如何用int划分列表中的每个元素？

Question

我只是想用一个int来划分列表中的每个元素.

myList = [10,20,30,40,50,60,70,80,90]
myInt = 10
newList = myList/myInt

这是错误:

TypeError: unsupported operand type(s) for /: 'list' and 'int'

我明白为什么我收到这个错误.但我很沮丧,我找不到解决方案.

还尝试过:

newList = [ a/b for a, b in (myList,myInt)]

错误:

ValueError: too many values to unpack

预期结果:

newList = [1,2,3,4,5,6,7,8,9]

编辑:

以下代码给出了我的预期结果:

newList = []
for x in myList:
    newList.append(x/myInt)

但有更容易/更快的方法吗？

Answer 1

惯用的方法是使用列表理解:

myList = [10,20,30,40,50,60,70,80,90]
myInt = 10
newList = [x / myInt for x in myList]

或者,如果您需要维护对原始列表的引用:

myList[:] = [x / myInt for x in myList]

Answer 2

您首先尝试的方式实际上可以直接使用numpy:

import numpy
myArray = numpy.array([10,20,30,40,50,60,70,80,90])
myInt = 10
newArray = myArray/myInt

如果你使用长列表进行此类操作,尤其是在任何类型的科学计算项目中,我真的建议使用numpy.

Answer 3

>>> myList = [10,20,30,40,50,60,70,80,90]
>>> myInt = 10
>>> newList = map(lambda x: x/myInt, myList)
>>> newList
[1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer 4

抽象版本可以是：

import numpy as np
myList = [10, 20, 30, 40, 50, 60, 70, 80, 90]
myInt = 10
newList  = np.divide(myList, myInt)

Answer 5

myList = [10,20,30,40,50,60,70,80,90]
myInt = 10
newList = [i/myInt for i in myList]