Mapping two list without looping

Question

I have two lists of equal length. The first list l1 contains data.

l1 = [2, 3, 5, 7, 8, 10, ... , 23]

The second list l2 contains the category the data in l1 belongs to:

l2 = [1, 1, 2, 1, 3, 4, ... , 3]

How can I partition the first list based on the positions defined by numbers such as 1, 2, 3, 4 in the second list, using a list comprehension or lambda function. For example, 2, 3, 7 from the first list belongs to the same partition as they have corresponding values in the second list.

The number of partitions is known at the beginning.

Answer 1

You can use a dictionary:

>>> l1 = [2, 3, 5, 7, 8, 10, 23] 
>>> l2 = [1, 1, 2, 1, 3, 4, 3]

>>> d = {}
>>> for i, j in zip(l1, l2):
...     d.setdefault(j, []).append(i)
... 
>>> 
>>> d
{1: [2, 3, 7], 2: [5], 3: [8, 23], 4: [10]}

Answer 2

If a dict is fine, I suggest using a defaultdict:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for number, category in zip(l1, l2):
...     d[category].append(number)
... 
>>> d
defaultdict(<type 'list'>, {1: [2, 3, 7], 2: [5], 3: [8, 23], 4: [10]})

Consider using itertools.izip for memory efficiency if you are using Python 2.

This is basically the same solution as Kasramvd's, but I think the defaultdict makes it a little easier to read.

Answer 3

This will give a list of partitions using list comprehension :

>>> l1 = [2, 3, 5, 7, 8, 10, 23] 
>>> l2 = [1, 1, 2, 1, 3, 4, 3]
>>> [[value for i, value in enumerate(l1) if j == l2[i]] for j in set(l2)]
[[2, 3, 7], [5], [8, 23], [10]]