| help with sorting vector elements in to subarrays [message #80234] |
Mon, 21 May 2012 07:33  |
Tito
Messages: 16
Registered: March 2012
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Junior Member |
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Hallo!
I have this vector: zz = [5,6,9,10,11]
Anybody can tell me how I can sort it like zz1 = [5,6], zz2 =[9,10,11]?
Here is the code:
zz =[]
f=indgen(n_elements(amp))
;;I have amp[i] on witch some of the elements are identical.
;I need to take this [i]indexies and
;1-st to take it out
;2nd to write it to another array [sub arrays]
for i=0L, n_elements(amp)-1 do begin
a = where(amp[i] eq amp ,count)
if count gt 0.0 then begin
if n_elements(a) gt 1 then begin
zz = [zz,i]
endif
endif
endfor
if n_elements(zz) ne 0.0 then begin
match2,f,zz,subf,subzz
f[subzz] = -1
f = f[where(f[*] NE -1)]
endif
zz = Long[5]
4 5 9 10 11
f = Int[13]
0 1 2 3 6 7 8 12 13 14
15 16 17
All the best,
Trifon
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| Re: help with sorting vector elements in to subarrays [message #80447 is a reply to message #80234] |
Thu, 31 May 2012 07:49  |
greg.addr
Messages: 160
Registered: May 2007
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Senior Member |
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If you want to record the sequences, the tricky part is to deal with their variable lengths. If you're using 8.x, you could try putting them in a list:
sequence=list()
then...
sequence.add,where(ss[i] eq amp)
print,sequence[i]
Or the older way was to use an array of pointers:
sequence=ptrarr(n_elements(ss))
...
sequence[i]=ptr_new(where(ss[i] eq amp))
print,*sequence[i]
greg
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| Re: help with sorting vector elements in to subarrays [message #80448 is a reply to message #80234] |
Thu, 31 May 2012 06:53 |
Tito
Messages: 16
Registered: March 2012
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Junior Member |
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On Thursday, May 31, 2012 2:36:29 PM UTC+2, (unknown) wrote:
> Would this do it?
>
> for i = 0L, n_elements(ss)-1 do print,where(ss[i] eq amp)
>
> greg
Thanks for the fast answer Greg!
I all ready did something similar with:
for i = 0L , n_elements(ss)-1 do begin
a = where(ss[i] eq amp,count)
dd[i,a] = a
endfor
I just wanted to record this indexies in seperate arrays and use it later on.
But you are right... why not use it inside the loop???
For example:
for i = 0L, n_elements(ss)-1 do begin
a = where(ss[i] eq amp,count)
for ii = 0L, n_elements(a)-1 do spec = spec - (amp[a[ii]]*exp((-1d0*(wave-line[a[ii]])^2d0)/var[a[ii]]))/n _elements(a)
endfor
I think I will be fine from now on....
Any way is there a way how to extract the neeeded data from ;; Now dd = Double[3, 20]
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
3.0000000 -1.0000000 -1.0000000
4.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 6.0000000 -1.0000000
-1.0000000 7.0000000 -1.0000000
-1.0000000 8.0000000 -1.0000000
-1.0000000 -1.0000000 9.0000000
-1.0000000 -1.0000000 10.000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
-1.0000000 -1.0000000 -1.0000000
??? Just want to know.....
All the best!
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