Question

我有一个类似于下面的嵌套列表，为此，我试图将与第2列中相同值相关的第1列的值相加，然后将一个具有总和的新子列表，另一个具有“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，我认为这将对列表a进行操作，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', '']       
    ]

如何执行此操作？谢谢

更新

输入列表，该列表具有4列，如左侧的一列，需要将col1，col3和col4求和，以得出一列数字。

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

使用... import { Router } from '@angular/router'; ... constructor( private router: Router ) {} ... ngOnInit() { this.router.events.subscribe((event) => { if(event instanceof NavigationEnd) { // Check if this.router.url is matching to the user page's url // pattern } }); }

itertools.groupby

输出：

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)

根据评论编辑

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

如果可以使用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, ""]])，则轴0的总和会更简单

例如：

numpy

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))

Answer 2

好像您已从电子表格中导入数据？一种非常干净的方法是使用func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) { switch indexPath.section { case 0: print("1") case 1: print("1") case 2: print("2") // I need to add it here case 3: print("3") default: break } }库，并让它为您处理数据：

pandas

应打印以下内容：

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)

Answer 3

一种非常原始的方法。这仅仅是出于知识目的。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', '']]

如何根据条件求和嵌套列表的元素？

3 个答案: