CODEWITHSUNDEEP
pythonpython-2.7
Wilsonwatson
Wilsonwatson

Reputation: 379

How to find the closest (x, y) position to (x,y) position in another list?

I have two 2D lists with x and y coordinates and I want to go through list1, and for each point find the closest (x, y) coordinates in list2. They're of different lengths, and it's okay if I don't use all the points of list2 or even if I reuse points, as long as I go through all the points only once in list1. I need the shift itself as well as the location in the lists of both points. Here's what I have done to find the shift:

s_x = ([1.0,2.0,3.0])
s_y = ([1.5,2.5,3.5])
SDSS_x = ([3.0,4.0,5.0])
SDSS_y = ([3.5,4.5,5.5])

list1 = zip(s_x,s_y)
list2 = zip(SDSS_x,SDSS_y)

shift = []
place_in_array = []

for num,val in enumerate(list1):
    guess = 9999999999999.0
    place_guess = 0
    for index,line in enumerate(list2):
        new_guess = math.hypot(line[0] - val[0], line[1] - val[1])
        if new_guess < guess:
            guess = new_guess
            place_guess = index
    shift.append(guess)
    place_in_array.append(place_guess)

print shift
print place_in_array

but the output is this:

[2.8284271247461903, 1.4142135623730951, 0.0]
[0, 0, 0]

These are wrong, and I can't figure out what the problem is.

Upvotes: 2

Views: 1514

Answers (2)

Joran Beasley
Joran Beasley

Reputation: 114098

def lazy_dist(p0):
    return lambda p1:(p0[0] -p1[0])**2 + (p0[1] - p1[1])**2
closest_matches = {p0:min(list2,key=lazy_dist(p0)) for p0 in list1}

I think will do what you want (not very fast but it should work

Upvotes: 2

Ami Tavory
Ami Tavory

Reputation: 76381

There are specialized data structures for these geometric queries, with implementations that are debugged and efficient. Why not use them? sklearn.neighbors.BallTree can do these types of queries, as well as sklearn.neighbors.KDTree.

Upvotes: 3

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