Question

我正在尝试计算子列表元素的唯一实例的数量，然后将每个唯一元素写入新列表，并将实例数附加到子列表。 list_1中的每个子列表只有两个元素，顺序无关紧要。

这样：

list_1 = [[a, b], [a, c], [a, c], [a, c], [b, e], [d, q], [d, q]]

变为：

new_list =  [[a, b, 1], [a, c, 3], [b, e, 1], [d, q, 2]]

我在想我需要使用套装，但我感谢任何指点我正确方向的人。

Answer 1

你想看看collections.Counter(); Counter个对象是多组的（也称为包）;他们将键映射到他们的计数。

您必须将您的子列表转换为元组才能用作键：

from collections import Counter

counts = Counter(tuple(e) for e in list_1)

new_list = [list(e) + [count] for e, count in counts.most_common()]

它为您提供按计数排序的new_list（降序）：

>>> from collections import Counter
>>> list_1 = [['a', 'b'], ['a', 'c'], ['a', 'c'], ['a', 'c'], ['b', 'e'], ['d', 'q'], ['d', 'q']]
>>> counts = Counter(tuple(e) for e in list_1)
>>> [list(e) + [count] for e, count in counts.most_common()]
[['a', 'c', 3], ['d', 'q', 2], ['a', 'b', 1], ['b', 'e', 1]]

如果您的出现时间是连续的，那么您也可以使用itertools.groupby()：

from itertools import groupby

def counted_groups(it):
    for entry, group in groupby(it, key=lambda x: x):
        yield entry + [sum(1 for _ in group)]

new_list = [entry for entry in counted_groups(list_1)]

我在这里使用了一个单独的生成器函数，你可以将循环内联到列表理解中。

这给出了：

>>> from itertools import groupby
>>> def counted_groups(it):
...     for entry, group in groupby(it, key=lambda x: x):
...         yield entry + [sum(1 for _ in group)]
... 
>>> [entry for entry in counted_groups(list_1)]
[['a', 'b', 1], ['a', 'c', 3], ['b', 'e', 1], ['d', 'q', 2]]

并保留原始订单。

Answer 2

如果相同的子列表是连续的：

from itertools import groupby

new_list = [sublist + [sum(1 for _ in g)] for sublist, g in groupby(list_1)]
# -> [['a', 'b', 1], ['a', 'c', 3], ['b', 'e', 1], ['d', 'q', 2]]

Answer 3

有点'围绕房屋'解决方案

list_1 = [['a', 'b'], ['a', 'c'], ['a', 'c'], ['a', 'c'], ['b', 'e'], ['d', 'q'], ['d', 'q']]
new_dict={}
new_list=[]
for l in list_1:
    if tuple(l) in new_dict:
        new_dict[tuple(l)] += 1
    else:
        new_dict[tuple(l)] = 1
for key in new_dict:
    entry = list(key)
    entry.append(new_dict[key])
    new_list.append(entry)
print new_list

在python中计算和追加子列表元素

3 个答案: