When checking if an item does not exist in a list, why doesn't this code work - if item in list == False:

Question

Consider this list:

list = [1,2,3,4,5]

I want to check if the number 9 is not present in this list. There are 2 ways to do this.

Method 1: This method works!

if not 9 in list: print "9 is not present in list"

Method 2: This method does not work.

if 9 in list == False: print "9 is not present in list"

Can someone please explain why method 2 does not work?

see http://stackoverflow.com/questions/7479808/python-operator-precedence-of-in-and-comparision — grc, Mar 11 '13 at 10:47

score 19 · Accepted Answer · answered Mar 11 '13 at 10:49

This is due to comparison operator chaining. From the documentation:

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

You are assuming that the 9 in list == False expression is executed as (9 in list) == False but that is not the case.

Instead, python evaluates that as (9 in list) and (list == False) instead, and the latter part is never True.

You really want to use the not in operator, and avoid naming your variables list:

if 9 not in lst:

Thanks! I was not aware of the concept of operator chaining. — Raghav Mujumdar, Mar 11 '13 at 10:54

Thanakron Tandavas · Answer 2 · 2013-03-11T10:54:36.883

4

It should be:

if (9 in list) == False: print "9 is not present in list"

edited Mar 11 '13 at 10:54

answered Mar 11 '13 at 10:49

Thanakron Tandavas

5,615
5
28
41

1

Outer parentheses are completely unnecessary. – grc Mar 11 '13 at 10:53
@grc Pretty sure they are invalid as well. – Syed M. Sannan May 01 '23 at 17:08

When checking if an item does not exist in a list, why doesn't this code work - if item in list == False:

2 Answers2