Python create new lists within a list based on the index within a list

Question

If I have the list

a = ['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc', '2 5 6 8', '2 7 3 9', '2 etc etc']

I want to be able to sort this based on what each element starts on. So the output I want is:

a = [['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc'], ['2 5 6 8', '2 7 3 9', '2 etc etc']]

But the thing is, for my real code, I won't know have many strings starts with a '1' or with a '2', so therefore I can't divide the list based on a fixed value, is there a way of comparing each element and combine them if they're the same?

Answer 1

You can use itertools.groupby() combined with a list comprehension:

>>> import itertools
>>> a = ['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc', '2 5 6 8', '2 7 3 9', '2 etc etc']
>>> [list(x[1]) for x in itertools.groupby(a, lambda i: i.split(" ")[0])]
[['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc'], ['2 5 6 8', '2 7 3 9', '2 etc etc']]

Note that .groupby() requires the iterable (ie a ) to be sorted, so you may have to sort it first if your real data looks different.

Answer 2

This works without using any package and independently of the type of object the 0th element may be:

a = ['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc', '2 5 6 8', '2 7 3 9', '2 etc etc']

already_sorted = []
new_a = []

for i in range(0, len(a)):
    if i in already_sorted:
        continue
    else:
        tmp = []
        for j in range(0, len(a)):

            if a[i][0] == a[j][0] and j not in already_sorted:
                tmp.append(a[j])
                already_sorted.append(j)

        new_a.append(tmp)

print(new_a)

Output:

[['1 2 3 4 5', '1 2 3 4 etc', '1 etc etc'], ['2 5 6 8', '2 7 3 9', '2 etc etc']]