On Tue, 25 Jan 2005 23:53:26 +1000, Nick Coghlan <[EMAIL PROTECTED]> wrote:
>Peter Otten wrote: >> Nick Coghlan wrote: >> >> >>>>Py> print islice((x for x in xrange(1, 996) if x % 2 == 0), 1, 2).next() >>>>4 >>> >>>Wouldn't it be nice if this could be spelt: >>> >>>print (x for x in xrange(1, 996) if x % 2 == 0)[2] >>> >>>Well, I just put a patch on SF to enable exactly that: >>>http://www.python.org/sf/1108272 >> >> >> I like it. Of course you always have to bear in mind that one giant leap for >> a list could be _many_ small steps for an iterator. > >Indeed. The main cases I am thinking of involve picking off the first few >items >of an iterator (either to use them, or to throw them away before using the >rest). > >And if an app actually *needs* random access, there's a reason lists still >exist ;) > You can bail out of a generator expression with a raise-StopIteration expression spelled iter([]).next() ;-) >>> def show(x): print x,; return x ... >>> list(show(x) for x in xrange(20) if x<8 or iter([]).next()) 0 1 2 3 4 5 6 7 [0, 1, 2, 3, 4, 5, 6, 7] Well, since >>> list(show(x) for x in xrange(20) if x<8) 0 1 2 3 4 5 6 7 [0, 1, 2, 3, 4, 5, 6, 7] this change due to adding iter([]).next() might be more convincing: >>> list(show(x) for x in xrange(20) if x<8 or True) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19] >>> list(show(x) for x in xrange(20) if x<8 or iter([]).next() or True) 0 1 2 3 4 5 6 7 [0, 1, 2, 3, 4, 5, 6, 7] Applied to example, >>> print list(x for i,x in enumerate(x for x in xrange(1, 996) if x % 2 ==0) >>> if i<3 or iter([]).next())[2] 6 or in case you just want one item as result, >>> print list(x for i,x in enumerate(x for x in xrange(1, 996) if x % 2 ==0) >>> if i==2 or i==3 and iter([]).next())[0] 6 Regards, Bengt Richter -- http://mail.python.org/mailman/listinfo/python-list
