我有一个以整数作为字符串的列表,所以如何将其转换回
lst = ["['1','2']", "['2','4']", "['1','4']", "['1','5']", "['3','5']", "['3','4']"]
我尝试使用列表理解
[j for j in i if j.isdigit() for i in lst ]
但返回
['3', '3', '3', '3', '3', '3', '4', '4', '4', '4', '4', '4']
所需的输出:
[[1,2],[2,4],[1,4],[1,5],[3,5],[3,4]]
有帮助吗?
答案 0 :(得分:8)
使用ast
模块。
例如:
import ast
lst = ["['1','2']", "['2','4']", "['1','4']", "['1','5']", "['3','5']", "['3','4']"]
res = [list(map(int, ast.literal_eval(i))) for i in lst]
print(res)
输出:
[[1, 2], [2, 4], [1, 4], [1, 5], [3, 5], [3, 4]]
答案 1 :(得分:2)
天真的解决方案:
>>> lst = ["['1','2']", "['2','4']", "['1','4']", "['1','5']", "['3','5']", "['3','4']"]
>>> [[int(x.strip("'")) for x in s[1:-1].split(',')] for s in lst]
[[1, 2], [2, 4], [1, 4], [1, 5], [3, 5], [3, 4]]
答案 2 :(得分:1)
您可以使用ast模块的literal_eval
函数并列出理解
>>>[[int(j) for j in ast.literal_eval(i)] for i in lst]
[[1, 2], [2, 4], [1, 4], [1, 5], [3, 5], [3, 4]]
答案 3 :(得分:1)
final_lst = []
for s in lst:
sublist = []
for item in eval(s):
sublist.append(int(item))
final_lst.append(sublist)
print(final_lst)
答案 4 :(得分:1)
您只需eval
主列表中的每个列表,然后解析子列表中的每个int
lst = ["['1','2']", "['2','4']", "['1','4']", "['1','5']", "['3','5']", "['3','4']"]
x = [[int(j) for j in eval(i)] for i in lst]
print (x)
输出
[[1, 2], [2, 4], [1, 4], [1, 5], [3, 5], [3, 4]]