CODEWITHSUNDEEP
pythonnumpyvectorcoordinatesbounding-box
Jivan
Jivan

Reputation: 23068

Determine if many points are in bounding box

The array x denotes a bounding box, delimited by two (x,y) coordinates:

x = np.array([[2, 1], [5, 3]])

And the array p denotes a collection of points by their (x,y) coordinates:

p = np.array([[3, 2], [6, 4], [3, 4], [4, 2]])

For each point in p, I want to determine (in one operation) whether it's in the bounding-box x, so as to get the following result.

result = np.array([1, 0, 0, 1])

I've though about np.where() but not sure how to process the whole collection of points in one operation. Also, wouldn't a purely numerical approach be faster?

EDIT:

I'd be interested as well in the case where multiple bounding boxes are present, such as:

# [[bbox_1], [bbox_2], [bbox_2]]
# with each bbox as [x1 y1 x2 y2]
x = np.array([[2, 1, 5, 3], [2, 2, 4, 6], [0, 4, 3, 2]])

with the result being of the form

# [p1_bbox_1, p2_bbox_1, p3_bbox_1, p4_bbox_1], [p1_bbox_2, p2_bbox_2 ...]]
# results not necessarily exact but shape is correct
result = np.array([[True, False, False, True], [True, False, False, False], [False, False, True, False]])

A point sitting exactly on a bbox boundary should be considered inside.

Upvotes: 1

Views: 2151

Answers (1)

Divakar
Divakar

Reputation: 221574

With slicing and leveraging broadcasting -

In [24]: ((p>=x[0]) & (p<=x[1])).all(1)
Out[24]: array([ True, False, False,  True])

This assumes bounding box array as :

[[x1,y1]
 [x2,y2]]

If the format is say [x1,y1,x2,y2] (1D array) that define the box corners, we could get it like so -

((p>=x[:2]) & (p<=x[2:])).all(1)

Also, edit >= to > and so on for exclusiveness on boundaries.

Extend to multiple boxes

For multiple boxes in x with a format of 2D array :

[[B1x1,B1y1,B1x2,B1y2],
 [B2x1,B2y1,B2x2,B2y2], ..
                      ]

So, B1 is box-1, B2 is box-2 and so on, while (x1,y1) and (x2,y2) are the bounding box corners for each box. The solution would be -

((p>=x[:,None,:2]) & (p<=x[:,None,2:])).all(2)

Upvotes: 2

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