如何对列表中的元素进行分组？

Question

I have a python list:

['AM43',
 'AM22',
 'AM51',
 'AM43',
 'AM22',
 'AM51',
 'AM43',
 'AM22',
 'AM51']

I want the output to be a list:

['AM43',
 'AM43',
 'AM43',
 'AM22',
 'AM22',
 'AM22',
 'AM51',
 'AM51',
 'AM51']

I tried sort() but that also rearranges the order. I don't want that. I want the output to be in the same order as the input list.

Answer 1

You can create a dict that stores the index of the first occurrence of each value, and use the dict to perform the sort:

lst = ["AM43", "AM22", "AM51", "AM43", "AM22", "AM51", "AM43", "AM22", "AM51"]

ix = {k: i for i, k in reversed(list(enumerate(lst)))}
res = sorted(lst, key=ix.get)
# ['AM51', 'AM51', 'AM51', 'AM22', 'AM22', 'AM22', 'AM43', 'AM43', 'AM43']

Edit: @emsimposon92 provides a 2-pass linear-time solution implementable as follows:

from collections import Counter

ctr = Counter(lst)
visited = set()
res2 = list()
for x in lst:
    if x in visited:
        continue
    res2.extend([x] * ctr[x])
    visited.add(x)

Answer 2

您可以为每个唯一编号创建一个列表，然后根据原始输入顺序将它们串联起来。

Answer 3

Following Yakym Pirozhenko's answer, to get the appropriate order you need to reverse-enumerate lst when building the index dict:

lst = ["AM43", "AM22", "AM51", "AM43", "AM22", "AM51", "AM43", "AM22", "AM43"]

ix = {k: i for i, k in zip(range(len(lst), -1, -1), reversed(lst))}

res = sorted(lst, key=ix.get)

Answer 4

Here's a solution using collections.Counter , itertools and toolz.unique . Note the last uses a 3rd party library, but the source code is just an itertools unique_everseen recipe .

from collections import Counter
from itertools import repeat, chain
from toolz import unique

lst = ['AM43', 'AM22', 'AM51', 'AM43', 'AM22',
       'AM51', 'AM43', 'AM22', 'AM51']

c = Counter(lst)
uniques = unique(lst)

res = list(chain.from_iterable(repeat(u, c[u]) for u in uniques))

res

['AM43', 'AM43', 'AM43',
 'AM22', 'AM22', 'AM22',
 'AM51', 'AM51', 'AM51']