| Re: remove duplicates WITHOUT sorting [message #52565 is a reply to message #52564] |
Mon, 12 February 2007 11:23   |
btt
Messages: 345
Registered: December 2000
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Senior Member |
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Brian Larsen wrote:
> In the spirit of histogram has anyone written code to turn the
> REVERSE_INDICES stuff (which I can never seem to really get my head
> around) into something less confusing like a really helpful structure
> with tags i_vec and o_vec?
>
> Just asking before I reinvent the wheel next time I need to use the
> REVERSE_INDICES.
>
Hi,
Geez, I know just what you mean. I put the whole mess into one function
and let it do the thinking. You might try the following which is just a
cooked down version of what is shown in the online help...
***BEGIN
;+
; NAME:
; hist_index
;
; PURPOSE:
; Use this routine to extract the ith bin elements in
; REVERSE_INDICES as returned by the HISTOGRAM function.
;
; CALLING SEQUENCE:
; result = hist_index(reverse_indices, i, [chunksize = value],[count =
variable])
;
; RETURNED VALUE: (lifted from IDL online manual)
; Returns subscripts of the original array elements
; falling in the ith bin.
;
; If none are in that bin then -1 is returned. Use COUNT to check.
;
; ARGUMENTS:
; REVERSE_INDICES See HISTOGRAM function description.
; You may supply this as a vector or a pointer to a vector.
; I set this equal to the ith bin from which to return the indices.
This maybe
; a vector of bins (contiguous or otherwise).
;
; KEYWORDS:
; CHUNKSIZE Set equal to the number of elements used in each
; chunking operation (default = 10000) This is useful
; for very large arrays. Memory is allocated in chunks.
; This maybe temporarily increased to more than chuncksize if
required.
; COUNT Set equal to a named variable to return the
; number of elements in the returned value.
; Set to 0 if returned value is -1.
;
; EXAMPLE:
; get all of the pixels in the 10th bin
;IDL> img=read_png( filepath('mineral.png', SUBDIR=['examples','data']))
;IDL> h = histogram(img, reverse_indices = r)
;IDL> index = hist_index(r,10, count= cnt)
;IDL> help, cnt, index
;CNT LONG = 25
;INDEX LONG = Array[25]
;
; COMMENT:
; Take care to consider the number of bins available in histogram.
; For example, in the example above there are 255 bins (it's a byte image)
; but it is possible to get a real (but meaningless) result.
; In the example below, fictious bin 300 is requested.
;IDL> index = hist_index(r,300, count= cnt)
;IDL> help, cnt, index
;CNT LONG = 2800
;INDEX LONG = Array[2800]
;
; MODIFICATION HISTORY;
; 25 MAY 2004, BT written
; 22 OCT 2004 BT added multiple bin requests with chunking
; added error handling
;-
FUNCTION hist_index, r, i, $
CHUNKSIZE = chunksize, $
COUNT = count
COMPILE_OPT IDL2
ON_ERROR, 2
ni = n_elements(i)
ctr = 0
If n_elements(chunkSize) EQ 0 then CS = 10000 else CS = chunksize[0]
currentSize = CS
index = lonarr(CS)
If SIZE(r,/TYPE) EQ 10 then Begin
; a pointer to R
For j = 0, ni-1 Do begin
;data supplied as a pointer to vector
if (*r)[i[j]] NE (*r)[i[j]+1] Then begin
idx = (*r)[(*r)[i[j]]:(*r)[i[j]+1]-1]
count = SIZE(idx,/N_ELEMENTS)
;increase the size of the indexkeeper if needed
If (ctr + count) GT currentSize Then Begin
index = [index, lonarr(CS > count)]
currentSize += (CS > count)
EndIf
index[ctr] = idx
ctr += count
EndIf
EndFor
Endif Else Begin
For j = 0, ni-1 Do begin
; data supplied as a vector
if r[i[j]] NE r[i[j]+1] Then begin
idx = r[r[i[j]]:r[i[j]+1]-1]
count = SIZE(idx,/N_ELEMENTS)
;increase the size of the indexkeeper if needed
If (ctr + count) GT currentSize Then Begin
index = [index, lonarr(CS > count )]
currentSize += (CS > count)
EndIf
index[ctr] = idx
ctr += count
EndIf
EndFor
EndElse
Count = ctr
If Count GT 0 then $
Return, index[0:ctr-1] Else $
Return, -1
END
***END
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