[python] Finding the indices of matching elements in list in Python

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, e.g. [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?

>>> 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]

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.