将匹配的列表项分组直到更改然后重复

Question

I am trying to group items in a list to display them. The logic is the following.

I have a list: list = [1,1,2,3,4,4,5,5,6] I want to turn this into: grouped_list = [[1,1],[2],[3],[4,4],[5,5],[6]]

I have tried the following:

for i in range(0, len(list)):
    if(list[i] == list[i+1]):
        temp_list.append(list[i])
    else:
        grouped_list.append(temp_list)
        temp_list.clear()
        grouped_list.append([list[i]])

However, this keeps resulting in the wrong output.

Answer 1

You can use itertools.groupby

>>> l = [1,1,2,3,4,4,5,5,6]
>>> res = [list(grp) for k, grp in itertools.groupby(l)]
>>> res
[[1, 1], [2], [3], [4, 4], [5, 5], [6]]

Answer 2

You could use a collections.defaultdict for that:

from collections import defaultdict

your_list = [1, 1, 2, 3, 4, 4, 5, 5, 6]

your_dict = defaultdict(list)
for i in your_list:
    your_dict[i].append(i)

result = sorted(your_dict.values())  # Skip this line if order doesn't matter
print(result)
# [[1, 1], [2], [3], [4, 4], [5, 5], [6]]

Answer 3

The worst error is the use of an unique temp_list . Each time you add temp_list to grouped_list it is the same list that you add. The clear method empty this unique list. Instead temp_list.clear() you should do temp_list = [] to have a new list.

You should iterate only to len() - 1 because you do an access to i + 1 .

There are other problems, but these two are the most important.

Also, don't use list for a variable name, because that redefine a standard item of Python. You could that, but it is a bad idea.

Answer 4

Without dependencies:

l = [1, 1, 2, 3, 4, 4, 5, 5, 6]

c = list()
for n in set(l):
    c.append( [n]*l.count(n) )

Result:

[[1, 1], [2], [3], [4, 4], [5, 5], [6]]   

Answer 5

An answer without use of a "special" function of the standard library. (It is to show to @Adam. I think that the answer of @abc with itertools.groupby() is the better answer.)

seq = [1, 1, 2, 3, 4, 4, 5, 5, 6]
print(seq)


temp_seq = []
grouped_seq = []
previous = None

for n in seq:
    if previous != n:
        previous = n
        if temp_seq:
            grouped_seq.append(temp_seq)
            temp_seq = []

    temp_seq.append(n)

if temp_seq:
    grouped_seq.append(temp_seq)


print(grouped_seq)

Answer 6

Use a list comprehension with some built-in structures (repeat, count and set) :

import numpy as np

lis = [1,1,2,3,4,4,5,5,6] 
grouped_list = [np.repeat(x, lis.count(x)).tolist() for x in set(lis)]
grouped_list

Result:

[[1, 1], [2], [3], [4, 4], [5, 5], [6]]

Hint: avoid use "list" as a name for your lists, since it is a reserved word for a Python structure. The same is valid for "dict" and "set".