Thanks. Here's how the inner loop should be: imgPaths2=map(lambda x: (x, re.sub( r"^(.+?)-s(\.[^.]+)$",r"\1\2", x)), imgPaths)
though, now i just need something like map( lambda x: os.path.exists(s)? x[1]:x[0],impPaths2) but Pyhton doesn't support the test ? trueResult : falseResult construct. (the semantic of this construct, of a conditional that RETURNS A EXPRESSION, all in one line, is important in functional programing. Perl supports it. In Mathematica, it's simply the form If[testExpr, trueExpr, falseExpr]. In lisp, similar: (if testExpr trueExpr falseExpr) ) is there a way to similate it? Xah [EMAIL PROTECTED] ∑ http://xahlee.org/ [EMAIL PROTECTED] wrote: > what do you mean by one line ? Using map/filter, I believe it is > possible. > > Somthing like: > > map(lambda (s,f): os.path.exists(f) and f or s, > map(lambda x: (x, re.replace(x, "-s","")), imgPaths) > > My regex is a bit rusty but I hope you got the idea of what I am trying > to do. If there is a way to make the re return without destroying x, > the outer map is not needed I believe(that is run it twice, once for > getting the filename to do the testing, then again based on the testing > result). > -- http://mail.python.org/mailman/listinfo/python-list
