Python - Finding elements in a list efficiently

Question

I have a list, list_a, that contains floating numbers:

list_a = [[[ 0 for i in range(40)] for j in range(1000)]for k in range(47)]

And I have a sorted version of this:

list_a_sorted = list_a
list_a_sorted[0].sort()

So list_a_sorted is sorted and contains values for list_a starting from the lowest first. Let's assume it is as follows:

[2.3,3.1.........9]

So 2.3 is the lowest value but how can I find out if this was the 8th element in list_a or the 15th or nth?

As my lists are quite large, I also need to do this as efficiently as possible? Any help would be appreciated, thanks

Answer 1

to find the position of an element in a list you can use l.index(something)

http://docs.python.org/library/stdtypes.html#typesseq

Answer 2

if you want to find the n smallest values in an unsorted list look at heapq.nsmallest() which may be more efficient if n isn't too large. To find the position of the smallest values try this:

>>> from heapq import nsmallest
>>> from random import random
>>> values = [random() for i in range(20)]
>>> values
[0.012227103410989537, 0.9782624648209769, 0.9896111545377924, 0.9033620518745159, 0.6767780103989406, 0.4595455061820246, 0.39814471642551696, 0.6904798136040561, 0.8727083752258934, 0.6680153337266017, 0.606044647078923, 0.5644656135679249, 0.934351848916147, 0.05955628567745763, 0.7236000566917332, 0.8303865367817055, 0.9671576336593124, 0.3164892315873573, 0.8416372881413415, 0.5009057933309073]
>>> nsmallest(4, range(len(values)), key=lambda i: values[i])
[0, 13, 17, 6]

Or faster but slightly less clear:

>>> nsmallest(4, range(len(values)), key=values.__getitem__)
[0, 13, 17, 6]

For your list you may want something like (untested code):

def indices():
    for k in range(47):
        for j in range(1000):
            for i in range(40):
                yield k, j, i
def keyfn(ind):
    k, j, i = ind
    return list_a[k][j][i]

print(nsmallest(4, indices(), key=keyfn))