Python sum 二维列表中具有相同第一个值的元素

2022-01-09 00:00:00 python sum

问题描述

我正在尝试找到一种有效的方法来执行以下操作:

I'm trying to find an efficient way to do the following:

我有这个样本:

sample = [['no',2, 6], ['ja',5,7], ['no',4,9], ['ja',10,11], ['ap',7,12]]

并且需要

res = [['no', 6, 15], ['ja', 15, 18], ['ap',7,12]]

即将第一个元素相同的子列表的对应值相加.

i.e. sum the corresponding values of the sublists where the first element is the same.

非常感谢

我的代码是:

codes = list(set([element[0] for element in sample]))
res=[]
for code in codes:
    aux=[code]
    res01 = 0
    res02 = 0
    for element in sample:
        if element[0] == code:
            res01 += element[1]
            res02 += element[2]
    aux += [res01, res02]
    res.append(aux)

解决方案

使用defaultdict:

>>> from collections import defaultdict

>>> d = defaultdict(lambda: [0,0], list())
>>> for a,b,c in sample: 
        d[a][0]+=b 
        d[a][1]+=c

#driver 值:

IN : sample = [['no',2, 6], ['ja',5,7], ['no',4,9], ['ja',10,11], ['ap',7,12]]

OUT : d = defaultdict(<function <lambda> at 0x7f4349f17620>, 
           {'no': [6, 15], 'ja': [15, 18], 'ap': [7, 12]})

由于输出的结构是这样的，我建议您使用 dict 类型来存储您的输出，因为将来处理它会更容易.

Since the output is structured as such, I would suggest you utilise the dict type for storing your output as future processing with it will be easier.

如果您仍然希望输出为 list，只需映射 dict，如下所示:

In case you still want the output as a list, just map the dict, as follows:

>>> [ [key]+ele for key,ele in d.items()]

=> [['no', 6, 15], ['ja', 15, 18], ['ap', 7, 12]]

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