我需要将由2个项目组成的字符串(用逗号分隔)转换为整数.
发件人:
[['(0,3)', '(1,2)', '(2,2)'], ['(0,3)', '(1,2)', '(2,2)']]
收件人:
[[(0,3), (1,2), (2,2)], [(0,3), (1,2), (2,2)]]
我需要将由2个项目组成的字符串(用逗号分隔)转换为整数.
发件人:
[['(0,3)', '(1,2)', '(2,2)'], ['(0,3)', '(1,2)', '(2,2)']]
收件人:
[[(0,3), (1,2), (2,2)], [(0,3), (1,2), (2,2)]]
您可以使用ast.literal_eval
执行此任务:
from ast import literal_eval
lst = [["(0,3)", "(1,2)", "(2,2)"], ["(0,3)", "(1,2)", "(2,2)"]]
lst = [[literal_eval(v) for v in l] for l in lst]
print(lst)
打印:
[[(0, 3), (1, 2), (2, 2)], [(0, 3), (1, 2), (2, 2)]]
编辑:另一种方法(感谢@S3DEV):
out = [list(map(literal_eval, sub_list)) for sub_list in lst]
快速基准测试:
from timeit import timeit
from ast import literal_eval
lst = [["(0,3)", "(1,2)", "(2,2)"], ["(0,3)", "(1,2)", "(2,2)"]]
def fn1():
return [[literal_eval(v) for v in l] for l in lst]
def fn2():
return [list(map(literal_eval, sub_list)) for sub_list in lst]
assert fn1() == fn2()
t1 = timeit(lambda: fn1(), number=1_000)
t2 = timeit(lambda: fn2(), number=1_000)
print(t1)
print(t2)
在我的机器上打印(AMD 3700X,Python 3.9.7):
0.040873110003303736
0.04002662200946361