On 07/03/07, David Perlman <[EMAIL PROTECTED]> wrote: > On Mar 6, 2007, at 11:03 AM, Alan Gauld wrote: > > It's doing the latter and since anything that's not 'empty' in > > Python evaluates to true we wind up checking whether > > true == (item in word) > > > > So if the item is in word we get true == true which is true. > Sorry, but this still doesn't make sense to me.
Hmm, ok, colour me confused as well. Let's try creating the conditions in the original poster's code. Python 2.4.3 (#1, Mar 30 2006, 11:02:16) [GCC 4.0.1 (Apple Computer, Inc. build 5250)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> item = 'b' >>> word2 = 'bees' The test is "item == item in word2". >>> item == item in word2 True Let's see how we can group this: >>> item == (item in word2) False That makes sense. It's not True though. >>> (item == item) in word2 Traceback (most recent call last): File "<stdin>", line 1, in ? TypeError: 'in <string>' requires string as left operand That also makes sense.. >>> if (item == item) in word2: ... print 'foo' ... Traceback (most recent call last): File "<stdin>", line 1, in ? TypeError: 'in <string>' requires string as left operand >>> if item == item in word2: ... print 'foo' ... foo So ... hmm. Not sure what's going on here. I also notice that, according to http://docs.python.org/ref/summary.html, '==' has higher precedence than 'in'. Which means that 'foo == bar in baz' should group as '(foo == bar) in baz'. But Luke's example contradicts this: >>> lst = [1,2,3,4] >>> 555 == 555 in lst False >>> (555 == 555) in lst True >>> 1 == 1 in lst True (this works because 1 == True) -- John. _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
