问题描述
我正在尝试计算元素与数组中起点之间的距离.
I am trying to compute a distance between an element and a starting point in an array.
这是一个数组
假设元素(0,1)是当前值最高的起点.
assume the element (0,1) is a starting point which has the highest value currently.
如果邻居具有一个共同点，并且在另一个轴上相差1个单位，则邻居是围绕特定点的元素.
a neighbors is an element around a specific point if they have one axis in common and different in another axis by 1 unit.
通常，邻居可以是数组内部特定点的顶部，底部，左侧，右侧.
generally, a neighbor could be the top, bottom, left, right of a specific point, which is inside the array.
任务是用距离值标记每个元素，该距离值指示距起点(0,1)的距离.
the task is to label every elements with a distance value indicate how far it is from the starting point (0,1).
ds = np.array([[1, 2, 1],
[1, 1, 0],
[0, 1, 1]])
dist = np.full_like(ds, -1)
p0 = np.where(ds == 2)
dist[p0] = 0
que = []
que.append(p0)
nghb_x = [0, 0, -1, 1]
nghb_y = [-1, 1, 0, 0]
while len(que):
x, y = que.pop()
d = dist[(x,y)]
for idx0, idx1 in zip(nghb_x, nghb_y):
tmp_x = x + idx0
tmp_y = y + idx1
if np.any(tmp_x >= 0) and np.any(tmp_x < ds.shape[0]) and np.any(tmp_y >= 0) and np.any(tmp_y < ds.shape[1]) and np.any(dist[(tmp_x,tmp_y)] == -1):
dist[(tmp_x,tmp_y)] = d + 1 # distance = distance(x) + 1
que.append((tmp_x, tmp_y))
print('dist:')
print(dist)
输出
dist:
[[1 0 1]
[2 1 2]
[3 2 3]]
是如预期的那样，我想知道是否有更有效的方法来做到这一点?
is as expected though, I would like to know if is there a more efficient way to do this?
推荐答案
您正在计算曼哈顿距离(每个点距目标点的x距离和y距离).
You're calculating the Manhattan distance (the x-distance plus the y-distance) from a target point for each point.
在给出目标坐标和数组形状的情况下，您可以使用numpy函数一步完成此操作:
You can use a numpy function to do it in one step, given the target coordinates and the shape of the array:
target = (0, 1)
np.fromfunction(lambda x,y: np.abs(target[0]-x) + np.abs(target[1]-y), ds.shape)
结果:
[[1. 0. 1.]
[2. 1. 2.]
[3. 2. 3.]]
演示: https://repl.it/repls/TrustyUnhappyFlashdrives
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