Find the first instance of a nonzero number in a list in Python

Question

I have a list like this:

myList = [0.0, 0.0, 0.0, 2.0, 2.0]

I would like to find the location of the first number in the list that is not equal to zero.

myList.index(2.0)

It works in this example, but sometimes the first nonzero number will be 1 or 3.

Is there a fast way of doing this?

just be cautious about comparison with floating point numbers — Ant, Oct 21 '13 at 18:57
I suspect you're getting a lot of down votes because your question does not demonstrate that you've met the ["what have you tried?"](http://mattgemmell.com/2008/12/08/what-have-you-tried/) requirement. Just a helpful tip for future questions. And also a helpful tip for down-voters, let the user know why you're down voting so that they can make corrections. — Bi Rico, Oct 21 '13 at 19:04
Related: [What is the best way to get the first item from an iterable matching a condition?](http://stackoverflow.com/questions/2361426/what-is-the-best-way-to-get-the-first-item-from-an-iterable-matching-a-condition), [find first list item that matches criteria](http://stackoverflow.com/questions/9868653/find-first-list-item-that-matches-criteria) — user2314737, May 05 '17 at 09:16

score 62 · Accepted Answer · answered Oct 21 '13 at 18:54

62

Use next with enumerate:

>>> myList = [0.0 , 0.0, 0.0, 2.0, 2.0]
>>> next((i for i, x in enumerate(myList) if x), None) # x!= 0 for strict match
3

answered Oct 21 '13 at 18:54

Ashwini Chaudhary

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What would be the equivalent for numpy arrays ? – Charly Empereur-mot Dec 15 '20 at 18:58
It begs questions about comparison of floating numbers. E.g., what would happen if `42.8/21.4` was used instead of `2.0`? Can you [address](https://stackoverflow.com/posts/19502403/edit) comparison of floating numbers in your answer (even if it is somehow automatically handled (rounding, etc.))? (But ***without*** "Edit:", "Update:", or similar - the answer should appear as if it was written today. And ***without*** reference to this comment.) – Peter Mortensen Feb 04 '22 at 12:58

Puffin GDI · Answer 2 · 2019-04-10T01:45:41.367

21

Use filter

Python 2:

myList = [0.0, 0.0, 0.0, 2.0, 2.0]
myList2 = [0.0, 0.0]

myList.index(filter(lambda x: x!=0, myList)[0])       # 3
myList2.index(filter(lambda x: x!=0, myList2)[0])     # IndexError

Python 3: (Thanks for Matthias's comment):

myList.index(next(filter(lambda x: x!=0, myList)))    # 3
myList2.index(next(filter(lambda x: x!=0, myList2)))  # StopIteration

# from Ashwini Chaudhary's answer
next((i for i, x in enumerate(myList) if x), None)    # 3
next((i for i, x in enumerate(myList2) if x), None)   # None

You have to handle special case.

edited Apr 10 '19 at 01:45

answered Oct 21 '13 at 19:08

Puffin GDI

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Great alternative too, main difference from `next` is that `next` provides a default value whereas this would raise an exception. – Wtower Mar 30 '15 at 12:18
3

for python 3 you can use `myList.index(next(filter(lambda x: x!=0, myList)))` – Matthias Sep 29 '16 at 13:48
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I like this answer best, but with the Matthias mod to use "next". Note that in either case, depending on the situation you should consider a test of any(myList) first, to make sure there is at least one nonzero value. It may help with program flow and avoiding exceptions. – Starman Apr 09 '19 at 21:01

score 11 · Answer 3 · edited Feb 04 '22 at 13:03

11

You can use numpy.nonzero:

myList = [0.0, 0.0, 0.0, 2.0, 2.0]
I = np.nonzero(myList)
# The first index is necessary because the vector is within a tuple
first_non_zero_index = I[0][0]
# 3

edited Feb 04 '22 at 13:03

Peter Mortensen

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answered Jan 16 '16 at 10:33

Rodrigo Ferreira

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score 6 · Answer 4 · answered Oct 21 '13 at 19:00

Here's a one liner to do it:

val = next((index for index,value in enumerate(myList) if value != 0), None)

Basically, it uses next() to find the first value, or return None if there isn't one. enumerate() is used to make an iterator that iterates over index,value tuples so that we know the index that we're at.

score 5 · Answer 5 · edited Feb 04 '22 at 13:04

5

Use this:

[i for i, x in enumerate(myList) if x][0]

edited Feb 04 '22 at 13:04

Peter Mortensen

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answered Jun 20 '17 at 14:17

SebMa

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score 3 · Answer 6 · edited Feb 04 '22 at 13:04

3

Using next with enumerate is excellent when the array is large. For smaller arrays, I would use argmax from NumPy so that you won't need a loop:

import numpy as np

myList = [0.0, 0.0, 0.0, 2.0, 2.0]
myArray = np.array(myList)
np.argmax(myArray > 0)
3

edited Feb 04 '22 at 13:04

Peter Mortensen

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answered Mar 07 '17 at 18:21

cjeiler

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This will not find the first nonzero element, e.g. for `[0, 0, 1, 2]`. – cheersmate Oct 08 '18 at 13:39
Yes it will, but it's not very intuitive. All elements in `myArray > 0` are either `False` or `True`, so the maximum of that array is `True`. According to the [documentation](https://numpy.org/doc/stable/reference/generated/numpy.argmax.html) for numpy.argmax, "In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned." So with your example, if `myArray` is `[0, 0, 1, 2]`, then `myArray > 0` is `[False, False, True, True]` and the `argmax` of that is 2. – PieterNuyts Feb 13 '23 at 14:23

score 1 · Answer 7 · edited May 16 '17 at 16:13

1

What about using enumerate? Check the enumerate documentation.

def first_non_zero(mylist):
  for index, number in enumerate(mylist):
    if number != 0: # or 'if number:'
      return index

edited May 16 '17 at 16:13

David Schumann

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answered Oct 21 '13 at 18:55

Eduardo

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score 1 · Answer 8 · edited Feb 04 '22 at 13:10

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Do the following:

print (np.nonzero(np.array(myList))[0][0])

This is more convenient, because along with finding nonzero values, this can also help to apply logic function directly. For example:

print (np.nonzero(np.array(myList)>1))

edited Feb 04 '22 at 13:10

Peter Mortensen

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answered Sep 14 '18 at 09:55

devil in the detail

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Perhaps elaborate a little bit in your answer how floating point comparisons are handled? Presumably it is not *exact* comparison. (But ***without*** "Edit:", "Update:", or similar - the answer should appear as if it was written today.) – Peter Mortensen Feb 04 '22 at 13:15

Rushy Panchal · Answer 9 · 2013-10-21T19:11:40.533

0

Simply use a list comprehension:

myDict = {x: index for index, x in enumerate(myList) if x}

The indices of the nonzero elements are myDict[element].

edited Oct 21 '13 at 19:11

answered Oct 21 '13 at 18:55

Rushy Panchal

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This doesn't give the index, it just filters out the non-zero elements from the list. – Ian McLaird Oct 21 '13 at 18:59

Find the first instance of a nonzero number in a list in Python

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