问题描述

我有一个元组列表:

[ ('A',100), ('B',50), ('A',50), ('B',20), ('C',10) ]

我试图总结所有具有相同字母的数字.IE.我要输出

I am trying to sum up all numbers that have the same letter. I.e. I want to output

[('A', 150), ('B', 70), ('C',10)] 

我尝试使用 set 来获取唯一值，但是当我尝试将第一个元素与 set 进行比较时，我得到了

I have tried using set to get the unique values but then when I try and compare the first elements to the set I get

TypeError: unsupported operand type(s) for +: 'int' and 'str'

有什么快速的方法可以按字母匹配数字吗?

Any quick solutions to match the numbers by letter?

推荐答案

这是一个(半?)-liner:按字母分组(您需要先排序)，然后取第二个的总和元组的条目.

Here is a one(and a half?)-liner: group by letter (for which you need to sort before), then take the sum of the second entries of your tuples.

from itertools import groupby
from operator import itemgetter

data = [('A', 100), ('B', 50), ('A', 50), ('B', 20), ('C', 10)]
res = [(k, sum(map(itemgetter(1), g)))
       for k, g in groupby(sorted(data, key=itemgetter(0)), key=itemgetter(0))]
print(res)
// => [('A', 150), ('B', 70), ('C', 10)]

上面是O(n log n) —排序是最昂贵的操作.如果您的输入列表确实很大，那么以下 O(n) 方法可能会更好地为您服务:

The above is O(n log n) — sorting is the most expensive operation. If your input list is truly large, you might be better served by the following O(n) approach:

from collections import defaultdict

data = [('A', 100), ('B', 50), ('A', 50), ('B', 20), ('C', 10)]

d = defaultdict(int)
for letter, value in data:
    d[letter] += value
res = list(d.items())
print(res)
// => [('B', 70), ('C', 10), ('A', 150)]

这篇关于在元组列表中按字母对数字求和的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！