Creating sublists from a list based on another list in python

Question

I have two lists: l1 and l2, which are shown below.

l1 = [0, 1, 1, 0, 1]
l2 = [0, 0, 1, 1, 0]

I want to sublist l1 into a list of lists based on values in l2, like below, one for '0' and one for '1'

[[0, 1, 1], [1, 0]]

this basically says, for 0 in l2, there are three values in l1 and for 1 in l2, there are two values in l1.

Is there a simple python way to do this?

EDIT: Here is what I have done to solve. This works well, but seems clunky and strange looking. I am looking for a more elegant solution.

d = [[],[]]
for i in range(len(l2)):
    d[l2[i]].append(l1[i])

What have you tried, and what precisely is the problem with it? This isn't a code writing service. — jonrsharpe
– jonrsharpe, Commented Mar 19, 2017 at 6:29

busybear · Accepted Answer · 2018-12-29 06:12:24Z

3

[[l1[n] for n, l2s in enumerate(l2) if l2s == element] for element in set(l2)]

Something like this might do the trick. set is just a bit faster.

edited Dec 29, 2018 at 6:12

answered Mar 19, 2017 at 6:50

busybear

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Mark Tolonen · Accepted Answer · 2017-03-19 07:04:41Z

2

Basically your same algorithm, but a little more Pythonic. zip iterates the lists in parallel, giving you the first items, then the 2nd items, etc.:

l1 = [0, 1, 1, 0, 1]
l2 = [0, 0, 1, 1, 0]

L = [[],[]]
for a,b in zip(l1,l2):
    L[b].append(a)

answered Mar 19, 2017 at 7:04

Mark Tolonen

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Comments

Kevin J. Chase · Accepted Answer · 2017-03-19 07:54:08Z

0

I've assumed that you might have an arbitrary number of subsets defined by their "keys" in l2, and also that your keys are both sortable and hashable:

def partition(keys, values):
    unique_keys = sorted(set(keys))
    for key in unique_keys:
        yield [val for idx, val in enumerate(values) if keys[idx] == key]
    return

Using your two lists, l1 and l2, you could produce your list-of-lists like this:

list(partition(keys=l2, values=l1))

(Note that, technically, a partition is a function of sets, not lists, but I think it gets the idea across.)

answered Mar 19, 2017 at 7:54

Kevin J. Chase

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