如何根据条件对嵌套列表的元素求和？

Question

我有一个如下所示的嵌套列表，我试图将第 1 列的值与第 2 列中的相同值相加，并添加一个带有总和的新子列表和另一个带有“总计”的子列表。

a = [
        ['45', '00128'], 
        ['88', '00128'], 
        ['87', '00128'], 
        ['50', '88292'], 
        ['69', '88292'], 
        ['70', '72415'], 
        ['93', '72415'], 
        ['79', '72415']
    ]

我当前的代码如下所示，我想它会在列表 a 上运行某些内容，但不会更改 a.

中的任何内容

for sl in a:
    x = sl[1]; c0=0
    if (sl[1] == x):
        c0 = c0 + int(sl[0])
    else:
        a.insert(a.index(sl)+1,[c0,''])
        a.insert(a.index(sl)+2,['Total',''])

我正在寻找的输出是这样的：

b = [
        ['45',  '00128'], 
        ['88',  '00128'], 
        ['87',  '00128'],
        ['220', ''],      # This is 45 + 88 + 87
        ['Total', ''],
        ['50', '88292'], 
        ['69', '88292'], 
        ['119', ''],      # This is 50 + 69
        ['Total', ''],      
        ['70', '72415'], 
        ['93', '72415'], 
        ['79', '72415'],
        ['242', ''],      # This is 70 + 93 + 79
        ['Total', '']       
    ]

如何做到这一点？谢谢

更新

包含 4 列的输入列表，例如左侧的一列，需要对 col1、col3 和 col4 求和以获得最右边的一列。

a = [                              >>  b = [
        ['45', '00128', '2','4'],  >>          ['45',    '00128', '2',     '4'    ], 
        ['88', '00128', '1','3'],  >>          ['88',    '00128', '1',     '3'    ], 
        ['87', '00128', '4','0'],  >>          ['87',    '00128', '4',     '0'    ], 
        ['50', '88292', '1','1'],  >>          ['220',   ''     , '7',     '7'    ],
        ['69', '88292', '9','5'],  >>          ['Total', '',      'Total', 'Total'],
        ['70', '72415', '8','9'],  >>          ['50',    '88292', '1',     '1'    ], 
        ['93', '72415', '3','2'],  >>          ['69',    '88292', '9',     '5'    ],
        ['79', '72415', '5','7']   >>          ['119',   '',      '10',    '6'    ],
    ]                              >>          ['Total', '',      'Total', 'Total'],
                                   >>          ['70',    '72415', '8',     '9'    ], 
                                   >>          ['93',    '72415', '3',     '2'    ], 
                                   >>          ['79',    '72415', '5',     '7'    ],
                                   >>          ['242',   '',      '16',    '18'   ],
                                   >>          ['Total', '',      'Total', 'Total'],
                                   >>       ]

Answer 1

您似乎从电子表格中导入了数据？一种非常简洁的方法是使用 pandas 库，让它为您处理数据：

a = [
        ['45', '00128'], 
        ['88', '00128'], 
        ['87', '00128'], 
        ['50', '88292'], 
        ['69', '88292'], 
        ['70', '72415'], 
        ['93', '72415'], 
        ['79', '72415']
]

# Import the library
import pandas as pd

# Put data into a pandas DataFrame and set column names
df = pd.DataFrame(a, columns=['value', 'category'])

# Change `value` column to integers
df['value'] = df['value'].astype(int)

# Group by the `category` column and sum
sum_df = df.groupby('category').sum()

# Show answer
print(sum_df)

应打印出以下内容：

          value
category
00128       220
72415       242
88292       119

Answer 2

使用itertools.groupby

from itertools import groupby

result = []
for m,n in groupby(a, lambda x: x[1]):
    n = list(n)
    result.extend(n + [[sum(int(i) for i, _ in n), ""]])
print(result)

输出：

[['45', '00128'],
 ['88', '00128'],
 ['87', '00128'],
 [220, ''],
 ['50', '88292'],
 ['69', '88292'],
 [119, ''],
 ['70', '72415'],
 ['93', '72415'],
 ['79', '72415'],
 [242, '']]

根据评论编辑

for m,n in groupby(a, lambda x: x[1]):
    n = list(n)
    val_1, val_2, val_3 = 0, 0, 0
    for i in n:
        val_1 += int(i[0])
        #val_2, val_3....
    result.extend(n + [[val_1, ""]])

如果可以用numpy那么0轴求和就更简单了

例如：

for m,n in groupby(a, lambda x: x[1]):
    n = np.array(list(n), dtype=int)
    print(np.delete(np.sum(n, axis=0), 1))

np.delete --> Delete element in index 1
np.sum with axis=0 --> sum element in column.

Answer 3

一种非常原始的方式。这仅用于知识目的。我会推荐 Rakesh

给出的答案

from collections import defaultdict
    
sums_dict = defaultdict(int)
a = [
        ['45', '00128'],
        ['88', '00128'],
        ['87', '00128'],
        ['50', '88292'],
        ['69', '88292'],
        ['70', '72415'],
        ['93', '72415'],
        ['79', '72415']
    ]
sums_dict = {v:sums_dict[v]+int(k) for k,v in a}

for k in sums_dict:
    index = next((len(a) - i - 1 for i, lst in enumerate(reversed(a)) if k in lst), -1)
    a.insert(index, [str(sums_dict[k]),""])
    a.insert(index+2, ['Total', ''])

print(a)

输出：

[['45', '00128'], ['88', '00128'], ['87', ''], ['87', '00128'], ['Total', ''], ['50', '88292'], ['69', ''], ['69', '88292'], ['Total', ''], ['70', '72415'], ['93', '72415'], ['79', ''], ['79', '72415'], ['Total', '']]

如何根据条件对嵌套列表的元素求和？

How to sum elements of nested list based on condition?

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python-3.x