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    1. 在元组列表中按字母对数字求和

      时间:2023-08-31
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                本文介绍了在元组列表中按字母对数字求和的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                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!

                上一篇:有没有办法在 Python 中获取元组或列表的差异和交集? 下一篇:无论元组顺序如何,在两个元组列表中查找交集

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