unexpected python if behaviour

You need to break down your code - bit by bit - to see what's going on, and your use of parens is changing the meaning of this code.

When Python see parens, it evaluates that code first, so:

('z' or 'a')

returns 'z', which isn't in b, so your expression is false. Note: 'or' short-circuits.

If you remove the parens:

if 'z' or 'a' in b:

things should work the way you expect, or use the parens like this:

if ('z' in b ) or ('a' in b ):

to do the same thing.

Main message: parens are *not* required around conditional expressions in Python in the same way that they are in other languages, and using them changes how things work.

Hope this helps.

Regards.

--Paul.

Thanks, that answers the question
>>> print ('z' or 'a')
z
>>> print ('a' or 'z')
a
>>>

Now I know why...

Another strange related thing..seems like the parenthesis thing was only part of the problem.

>>> fmt.keys()
['0', '3', '2', 'description']
>>> if 'date' or 'description' in fmt.keys():
... print fmt
...
{'0': 0, '3': 3, '2': 2, 'description': 1}
>>> if 'description' or 'date' in fmt.keys():
... print fmt
...
{'0': 0, '3': 3, '2': 2, 'description': 1}
>>> if 'description' and 'date' in fmt.keys():
... print fmt
...
>>> if 'date' and 'description' in fmt.keys():
... print fmt
...
{'0': 0, '3': 3, '2': 2, 'description': 1}
>>>

Yeah - the solution is to explicitly state each test - but a shortcut would have been nice... (Python 2.5.2 BTW)

I think you are failing to understand the meaning of your tests.

What

if 'date' or 'description' in fmt.keys()

means is if 'date' is True, or if ('description' in fmt.keys()) is true, then do what is specified. As a string literal will always test true, your first two tests are true (True or True, True or False). Your third test is false (True and False) and fourth is true (True and True).

I'm not sure how to solve it other than to explicitly state each tests. With the set type in python 2.6 and above, it could be solveable by using set interections:

if set(['date','description']).intersection(fmt):
#at least one in fmt.keys()
print fmt

if set(['date','description']).issubset(fmt):
#all are in fmt.keys()
print fmt