我有一个点位置列表(以元组的形式),我正在尝试收集点N之前和之后的点。
points = [
(0, 0),
(1, 1),
(2, 2),
(3, 3),
(4, 4),
(5, 5),
(6, 6),
]
例如。如果我将第三点(3, 3)
作为输入,则预期输出为[(2,2), (4,4)]
。
对此进行测试,效果很好:
index = 3
a, n, b = points[ index-1 : index+2 ]
print(a, b)
# Returns: (2, 2) (4, 4)
只要索引不小于2或大于5,它就会按预期工作。
以下是一些有关元组列表按预期方式工作的示例:
# Tuple at index zero
print(points[0])
# Returns: (0, 0)
# Tuples from index zero to (but not including) index three
print(points[0:3])
# Returns: [(0, 0), (1, 1), (2, 2)]
# Tuple at index negative one, (6,6) in this case
print(points[-1])
# Returns: (6, 6)
但是,当我将负指数和正指数结合起来时,为什么呢?
# Tuples from index negative one, up to but not including index two
print(points[-1:2])
# Returns: []
预期输出为[(6, 6), (0, 0), (1, 1)]
,而python为[]
。
感谢Dev Khadka。
points = [
(0,0),
(1,1),
(2,2),
(3,3),
(4,4),
(5,5),
(6,6),
]
index = 0
a, _, b = np.roll(points, -(index-1), axis=0)[:3]
sel_points = zip(a,b)
print(list(zip(*sel_points)))
Result: [(6, 6), (1, 1)]
答案 0 :(得分:0)
此points[-1:2]
等效于points[-1:2:1]
,这意味着您正尝试从最后一个元素开始编制索引,并通过每次向前1进行索引来获得索引2。如果您将步骤-1 points[-1:2:-1]
设为该索引,则此索引将起作用,但这不是您想要的。您可以使用下面的滚动方法达到您想要的
points = [
(0, 0),
(1, 1),
(2, 2),
(3, 3),
(4, 4),
(5, 5),
(6, 6),
]
indx = 3
a,_,b = np.roll(points, -(indx-1), axis=0)[:3]
a,b