我知道有很多笨重的方法可以做到这一点,但我正在寻找一种更加优雅的Pythonic方式来完成以下任务。给定一个数字列表:
a = [0,1,2,3,4,5,6,7,8,9]
将此列表拆分成两个列表,对应于每个其他元素:
b = [0,2,4,6,8]
c = [1,3,5,7,9]
我知道有很多笨重的方法可以做到这一点,但我正在寻找一种更加优雅的Pythonic方式来完成以下任务。给定一个数字列表:
a = [0,1,2,3,4,5,6,7,8,9]
将此列表拆分成两个列表,对应于每个其他元素:
b = [0,2,4,6,8]
c = [1,3,5,7,9]
你想要:
b = a[::2] # Start at first element, then every other.
而且:
c = a[1::2] # Start at second element, then every other.
所以现在我们拥有了想要的东西:>>> print(b)
[0, 2, 4, 6, 8]
>>> print(c)
[1, 3, 5, 7, 9]
b = a[::2]
c = a[1::2]
>>> a = [0,1,2,3,4,5,6,7,8,9]
>>> b = a[::2]
>>> c = a[1::2]
>>> print b
[0,2,4,6,8]
>>> print c
[1,3,5,7,9]
[::]
语法如下:[start:end:step]
。如果您不为起始和结束指定任何参数,则它将与整个列表一起使用。因此,上面的代码正在执行以下操作:b
:从索引0开始,在a
中每隔一个元素取一个c
:从索引1开始,在a
中每隔一个元素取一个Try This :
a = [0,1,2,3,4,5,6,7,8,9]
>>> b=[i for x,i in enumerate(a) if x%2==1]
>>> c=[i for x,i in enumerate(a) if x%2==0]
>>> b
[1, 3, 5, 7, 9]
>>> c
[0, 2, 4, 6, 8]