CODEWITHSUNDEEP
pythonpython-3.xeuclidean-distance
mb567
mb567

Reputation: 681

Remove point from list if within certain threshold of any other point

I have code that filters points from a list where the distance between them is below 3:

import numpy
lst_x = [0,1,2,3,4,5,6,7,8,9,10]
lst_y = [9,1,3,2,7,6,2,7,2,3,8]
lst = numpy.column_stack((lst_x,lst_y))

diff = 3

new = []
for n in lst:
    if not new or ((n[0] - new[-1][0]) ** 2 + (n[1] - new[-1][1]) ** 2) ** .5 >= diff:
    new.append(n)

The problem is that the output is:

[array([0, 9]), array([1, 1]), array([4, 7]), array([6, 2]), array([7, 7]), array([8, 2]), array([10,  8])]

point [6,2] and [8,2] are closer than 3 to each other yet they are in the results, I believe this is because the for loop is only checking one point then jumping to the next.

How can I make it check all numbers at every point.

Upvotes: 2

Views: 213

Answers (1)

Prune
Prune

Reputation: 77837

Your algorithm very carefully checks against only the point most recently added to the list, new[-1]. This is insufficient. You need another loop to make sure that you check against each element of new. Something like this:

for n in lst:
    too_close = False
    for seen_point in new:  
        # Is any previous point too close to this one (n)?
        if not new or ((n[0] - seen_point[0]) ** 2 +     \
                       (n[1] - seen_point[1]) ** 2) ** .5 < diff:
            too_close = True
            break

    # If no point was too close, add this one to the "new" list.
    if not too_close:
        new.append(n)

Upvotes: 1

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