Question

Consider the statement below:

Prob 1:

>>> a="abc"
>>> 'd' or 'e' in a
'd'

Someone please explain this . I was expecting a True or False …

Prob 2:

>>> print any(c in a for c in 'da')
True

Whats happening here ? If i do this ,

>>> (c in a for c in 'da')
<generator object <genexpr> at 0x011E4300>

As you can see , it gives generator object…What role does ‘any’ (method,function ??) play
here? And the result ?

Prob 3:

>>> Pattern="sdfdfg"
>>> if '\\'or '^' or '.' in Pattern:
    print "yes"
else:print "no"

yes

How on earth is this “YES” ??

Show me the light someone plz………..

Answer 1

The results you are seeing are the result of the precedence of the or and in operators being different than you expect.

• 'd' or 'e' in a

This should be read as 'd' or ('e' in a), that is, the 2 operands to or are 'd' and ('e' in a). Since d is considered a true value in python, or doesn’t look at its next operand, and just returns its first operand, 'd'. Note that this means that or isn’t just a logical OR, it can deal with other types, and return other types.

• print any(c in a for c in 'da')

Read this print any((c in a) for c in 'da'), ie, go over the elementns of 'da', test if that letter is also in a, and then see if that held true for any of them. The reason you see <generator object <genexpr> at 0x011E4300> is that Python does not show you the elements of a generator unless it has cause to step over them. If you want to see the individual elements, write:

[(c in a) for c in 'da']

• if '\\'or '^' or '.' in Pattern:

Again, read this if ('\\' or '^') or ('.' in Pattern:). Since '\\' is true, the result is true.

Table of precedence