识别列表中的重复项并总结其最后一项 [英] Identify duplicates in a list of lists and sum up their last items

问题描述

我有一个列表列表，我希望从这些列表中删除重复项并总结重复项的最后一个元素.如果项目的前两个元素相同，则该项目为重复项.用一个例子可以更好地说明这一点:

I have a list of lists from which I would like to remove duplicates and sum up duplicates' last elements. An item is a duplicate if its first 2 elements are the same. This is better illustrated with an example:

input = [['a', 'b', 2], ['a', 'c', 1], ['a', 'b', 1]]

# Desired output
output = [['a', 'b', 3], ['a', 'c', 1]]

这里也有类似的问题，但是我还没有找到一个可以同时处理列表和汇总列表项的问题.

There are similar questions here on SO but I haven't found one which would deal with list of lists and summing up list items at the same time.

我尝试了几种方法，但无法使其起作用:

I tried several approaches but couldn't make it work:

• 创建输入列表的副本，进行嵌套循环，如果找到第二个重复项，则将其最后一项添加到原始项目中->这样会与太多嵌套混淆不清
• 我查看了Counter集合，但它似乎不适用于列表列表
• itertools

您能给我一些有关如何解决此问题的建议吗?

Could you give me any pointers on how to approach this problem?

推荐答案

我认为列表不是最佳的数据结构.我会用带元组键的字典.我确实需要列表，以后可以创建一个列表:

I don't think lists are the best data structure for it. I would use dictionaries with tuple key. I you really need list, you can create one later:

from collections import defaultdict

data = [['a', 'b', 2], ['a', 'c', 1], ['a', 'b', 1]]

result = collections.defaultdict(int) # new keys are auto-added and initialized as 0
for item in data:
    a, b, value = item
    result[(a,b)] += value
print result
# defaultdict(<type 'int'>, {('a', 'b'): 3, ('a', 'c'): 1})
print dict(result)
# {('a', 'b'): 3, ('a', 'c'): 1}
print [[a, b, total] for (a, b), total in result.items()]
# [['a', 'b', 3], ['a', 'c', 1]]

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