Question

我有一个键/值对列表

我想成对计算唯一值

列表示例

[(12, 0), (10, 1), (11, 777), (11, 0) ,(10, 1)]

结果如下：

[(12, 0, 1), (10, 1 , 2), (11, 777, 1), (11, 0, 1)]

我尝试了地图操作，但没有成功谢谢！

Answer 1

使用collections.Counter：

>>> from collections import Counter
>>>
>>> lst = [(12, 0), (10, 1), (11, 777), (11, 0) ,(10, 1)]
>>> [key + (cnt,) for key, cnt in Counter(lst).items()]
[(11, 0, 1), (11, 777, 1), (10, 1, 2), (12, 0, 1)]

Answer 2

使用collections.Counter：

>>> from collections import Counter
>>> a =[(12, 0), (10, 1), (11, 777), (11, 0) ,(10, 1)] 

>>> [(*i,j) for (i,j) in Counter(a).items()]
# [(11, 0, 1), (11, 777, 1), (10, 1, 2), (12, 0, 1)]

>>> [i for i in Counter(a).items()]
# [((11, 0), 1), ((11, 777), 1), ((10, 1), 2), ((12, 0), 1)]

Answer 3

这样的事情对你有用吗？

your_list = [(12, 0), (10, 1), (11, 777), (11, 0) ,(10, 1)]
result = [(a,b,your_list.count((a,b))) for (a,b) in your_list]
# [(12, 0, 1), (10, 1, 2), (11, 777, 1), (11, 0, 1), (10, 1, 2)]

如果您希望每个条目只出现一次

set(result)
# set([(12, 0, 1), (11, 0, 1), (11, 777, 1), (10, 1, 2)])

Answer 4

您也可以使用内置map方法，这样：

>>> l = [(12, 0), (10, 1), (11, 777), (11, 0) ,(10, 1)]
>>> result = []
>>> map(lambda s: result.append((s[0],s[1],l.count(s))), set(l)) 
[None, None, None, None]
>>> result
[(11, 0, 1), (11, 777, 1), (10, 1, 2), (12, 0, 1)]

python2.7：计算键/值对中的值

4 个答案: