在python中将带小数位的字符串列表转换为int

Question

example_list = ['21.48678', '21.46552', '21.45145', '21.43822',
                '21.42734', '21.41222', '21.40132', '21.37679']

我在将这个列表从字符串转换为整数时遇到了一些麻烦，我想将它作为整数。谢谢:)

Answer 1

最简单的事情是

[int(float(x)) for x in your_list]

这会截断所有数字

如果要对数字进行舍入，请使用此代替

[int(float(x)+.5) for x in your_list]

Answer 2

首先转换为float

>>> lst = ['21.48678', '21.46552', '21.45145', '21.43822', '21.42734', '21.41222', '21.40132', '21.37679']
>>> ints = [int(float(num)) for num in lst]
[21, 21, 21, 21, 21, 21, 21, 21]

Answer 3

foo = ['21.48678', '21.46552', '21.45145', '21.43822', '21.42734', '21.41222', '21.40132', '21.37679']

map(lambda x: int(float(x)), foo)

Answer 4

[int(round(float(i))) for i in example_list]

会将列表中的项目转换为float，围绕，然后将其转换为int。

Answer 5

new_list=[]
for each in example_list:
    Integer = int(each)
    new_list.append(Integer)
print new_list

初学者很容易理解：D