Question

如果我有大名单A = [1,2,3,4,5,6,7,8,9,10] 我有3个元素的子列表，例如

B = [1,2,3]

我想根据A列表向前滑动1步，因此B变为[2,3,4] - 是否有顺利的方法可以做到这一点？或者我只需要弹出B的第一个元素然后从A？

添加适当的元素

谢谢！

编辑：我的回答 B = A [i：i + 3] 如果你想通过'i'步骤向前看，你可以增加'i'。

Answer 1

您可以A成为deque：

from collections import deque
A = deque(range(1,11))

B可以是A的前3个元素的视图。当您需要“滑动”时，请向左旋转A。

A
Out[71]: deque([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

from itertools import islice #deques do not support slicing notation
B = list(islice(A,3))

B
Out[74]: [1, 2, 3]

A.rotate(-1)
B = list(islice(A,3))

B
Out[77]: [2, 3, 4]

Answer 2

for i in range(len(a)-2):
    b = a[i:i+3]
    print b 

[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 10]

你可以更普遍地做。

Answer 3

>>> A = [1,2,3,4,5,6,7,8,9,10]
>>> for B in (A[i - 3: i] for i in range(3, len(A) + 1)):
...     print B
... 
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 10]

Answer 4

使用itertools：

>>> from itertools import islice, izip
>>> A = [1,2,3,4,5,6,7,8,9,10]
>>> for l in izip(*(islice(A, x, None) for x in xrange(3))):
...     print list(l)
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]
[4, 5, 6]
[5, 6, 7]
[6, 7, 8]
[7, 8, 9]
[8, 9, 10]

islice(A, x, None)在A[x:]上创建一个迭代器。

在列表中移动头部和尾部1步

4 个答案: