pro match, a, b, suba, subb, COUNT = count, SORT = sort, epsilon=epsilon ;+ ; NAME: ; MATCH ; PURPOSE: ; Routine to match values in two vectors. ; ; CALLING SEQUENCE: ; match, a, b, suba, subb, [ COUNT =, /SORT, EPSILON = ] ; ; INPUTS: ; a,b - two vectors to match elements, numeric or string data types ; ; 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] ; suba and subb are set to !NULL if there are no matches (or set to -1 ; if prior to IDL Version 8.0) ; ; OPTIONAL INPUT KEYWORD: ; /SORT - By default, MATCH uses two different algorithm: (1) the ; /REVERSE_INDICES keyword to HISTOGRAM is used for integer data, ; while (2) 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. ; epsilon - if values are within epsilon, they are considered equal. Used only ; only for non-integer matching. Note that input vectors should ; be unique to within epsilon to provide one-to-one mapping.. ; Default=0. ; ; 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 a[suba] = b[subb] = [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 ; Use a HISTOGRAM algorithm for integer vector inputs for improved ; performance W. Landsman March 2000 ; Work again for strings W. Landsman April 2000 ; Use size(/type) W. Landsman December 2002 ; Work for scalar integer input W. Landsman June 2003 ; Assume since V5.4, use COMPLEMENT to WHERE() W. Landsman Apr 2006 ; Added epsilon keyword Kim Tolbert March 14, 2008 ; Fix bug with Histogram method with all negative values W. Landsman/ ; R. Gutermuth, return !NULL for no matches November 2017 ; Added epsilon test in na=1||nb=1 section (missed that when added ; epsilon in 2008) Kim Tolbert July 10, 2018 ; ;- ;------------------------------------------------------------------------- compile_opt idl2 Catch, theError IF theError NE 0 then begin Catch,/Cancel void = cgErrorMsg(/quiet) RETURN ENDIF if N_elements(epsilon) EQ 0 then epsilon = 0 if N_params() LT 3 then begin print,'Syntax - match, a, b, suba, subb, [ COUNT =, EPSILON=, /SORT]' print,' a,b -- input vectors for which to match elements' print,' suba,subb -- output subscript vectors of matched elements' return endif da = size(a,/type) & db =size(b,/type) if keyword_set(sort) then hist = 0b else $ hist = (( da LE 3 ) || (da GE 12)) && ((db LE 3) || (db GE 12 )) if ~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) || (nb EQ 1) then begin if (nb GT 1) then begin if epsilon eq 0. then subb = where(b EQ a[0], nw) else $ subb = where(abs(b - a[0]) lt epsilon, nw) if (nw GT 0) then suba = replicate(0,nw) else suba = [-1] endif else begin if epsilon eq 0. then suba = where(a EQ b[0], nw) else $ suba = where(abs(a - b[0]) lt epsilon, 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 if epsilon eq 0. then $ firstdup = where( (c EQ shift(c,-1)) and (vec NE shift(vec,-1)), Count ) $ else $ firstdup = where( (abs(c - shift(c,-1)) lt epsilon) 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 subb = ind[ where( vec, complement = vzero) ] ;b subscripts suba = ind[ vzero] 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 if !VERSION.RELEASE GE '8.0' then begin suba = !NULL subb = !NULL endif else begin suba = -1 subb = -1 endelse COUNT = 0L if maxab lt minab then return ;No overlap 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