[Tutor] trouble with function-- trying to check differences btwn 2 strings

[John Fouhy]

On 07/03/07, David Perlman <dperlman at wisc.edu> 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.