How do I find the index of an unknown value in python?

Question

So basically I have a list like this -

[None,None,None,val]

and I don't know what val is. So how do I get the index of val (3) only knowing that it is not None

`max(a, key=lambda i: 0 if i is None else i)`. There may still be a better way of writing this or another way. — shahkalpesh, Dec 30 '21 at 05:09
@shahkalpesh That doesn't work. WIth input `[None,None,None,3,5,2]` it returns 5, not 3. — Barmar, Dec 30 '21 at 05:10
It looks like that's the index of the last non-None value. I suppose that could be what the OP wants. — Barmar, Dec 30 '21 at 05:12
@sr0812: You should now try to find the index of the value. You have the list functions to do it. — shahkalpesh, Dec 30 '21 at 05:14
Based on https://stackoverflow.com/a/1701404/786593 you can do `next(i for i,v in enumerate(a) if v is not None)` — hostingutilities.com, Dec 30 '21 at 05:17
A pretty bad one liner for you `[item is not None for item in [None,None,None,val]].index(True)` - this returns index of first non-None value. — Kris, Dec 30 '21 at 05:19

Ali Hejazizo · Accepted Answer · 2021-12-31T07:16:36.220

0

This one line solution gives you the first value that is not None.

values = filter(lambda item: item is not None, mylist)
target_value = next(values)

print(target_value)

This returns the actual value, to get the index, look into this answer first comment which is:

next(filter(lambda item: item[1] is not None, enumerate(mylist)))[0]

edited Dec 31 '21 at 07:16

answered Dec 30 '21 at 06:51

Ali Hejazizo

2

OP wants the index not the item itself, so `next(filter(lambda item: item[1] is not None, enumerate(mylist)))[0]` seems more apt. – Dec 30 '21 at 06:58
1

Yes. Didn't see that part. Will edit the response so that they look into your comment. – Ali Hejazizo Dec 30 '21 at 07:01

1 Answers1