Python - 将一串数字转换为int列表

Question

我有一串数字,如:

example_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'

我想将其转换为列表:

example_list = [0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11]

我尝试过类似的东西:

for i in example_string:
    example_list.append(int(example_string[i]))

但这显然不起作用,因为字符串包含空格和逗号.但是,删除它们不是一种选择,因为像"19"这样的数字会被转换为1和9.你能帮我解决这个问题吗？

Answer 1

在逗号上拆分,然后映射到整数:

map(int, example_string.split(','))

或者使用列表理解:

[int(s) for s in example_string.split(',')]

如果你想要一个列表结果,后者在Python 3上效果更好.

这是因为int()容忍空格:

>>> example_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'
>>> map(int, example_string.split(','))  # Python 2, in Python 3 returns iterator object
[0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11]
>>> [int(s) for s in example_string.split(',')]
[0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11]

仅对逗号进行拆分也更容忍变量输入; 如果在值之间使用0,1或10个空格并不重要.

Answer 2

我想最脏的解决方案是:

list(eval('0, 0, 0, 11, 0, 0, 0, 11'))

Answer 3

您还可以对拆分的字符串使用列表理解

[ int(x) for x in example_string.split(',') ]

Answer 4

number_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'

number_string = number_string.split(',')

number_string = [int(i) for i in number_string]

Answer 5

它应该工作

example_string = '0, 0, 0, 11, 0, 0, 0, 0, 0, 19, 0, 9, 0, 0, 0, 0, 0, 0, 11'
example_list = [int(k) for k in example_string.split(',')]