Python：如何从包含重复项的列表列表中返回具有计数的不同项目列表？

Question

我有嵌套在列表中的列表。这是我尝试使用列表的迷你预览来绘制单个嵌套列表的计数的代码：

count_list = [['a', 'b'],['d', 'e'], ['e'], ['c'], ['a', 'b'], ['c']]
    distinct_list = []
    count_list.sort()
    for x in count_list:
        if x not in distinct_list: 
            distinct_list.append(x)       
    for z in distinct_list:
        for y in count_list:
            if y == z:
                print(z, count_list.count(z))

我试图让输出成为一个字典/列表，其中只有不同的嵌套列表及其计数，以便我可以按计数对结果进行排序。列表不采用两个值，因此我正在考虑将其作为字典以便于存储。 这些是我从上面的代码中得到的结果。

(['a', 'b'], 2)
(['a', 'b'], 2)
(['c'], 2)
(['c'], 2)
(['d', 'e'], 1)
(['e'], 1)

我希望输出只为每个不同的列表及其各自的计数提供一行（而不是在上面的当前输出中看到的同一列表的多行）。有什么想法吗？

Answer 1

如果您可以将嵌套列表转换为元组以便将它们用作键，那么将结果作为 dict（与您关于如何解决此问题的初始想法之一内联）也将相当简单。例如：

data = [['a', 'b'],['d', 'e'], ['e'], ['c'], ['a', 'b'], ['c']]

results = {}
for d in data:
    t = tuple(d)
    if t not in results:
        results[t] = data.count(d)

print(results)

# OUTPUT
# {('a', 'b'): 2, ('d', 'e'): 1, ('e',): 1, ('c',): 2}

Answer 2

如果你转换成元组，你可以使用字典理解

count_list = [tuple(i) for i in count_list]

d = {i: count_list.count(i) for i in set(count_list)}

{('a', 'b'): 2, ('d', 'e'): 1, ('c',): 2, ('e',): 1}

Answer 3

如果您不想使用任何库，即使 Counter 是 python 原生的，您也可以这样做...

arr = [[1, 2], [3], [1, 2], [4], [3]]
new_arr = []
for elem in arr:
    # You are wrapping your element in a list, at the same scale
    # than the list you wish to output (meaning 3 dimensions depth list).
    # In other words, you have a list... that contains a list, containing
    # your element and its count. The first depth seems useless at first glance,
    # but it's just here to keep the same depth you will have appending to new_arr.

    # You can try by yourself doing "obj = [elem, arr.count(elem)]" if you don't
    # believe me.
    obj = [ [elem, arr.count(elem)] ]
    if obj[0] not in new_arr:
        # checking on first index to check our elem and not our ""wrapper""
        new_arr += obj

print(new_arr)
# [[[1, 2], 2], [[3], 2], [[4], 1]]

Answer 4

你应该使用collections.Counter():

>>> count_list = [['a', 'b'],['d', 'e'], ['e'], ['c'], ['a', 'b'], ['c']]
>>> import collections
>>> result = collections.Counter(tuple(x) for x in count_list)
>>> print(result)
Counter({
    ('c',): 2,
    ('a', 'b'): 2,
    ('d', 'e'): 1,
    ('e',): 1
})