python - 如何将列表中的每个元素乘以一个数字？

Question

我有一个 list :

my_list = [1, 2, 3, 4, 5]

如何将 my_list 中的每个元素乘以 5？输出应该是:

[5, 10, 15, 20, 25]

Answer 1

您可以使用 list comprehension :

my_list = [1, 2, 3, 4, 5]
my_new_list = [i * 5 for i in my_list]

>>> print(my_new_list)
[5, 10, 15, 20, 25]

请注意，列表推导式通常是执行 for 循环的更有效方式:

my_new_list = []
for i in my_list:
    my_new_list.append(i * 5)

>>> print(my_new_list)
[5, 10, 15, 20, 25]

作为替代方案，这里有一个使用流行的 Pandas 包的解决方案:

import pandas as pd

s = pd.Series(my_list)

>>> s * 5
0     5
1    10
2    15
3    20
4    25
dtype: int64

或者，如果您只想要列表:

>>> (s * 5).tolist()
[5, 10, 15, 20, 25]

最后，一个可以使用map，尽管这通常是不被接受的。

my_new_list = map(lambda x: x * 5, my_list)

但是，使用 map 通常效率较低。根据 ShadowRanger 的评论关于这个问题的已删除答案:

The reason "no one" uses it is that, in general, it's a performancepessimization. The only time it's worth considering map in CPython isif you're using a built-in function implemented in C as the mappingfunction; otherwise, map is going to run equal to or slower than themore Pythonic listcomp or genexpr (which are also more explicit aboutwhether they're lazy generators or eager list creators; on Py3, yourcode wouldn't work without wrapping the map call in list). If you'reusing map with a lambda function, stop, you're doing it wrong.

他的另一条评论发布到 this reply :

Please don't teach people to use map with lambda; the instant youneed a lambda, you'd have been better off with a list comprehensionor generator expression. If you're clever, you can make map workwithout lambdas a lot, e.g. in this case, map((5).__mul__, my_list), although in this particular case, thanks to someoptimizations in the byte code interpreter for simple int math, [x * 5 for x in my_list] is faster, as well as being more Pythonic and simpler.