Elegant way to slice a list with a condition

Question

Given a list [2,8,13,15,24,30], all the elements of which are supposed to be in range(31). Now I want to slice it into 3 lists, first list with numbers from 0 to 10, the second one with numbers from 11 to 20, and the others into the rest.

Here is my ugly code :

numbers = [2,8,13,15,24,30]
mylist = [[],[],[]] # I hate this the most...
for i in numbers:
    if i <= 10 :
        mylist[0].append(i)
    elif i > 10 and i <= 20:
        mylist[1].append(i)
    else:
        mylist[2].append(i)

print mylist

I don't think this is a good way to do it. Any suggestions please?

Answer 1

Since your input is sorted you can do this in one pass with itertools.groupby:

from itertools import groupby

[list(g) for _,g in groupby(numbers, lambda x: x//10)]
Out[3]: [[2, 8], [13, 15], [24, 26]]

No need to initialize a bunch of lists this way, groupby yields them on the fly.

This may be off-by-one in terms of how you wanted to treat the modulo-10 boundaries; if it's unclear you can always define your own grouper function:

def grouper(x):
    '''bins matching the semantics:
    [0,10] (10,20] (20, 30]'''
    return (x-1)//10 if x > 0 else 0

and use it thusly:

numbers = [2,8,13,15,24,30]

[list(g) for _,g in groupby(numbers, grouper)]
Out[5]: [[2, 8], [13, 15], [24, 30]]