如何将字符串列表转换为列表列表

Question

我正在尝试使用 Python 将字符串列表转换为列表列表。

shopping_lists=['apple pie,corn bread,red wine\ready soups,milk,coffee']

我想得到

[{'apple pie','corn bread','red wine'},{'ready soups', 'milk', 'coffee'}]

我尝试在 Python 中使用split()函数

s=shopping_lists.split()
L=[set(i.split(',')) for i in s]

输出如下。

[{'applepie','cornbread','redwine'},{'readysoups', 'milk', 'coffee'}]

但是，我想保留单词之间的空格。

[{'apple pie','corn bread','red wine'},{'ready soups', 'milk', 'coffee'}]

Answer 1

有时将其分解为 for 循环更容易； 另外，我认为您的原始列表中可能有错字。

假设一个错字：

shopping_lists=['apple pie,corn bread,red wine','ready soups,milk,coffee']
L = []
for row in shopping_lists:
    L.append(set(row.split(',')))
print(L)

如果没有错别字：

shopping_lists='apple pie,corn bread,red wine/ready soups,milk,coffee'
L = []
shopping_lists=shopping_lists.split('/')
for row in shopping_lists:
    L.append(set(row.split(',')))
print(L)

两者都产生这个结果；

[{'红酒'、'玉米面包'、'苹果派'}、{'牛奶'、'咖啡'、'现成汤'}]

如果你真的是指一个列表列表，而不是一个集合列表......

shopping_lists=['apple pie,corn bread,red wine','ready soups,milk,coffee']
L = []
for row in shopping_lists:
    L.append(row.split(','))
print(L)

产量；

[['苹果派'、'玉米面包'、'红酒']、['现成汤'、'牛奶'、'咖啡']]

Answer 2

使用分割线代替 split()

Python 3.8.0b4 (default, Sep  5 2019, 14:10:43)
[GCC 8.2.1 20180831] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> s = 'apple pie,corn bread,red wine\nready soups,milk,coffee'
>>> [set(i.split(',')) for i in s.splitlines()]
[{'apple pie', 'corn bread', 'red wine'}, {'ready soups', 'milk', 'coffee'}]

Answer 3

您可以使用：

shopping_lists=['apple pie,corn bread,red wine\ready soups,milk,coffee']

[set(item.strip("'").split(',')) for e  in shopping_lists for item in ('%r' % e).split('\\')]

输出：

[{'apple pie', 'corn bread', 'red wine'}, {'coffee', 'milk', 'ready soups'}]

你有一个反斜杠，它是你的 2 个集合的分隔符，所以首先，你必须用反斜杠分割，然后你必须用逗号分割