-the-first-n-columns-based-on-duplicate-values-of-the -n1
>
> Cheers,
> B.R.
Ok, this might not be instructive. But it was fun to look into.
I basically shortened the whole thing into two instructions:
u = [uniq(a[2,*],sort(a[2,*])),n_elements(a[2,*])-1]
for i=0,n_elements(u)-2 do print, [total(reform(a[0:1,lindgen(u[i+1]-u[i]+1,start=u[i])],2,u[i +1]-u[i]+1),2)/float(u[i+1]-u[i]+1),a[2,u[i]]]
This works if a is defined as:
a = [[ 1, 2, 1],$
[ 4, 4, 2],$
[ 5, 4, 2],$
[ 4, 5, 2],$
[ 5, 5, 3],$
[10, 3, 3]]
This is what I get:
2.50000 3.00000 1.00000
4.50000 4.50000 2.00000
7.50000 4.00000 3.00000
Similar to Markus version, but it does not use the where().
Anyway, this was already solved, so it was a just for fun thing to do.
Cheers,
Helder
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