find indexes of all None items in a list in python

Question

I have a list of strings, some of them are None. i want to get a new list of all the Indexes of None.

list = ['a', 'b', None, 'c' ,None, 'd']

using the function index

n = list.index(None)

will only return the first appearance, n= 2, while i want to see n= [2,4]. thanks you.

dir(builtins) - will show that using list as var name is bad idea — Drako, Oct 18 '18 at 08:19
Possible duplicate of [How to find all occurrences of an element in a list?](https://stackoverflow.com/questions/6294179/how-to-find-all-occurrences-of-an-element-in-a-list) — Nouman, Oct 18 '18 at 08:52

U13-Forward · Accepted Answer · 2018-10-18T08:20:08.847

15

Try enumerate:

l=[i for i,v in enumerate(list) if v == None]

Or range:

l=[i for i in range(len(list)) if list[i] == None]

Both cases:

print(l)

Is:

[2,4]

Big Note: it is not good to name variables a existing method name, that overwrites it, (now it's list), so i would prefer it as l (or something)

I recommend the first example because enumerate is easy, efficient.

edited Oct 18 '18 at 08:20

answered Oct 18 '18 at 08:14

U13-Forward

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@MorZamir Now you've got more than 15 reputation, so you can up-vote if you want :-) – U13-Forward Oct 18 '18 at 23:39
1

funny to see your comment today for having 15 reputation, after receiving a badge for first asked question with 1k views. – Zusman Jun 13 '19 at 08:26
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@MorZamir Wow congratulations for getting it, i have it as well. – U13-Forward Jun 13 '19 at 08:30

LMSharma · Answer 2 · 2018-10-18T09:05:04.453

2

Faster way. Very useful in case of long list.

list = ['a', 'b', None, 'c' ,None, 'd']
import numpy as np
print(np.where(np.array(list) == None)[0])

Output :

[2 4]

In case you need list of index :

print(np.where(np.array(list) == None)[0].tolist())
>>> [2, 4]

edited Oct 18 '18 at 09:05

answered Oct 18 '18 at 08:29

LMSharma

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Note, it has a warning ``FutureWarning: comparison to `None` will result in an elementwise object comparison in the future.``, in the future, this will not work. – U13-Forward Oct 18 '18 at 08:39
1

No warnings in python 2.7 – MEdwin Oct 18 '18 at 08:41
@MEdwin Oh i didn't test on python 3, so? – U13-Forward Oct 18 '18 at 08:43
@Black Thunder . I don't know why but numpy arrays are printed without commas. That's why I have used .tolist() method – LMSharma Oct 18 '18 at 09:03
The first method gives me nothing except an empty list. – Nouman Oct 18 '18 at 09:05
@Lalit For python 3.6 it does. – U13-Forward Oct 18 '18 at 23:36

score 2 · Answer 3 · answered Oct 18 '18 at 08:41

2

Here's something different but it doesn't use list comprehension:

>>> l = ['a', 'b', None, 'c' ,None, 'd']
>>> out = []
>>> for _ in range(l.count(None)):
    out.append(l.index(None))
    l[l.index(None)] = "w"


>>> out
[2, 4]
>>>

answered Oct 18 '18 at 08:41

Nouman

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find indexes of all None items in a list in python

3 Answers3