我想将任意列表的字符串表示形式转换回类似原始的列表对象，如下面的代码所述:
list1 = [1,2,3,4]
list1str = str(list1)
str2list1 = list1str[1:-1].split(",") #stripping off square brackets from [1,2,3,4] and then splitting
print(list1) #[1,2,3,4]
print(str(list1)) #[1,2,3,4]
print(str2list1) #['1', ' 2', ' 3', ' 4']
#Notice that the elements are of string types.
#Also there is a prefix empty space in ' 2' (and also in ' 3' and ' 4')
# --> I will like str2list1 to be ['1', ' 2', ' 3', ' 4']
此外，如果我们正在处理嵌套列表:
list2 = [1,2,['a','b'],4]
list2str = str(list2)
str2list2 = list2str[1:-1].split(",")
print(list2) #[1, 2, ['a', 'b'], 4]
print(str(list2)) #[1, 2, ['a', 'b'], 4]
print(str2list2) #['1', ' 2', " ['a'", " 'b']", ' 4']
# --> I will like str2list2 to be [1, 2, ['a', 'b'], 4]
如何从其字符串表示中准确获取原始列表？
请您参考如下方法:
你可以试试:
>>> import ast
>>> ast.literal_eval("[1,2,['a','b'],4]")
[1, 2, ['a', 'b'], 4]
