在python中的同一项内出现计数器

Question

对不起，我很确定其他人可能已经问过这个问题，但我没找到。我想跟踪我看过这个特定项目的次数

输入：

[88,88,27,0,88]

期望的输出：

[1,2,1,1,3]

我正在寻找在性能方面特别好的东西。我对Numpy或Pandas解决方案没问题。

Answer 1

lst = [8,1,2,3,1,3,3,1,2,99]

cnt = {}
res = []

for x in lst:
    cnt[x] = cnt.get(x,0)+1
    res += [cnt[x]]

print(res)

产量

Answer 2

这是使用列表理解的简单方法：

x = [8,1,2,3,1,3,3,1,2,99]
y = [x[:i].count(el) + 1 for i, el in enumerate(x)]
print(y)

输出：

[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

Answer 3

>>> from collections import defaultdict
... 
... 
... def solution(lst):
...     result = []
...     seen = defaultdict(int)
...     for num in lst:
...         seen[num] += 1
...         result.append(seen[num])
...     return result
... 
>>> solution([88, 88, 27, 0, 88])
[1, 2, 1, 1, 3]
>>> solution([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])
[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

没有进口：

>>> def solution(lst):
...     result = []
...     seen = {}
...     for num in lst:
...         try:
...             seen[num] += 1
...         except KeyError:
...             seen[num] = 1
...         result.append(seen[num])
...     return result
... 
>>> solution([88, 88, 27, 0, 88])
[1, 2, 1, 1, 3]
>>> solution([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])
[1, 1, 1, 1, 2, 2, 3, 3, 2, 1]

Answer 4

一个带发电机：

def return_count(l):
    cnt = {}
    for x in l:
        cnt[x] = cnt.get(x, 0) + 1
        yield cnt[x]

print(list(return_count([8, 1, 2, 3, 1, 3, 3, 1, 2, 99])))