"if not in" in python to get the index

Question

I want to use "if not in" syntax in python but what I want to do is to if the element is found in list I want to get the index.

for example

if 3 not in [2,3,4]:
    print "hello"

If 3 is found in list I want to get the index which in this case is 1.

`else:at_index = [2,3,4].index(3)` – Steven Rumbalski Sep 24 '14 at 18:16 — Steven Rumbalski, Sep 24 '14 at 18:16

g4ur4v · Answer 1 · 2014-09-24T18:17:46.630

1

>>> a=[2,3,4]
>>> a.index(3)
1

EDIT

>>> a=[2,3,4]
>>> def check(n):
...     try:
...             print a.index(n)
...     except ValueError:
...             print "Element not found in list a"
... 
>>> check(5)
Element not found in list a
>>> check(3)
1

edited Sep 24 '14 at 18:17

answered Sep 24 '14 at 18:12

g4ur4v

3,210
5
32
57

Better use it with try-except. – Ashwini Chaudhary Sep 24 '14 at 18:13
This raises a ValueError exception if the element sought is not found. – Russell Borogove Sep 24 '14 at 18:13

"if not in" in python to get the index

1 Answers1