根据子列表的第一个元素拆分子列表列表

Question

if i have a list that looks something like (list a)

my_list=[[3, 1], [3, 0], [3, 0], [3, 1], [4, 1], [4, 0]]

i want to split that list into something like this (list b)

my_list = [ [[3, 1], [3, 0], [3, 0], [3, 1]] , [[4, 1], [4, 0]] ]

as you can see, list b is sorted and grouped together by the values of the first element of the sublists. The order of the pairs doesn't change.

thank you!

Answer 1

You can use itertools.groupby to group on the first element:

my_list = itertools.groupby(my_list, key = lambda e: e[0])

This will give you an itertools.groupby generator object of (key, list) pairs. Ignore the keys and convert it into a list by doing

[list(e[1]) for e in my_list]

This gives:

[[[3, 1], [3, 0], [3, 0], [3, 1]], [[4, 1], [4, 0]]]

Answer 2

itertools.groupby(...) should do the trick:

import itertools

my_list=[[3, 1], [3, 0], [3, 0], [3, 1], [4, 1], [4, 0]]
#although your input seems to be sorted by 1st element I'll put it in here, in case if it wouldn't be
my_list=sorted(my_list, key=lambda x: x[0])

my_list=list(list(el) for k, el in itertools.groupby(my_list, key=lambda x: x[0]))

Output:

[[[3, 1], [3, 0], [3, 0], [3, 1]], [[4, 1], [4, 0]]]

Ref: https://docs.python.org/2/library/itertools.html#itertools.groupby

Answer 3

my_list=[[3, 1], [3, 0], [3, 0], [3, 1], [4, 1], [4, 0]]
new_list =[]

j =0
for i in list(set([ value[0] for value in my_list])) :
    temp_list = []
    while (j < len(my_list)) and (my_list[j][0] == i):
        temp_list.append(my_list[j])
        j=j+1
    new_list.append(temp_list)


print (new_list)

[[[3, 1], [3, 0], [3, 0], [3, 1]], [[4, 1], [4, 0]]]