如何从列表中提取索引范围

Question

mylist1=[11,4,5,6,8,6,3]
for i in mylist1:
a=list(range(i))
print(a)
print (a[-2:2])

我知道(a[-2:2])是错的，但是我可以用a[0:4]做点什么。我希望结果是:

输入列表的列表:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[0, 1, 2, 3]
[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5]
[0, 1, 2]

:

[ 9, 10, 0, 1]
[2, 3, 0, 1]
[3, 4, 0, 1, ]
[4, 5, 0, 1 ]
[6, 7, 0, 1,]
[4, 5, 0, 1, ]
[1, 2, 0, 1]

例如，我想这样写:

[a[-2],a[-1],a[0],a[1],a[2]]

但是它很笨拙。有什么干净利落的方法吗?

Answer 1

另一种方法是:

mylist1 = [11, 4, 5, 6, 8, 6, 3]
for i in mylist1:
a = list(range(i))
print(a[-2:] + a[:2])

[9, 10, 0, 1]
[2, 3, 0, 1]
[3, 4, 0, 1]
[4, 5, 0, 1]
[6, 7, 0, 1]
[4, 5, 0, 1]
[1, 2, 0, 1]

Answer 2

你可以尝试这样做:

list1 = [0, 1, 2, 3, 4, 5]
for ln in range(1, len(list1)):
list2 = [list1[n] for n in range(-ln, ln)] #magic line
print(list2)

或者你也可以试着用下面这行来代替:

list2 = list1[ln:] + list1[:ln]

其中ln是列表开始和结束的元素个数。

Answer 3

也可以使用列表推导式

mylist1 = [11,4,5,6,8,6,3]
# [range(i) for i in mylist1] creates an array of ranges based on mylist
# [a[j] for j in range(-2,2)] for a in ... selects elements
#     a[-2],a[-1],a[0],a[1]  if you need a[2] then
#     change range(-2,2) to range(-2,3)
mylist2 = [[a[j] for j in range(-2,2)] for a in [range(i) for i in mylist1]]