How to build index map with multiple value from list in python with elegant way?

Question

I want to build a dict in python from a list, the key of the dict is the first column and the value of the map is the second column. Is there any elegant way to implement this?

For example, A= [[1,2], [1,3], [2,3]] The result is M={1:[2,3],2:[3]} My own way is for row in A: if row[0] not in M: M[row[0]] = [] M[row[0]].append(row[1]) Is there any better solution? Functional Programming methods are preferred.

Answer 1

One way to do this is with a defaultdict

from collections import defaultdict

A= [[1,2], [1,3], [2,3]]
d = defaultdict(list)
for k, v in A:
    d[k].append(v)

Answer 2

You can use setdefault:

A= [[1,2], [1,3], [2,3]]
M = {}                                                                 
for k,v in A:
    M.setdefault(k, []).append(v)

M 
# {1: [2, 3], 2: [3]}

If it must be functional (Disclaimer: wasteful, not recommended):

import operator, itertools
grps = itertools.groupby(sorted(A, key=operator.itemgetter(0)), operator.itemgetter(0))
M = {k: list(map(operator.itemgetter(1), v)) for k, v in grps}
M
# {1: [2, 3], 2: [3]}