If statement ordering and key error python

Question

Here I get a key error even I check if the key exists in the dict :

def foo(d):
    if (('element' in d.keys()) & (d['element'] == 1)):
        print "OK"

foo({})

In the documentation we can read :

The expression x and y first evaluates x; if x is false, its value is returned; otherwise, y is evaluated and the resulting value is returned.

Can anybody explain me this behavior?

You are looking at the documentation for `and` and executing `&`, the documentation for which is [here](https://docs.python.org/2/reference/expressions.html#binary-bitwise-operations) (on the same page). — Two-Bit Alchemist, Jun 20 '16 at 14:55
Actually I already CV'd but this is probably a duplicate: http://stackoverflow.com/questions/22646463/difference-between-and-boolean-vs-bitwise-in-python-why-difference-i — Two-Bit Alchemist, Jun 20 '16 at 14:57

score 6 · Accepted Answer · answered Jun 20 '16 at 14:51

6

& is "bitwise and", and is the logical "and" operator, they are not the same thing.

You should use and, and you can also remove the unneeded parenthesis for readability.

You also don't even have to call the keys method:

if 'element' in d and d['element'] == 1:

answered Jun 20 '16 at 14:51

DeepSpace

3

Or condense to `if d.get('element'):` since get automatically returns `None` (which is falsy). – Two-Bit Alchemist Jun 20 '16 at 14:56
2

Perhaps `if d.get('element') == 1:` to preserve the other part of the `if` statement – Simon Fraser Jun 20 '16 at 15:02

1 Answers1