python checking strings in a list?

There are two issues with your code. The first is that you are modifying the list while iterating over it. The second is that you are using the or operator in the wrong way – the condition in the if statement will always be True.

Here's a fixed version:

test = [i for i in range(100) if set("0468").isdisjoint(str(i))]

Consider this:

(Pdb) i=2
(Pdb) i='2'
(Pdb) '0' or 'beer' in str(i)
'0'
(Pdb) bool('0')
True

Do you see why '0' or '4' or '6' or '8' in str(i) is not always a boolean value?

Now, consider what you're removing:

>>> l=[str(i) for i in range(10)]
>>> for i in l:
...   print i
...   print l
...   l.remove(i)
... 
0
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
2
['1', '2', '3', '4', '5', '6', '7', '8', '9']
4
['1', '3', '4', '5', '6', '7', '8', '9']
6
['1', '3', '5', '6', '7', '8', '9']
8
['1', '3', '5', '7', '8', '9']

As you remove elements in-place while you're iterating over the list, you're shortening the list and skipping each next value.

Sven has the right pythonic solution, but I felt a more lengthy explanation might be worthwhile.

When you execute something like '0' or '4', the or operator will verify if the first parameter ('0') is false-ish and, if it is not false, it returns the first value. Since '0' is not false (it is a non-empty string, which yields True when used as boolean), it will return the first value and the following one will not even be considered:

>>> '0' or '4'
'0'

If you repeat it, you will get the same result:

>>> '0' or '4' or '6' or '8'
'0'

Also, the in operator has greater precedence, so it will be executed before each or:

'0' or '4' or '6' or '8' in '88' '0'

What you want to do in your condition is to verify if any of the values is in the result:

>>> '0' in str(i) or '4' in str(i) or '6' in str(i) or '8' in str(i)
True
>>> i = 17
>>> '0' in str(i) or '4' in str(i) or '6' in str(i) or '8' in str(i)
False

This is not the most elegant solution, but it is the best translation of your intentions.

So, what would be an elegant solution? As sugested by @sven, you can create a set (more about it) of the sought chars:

>>> sought = set("0468")
>>> sought
set(['0', '8', '4', '6'])

and then create a set of the digits in your number:

>>> i = 16
>>> set(str(i))
set(['1', '6'])

Now, just see if they are disjoint:

>>> i = 16
>>> sought.isdisjoint(set(str(i)))
False
>>> i = 17
>>> sought.isdisjoint(set(str(i)))
True

In this case, if the set are not disjoint, then you want to preserve it:

>>> found = []
>>> for i in range(100):
...     if sought.isdisjoint(set(str(i))):
...         found.append(i)
... 
>>> found
[1, 2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 22, 23, 25, 27, 29, 31, 32, 33, 35, 37, 39, 51, 52, 53, 55, 57, 59, 71, 72, 73, 75, 77, 79, 91, 92, 93, 95, 97, 99]

Most of the time, every time you get yourself creating a for loop for filtering an iterator, what you really want is a list comprehension:

>>> [i for i in range(100) if sought.isdisjoint(set(str(i)))]
[1, 2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 22, 23, 25, 27, 29, 31, 32, 33, 35, 37, 39, 51, 52, 53, 55, 57, 59, 71, 72, 73, 75, 77, 79, 91, 92, 93, 95, 97, 99]

Or, using a clumsier but more novice-friendly construct:

>>> [i for i in range(100) if not ( '0' in str(i) or '4' in str(i) or '6' in str(i) or '8' in str(i) )]
[1, 2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 22, 23, 25, 27, 29, 31, 32, 33, 35, 37, 39, 51, 52, 53, 55, 57, 59, 71, 72, 73, 75, 77, 79, 91, 92, 93, 95, 97, 99]

itms = ['0','4','6','8']
test = [str(i) for i in range(100) if not  any([x in str(i) for x in itms])]

maybe .... thats untested but i think it will work

The problem is here with the line:

if '0' or '4' or '6' or '8' in str(i):

It is incorrect. It would be better to write it so:

if any(x in i for x in '0468'):

The second problem (as mentioned Sven) is that you modify the list while iterating over it.

The best solution is:

[i for i in test if not any(x in i for x in '0468')]