如何在列表中查找字符串的总和

Question

我有一个这样的列表

['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']

在 python 中。我想找到像这样的字符串列表的总和：

['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']

我应该先将它们转换为列表中的列表吗？如果是这样，我将如何处理？

尽我所能学习如何为科学编写代码。

Answer 1

给定：

s = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]

以下将第二项转换为 Python 对象：

import ast
for sublist in s:
    sublist[1] = ast.literal_eval(sublist[1])

结果：

[['MGM', 1], ['MGD', 1], ['V1', [0, 2, 0, 1]], ['AuD', [0, 0, 0, 1]]]

然后通过对列表的特殊处理将它们转换回来：

for sublist in s:
    if isinstance(sublist[1],list):
        sublist[1] = sum(sublist[1])
    sublist[1] = str(sublist[1])

结果：

[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']]

Answer 2

这应该做你想做的事。说明在 cmets 中。

import ast

# Create the list
myList = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]

# Loop through each sublist
for num in myList:
    # try to convert the string to a list and sum it
    try:
        # This works by evaluating the string into a list object
        # Then summing the numbers in the list
        # then turning that number back into a string so it's like the rest
        num[1] = str(sum(ast.literal_eval(num[1])))
    # If it fails, it must just be a number, so ignore
    except TypeError:
        pass

print myList

Answer 3

>>> s = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]

由于您的数据是一场灾难，我们将使用辅助函数进行求和

>>> def mysum(L):
...     try:
...         return sum(L)
...     except TypeError:
...         return L
... 

可以使用 JSON 或 literal_eval 来解码字符串

>>> import json
>>> [[i, str(mysum(json.loads(j)))] for i, j in s]
[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']]

Answer 4

假设您有一个字符串列表，您可以在一行中执行此操作，方法是使用ast.literal_eval()（将字符串评估为列表），然后在其上使用sum()。示例 -

lst1 = [[a, (str(sum(ast.literal_eval(b))) if b.startswith('[') else b)] for a,b in lst]

虽然它有点不可读。使用for 循环的更易读的版本 -

lst1 = []
for a,b in lst:
    if b.startswith('['):
        lst1.append([a, (str(sum(ast.literal_eval(b)))])
    else:
        lst1.append([a,b])

演示 -

>>> lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]
>>> import ast
>>> lst1 = [[a, (str(sum(ast.literal_eval(b))) if b.startswith('[') else b)] for a,b in lst]
>>> lst1
[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']]

Answer 5

代码：

from ast import literal_eval
los=[['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]
for lst in los :
    if '[' in lst[1]:
        lst[1]=sum(literal_eval(lst[1]))
print los

Answer 6

谢谢马克，没有人提到eval！

lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]
for sublist in lst:
    sublist[1] = eval(sublist[1])
for sublist in lst:
    if isinstance(sublist[1],list):
        sublist[1] = sum(sublist[1])
    sublist[1] = str(sublist[1])
print lst

另一个可以是-

lst = [['MGM', '1'], ['MGD', '1'], ['V1', '[0,2,0,1]'], ['AuD', '[0,0,0,1]']]

for x in lst:
    x[1] = str(sum([int(i) for i in x[1].replace("[","").replace("]","").split(',')]))

print lst

打印出来

[['MGM', '1'], ['MGD', '1'], ['V1', '3'], ['AuD', '1']]