如何对列表的元素进行分组？

Question

I have a list of lists. Each list contains three elements, call them [number, letter, time] . It looks like this:

my_list = [[0, 'a', '0:00'],
           [1, 'a', '1:00'],
           [2, 'b', '2:00'],
           [3, 'c', '3:00'],
           [4, 'b', '4:00'],
           [5, 'a', '5:00'],
           [6, 'd', '6:00'],
           [7, 'c', '7:00'],
           [8, 'b', '8:00'],
           [9, 'e', '9:00']]

I want to format this list such that element[1] , the letter of each nested list becomes element[0] of a new nested list, with the occurrences of number and time as element[1] and element[2] .

The output I want should look like this

my list = [['a', [0, 1, 5], ['0:00', '1:00', '5:00']],
           ['b', [2, 4, 8], ['2:00', '4:00', '8:00']],
           ['c', [3, 7], ['3:00', '7:00']],
           ['d', [6], ['6:00']],
           ['e', [9], ['9:00']]]

So far I have used collections.Counter() to get the frequency of letter

>>> letter_count = Counter([element[1] for element in my_list])
>>> letter_count
Counter({'a': 3, 'b': 3, 'c': 2, 'd': 1, 'e': 1})

I believe I can use a list comprehension to get the rest of the way, using letter_count.keys() and a conditional, but I'm stuck. Any suggestions?

Also, any suggestions on a better, more descriptive title (for posterity's sake) are welcome.

Answer 1

out = {}
for a, b, c in my_list:
    out.setdefault(b, []).append((a, c))

out = [[k, [i for i, _ in v], [i for _, i in v]]  for k, v in out.items()]

print(out)

Prints:

[['a', [0, 1, 5], ['0:00', '1:00', '5:00']], 
 ['b', [2, 4, 8], ['2:00', '4:00', '8:00']], 
 ['c', [3, 7], ['3:00', '7:00']], 
 ['d', [6], ['6:00']], 
 ['e', [9], ['9:00']]]

Answer 2

Another approach is to use itertools.groupby

import itertools

new_list = []
my_list = sorted(my_list, key=lambda row: row[1])
for name, groups in itertools.groupby(my_list, key=lambda row: row[1]):
    groups = list(zip(*groups))
    
    grouped_item = [name, groups[0], groups[2]]
    new_list.append(grouped_item)
    
print(new_list)
[['a', (0, 1, 5), ('0:00', '1:00', '5:00')],
 ['b', (2, 4, 8), ('2:00', '4:00', '8:00')],
 ['c', (3, 7), ('3:00', '7:00')],
 ['d', (6,), ('6:00',)],
 ['e', (9,), ('9:00',)]]

Answer 3

Here's a quick n' dirty approach (same efficiency as current other answers, but more code)

my_list = [[0, 'a', '0:00'],
           [1, 'a', '1:00'],
           [2, 'b', '2:00'],
           [3, 'c', '3:00'],
           [4, 'b', '4:00'],
           [5, 'a', '5:00'],
           [6, 'd', '6:00'],
           [7, 'c', '7:00'],
           [8, 'b', '8:00'],
           [9, 'e', '9:00']]

tmp_dict = {}

for first, second, third in my_list:
    if second not in tmp_dict:
        tmp_dict[second] = [[],[]]

    tmp_dict[second][0].append(first)
    tmp_dict[second][1].append(third)

new_list = []

for key, val in tmp_dict.items():
    new_list.append([key] + val)

print(new_list)