pro match, a, b, suba, subb, COUNT = count, SORT = sort ;+ ; NAME: ; MATCH ; PURPOSE: ; Routine to match values in two vectors. ; ; CALLING SEQUENCE: ; match, a, b, suba, subb, [ COUNT =, /SORT ] ; ; INPUTS: ; a,b - two vectors to match elements, numeric or string datatype ; ; OUTPUTS: ; suba - subscripts of elements in vector a with a match ; in vector b ; subb - subscripts of the positions of the elements in ; vector b with matchs in vector a. ; ; suba and subb are ordered such that a[suba] equals b[subb] ; ; OPTIONAL INPUT KEYWORD: ; /SORT - By default, MATCH uses two different algorithm: (1) the ; /REVERSE_INDICES keyword to HISTOGRAM is used for integer data, ; while a sorting algorithm is used for non-integer data. The ; histogram algorithm is usually faster, except when the input ; vectors are sparse and contain very large numbers, possibly ; causing memory problems. Use the /SORT keyword to always use ; the sort algorithm. ; ; OPTIONAL KEYWORD OUTPUT: ; COUNT - set to the number of matches, integer scalar ; ; SIDE EFFECTS: ; The obsolete system variable !ERR is set to the number of matches; ; however, the use !ERR is deprecated in favor of the COUNT keyword ; ; RESTRICTIONS: ; The vectors a and b should not have duplicate values within them. ; You can use rem_dup function to remove duplicate values ; in a vector ; ; EXAMPLE: ; If a = [3,5,7,9,11] & b = [5,6,7,8,9,10] ; then ; IDL> match, a, b, suba, subb, COUNT = count ; ; will give suba = [1,2,3], subb = [0,2,4], COUNT = 3 ; and suba[a] = subb[b] = [5,7,9] ; ; ; METHOD: ; For non-integer data types, the two input vectors are combined and ; sorted and the consecutive equal elements are identified. For integer ; data types, the /REVERSE_INDICES keyword to HISTOGRAM of each array ; is used to identify where the two arrays have elements in common. ; HISTORY: ; D. Lindler Mar. 1986. ; Fixed "indgen" call for very large arrays W. Landsman Sep 1991 ; Added COUNT keyword W. Landsman Sep. 1992 ; Fixed case where single element array supplied W. Landsman Aug 95 ; Converted to IDL V5.0 W. Landsman September 1997 ; Use a HISTOGRAM algorithm for integer vector inputs for improved ; performance W. Landsman March 2000 ; Work again for strings W. Landsman April 2000 ;- ;------------------------------------------------------------------------- On_error,2 if N_params() LT 3 then begin print,'Syntax - match, a, b, suba, subb, [ COUNT = ]' print,' a,b -- input vectors for which to match elements' print,' suba,subb -- output subscript vectors of matched elements' return endif sa = size(a) & sb = size(b) da = sa[sa[0]+1] & db = sb[ sb[0]+1] if keyword_set(sort) then hist = 0b else $ hist = (( da LE 3 ) or (da GE 12)) and ((db LE 3) or (db GE 12 )) if not hist then begin ;Non-integer calculation na = N_elements(a) ;number of elements in a nb = N_elements(b) ;number of elements in b ; Check for a single element array if (na EQ 1) or (nb EQ 1) then begin if (nb GT 1) then begin subb = where(b EQ a[0], nw) if (nw GT 0) then suba = replicate(0,nw) else suba = [-1] endif else begin suba = where(a EQ b[0], nw) if (nw GT 0) then subb = replicate(0,nw) else subb = [-1] endelse count = nw return endif c = [ a, b ] ;combined list of a and b ind = [ lindgen(na), lindgen(nb) ] ;combined list of indices vec = [ bytarr(na), replicate(1b,nb) ] ;flag of which vector in combined ;list 0 - a 1 - b ; sort combined list sub = sort(c) c = c[sub] ind = ind[sub] vec = vec[sub] ; find duplicates in sorted combined list n = na + nb ;total elements in c firstdup = where( (c EQ shift(c,-1)) and (vec NE shift(vec,-1)), Count ) if Count EQ 0 then begin ;any found? suba = lonarr(1)-1 subb = lonarr(1)-1 return end dup = lonarr( Count*2 ) ;both duplicate values even = lindgen( N_elements(firstdup))*2 ;Changed to LINDGEN 6-Sep-1991 dup[even] = firstdup dup[even+1] = firstdup+1 ind = ind[dup] ;indices of duplicates vec = vec[dup] ;vector id of duplicates suba = ind[ where( vec EQ 0) ] ;a subscripts subb = ind[ where( vec) ] ;b subscripts endif else begin ;Integer calculation using histogram. minab = min(a, MAX=maxa) > min(b, MAX=maxb) ;Only need intersection of ranges maxab = maxa < maxb ;If either set is empty, or their ranges don't intersect: ; result = NULL (which is denoted by integer = -1) !ERR = -1 suba = -1 subb = -1 COUNT = 0L if (maxab lt minab) or (maxab lt 0) then return ha = histogram(a, MIN=minab, MAX=maxab, reverse_indices=reva) hb = histogram(b, MIN=minab, MAX=maxab, reverse_indices=revb) r = where((ha ne 0) and (hb ne 0), count) if count gt 0 then begin suba = reva[reva[r]] subb = revb[revb[r]] endif endelse return end