Python find first instance of non zero number in list

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 e

Answer 1

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].

Answer 2

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.

Answer 3

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

Answer 4

How'bout this :

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

Answer 5

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

Answer 6

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