Finding the indices of matching elements in list in Python

Question

I have a long list of float numbers ranging from 1 to 5, called "average", and I want to return the list of indices for elements that are smaller than a or larger than b

def find(lst,a,b):
    result = []
    for x in lst:
        if x<a or x>b:
            i = lst.index(x)
            result.append(i)
    return result

matches = find(average,2,4)

But surprisingly, the output for "matches" has a lot of repetitions in it, eg [2, 2, 10, 2, 2, 2, 19, 2, 10, 2, 2, 42, 2, 2, 10, 2, 2, 2, 10, 2, 2, ...] .

Why is this happening?

Answer 1

You are using .index() which will only find the first occurrence of your value in the list. So if you have a value 1.0 at index 2, and at index 9, then .index(1.0) will always return 2 , no matter how many times 1.0 occurs in the list.

Useenumerate() to add indices to your loop instead:

def find(lst, a, b):
    result = []
    for i, x in enumerate(lst):
        if x<a or x>b:
            result.append(i)
    return result

You can collapse this into a list comprehension:

def find(lst, a, b):
    return [i for i, x in enumerate(lst) if x<a or x>b]

Answer 2

if you're doing a lot of this kind of thing you should consider using numpy .

In [56]: import random, numpy

In [57]: lst = numpy.array([random.uniform(0, 5) for _ in range(1000)]) # example list

In [58]: a, b = 1, 3

In [59]: numpy.flatnonzero((lst > a) & (lst < b))[:10]
Out[59]: array([ 0, 12, 13, 15, 18, 19, 23, 24, 26, 29])

In response to Seanny123's question, I used this timing code:

import numpy, timeit, random

a, b = 1, 3

lst = numpy.array([random.uniform(0, 5) for _ in range(1000)])

def numpy_way():
    numpy.flatnonzero((lst > 1) & (lst < 3))[:10]

def list_comprehension():
    [e for e in lst if 1 < e < 3][:10]

print timeit.timeit(numpy_way)
print timeit.timeit(list_comprehension)

The numpy version is over 60 times faster.

Answer 3

>>> average =  [1,3,2,1,1,0,24,23,7,2,727,2,7,68,7,83,2]
>>> matches = [i for i in range(0,len(average)) if average[i]<2 or average[i]>4]
>>> matches
[0, 3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15]