Python "round robin"

Question

Given multiple (x,y) ordered pairs, I want to compare distances between each one of them. So pretend I have a list of ordered pairs:

pairs = [a,b,c,d,e,f]

I have a function that takes two ordered pairs and find the distance between them:

def distance(a,b):
    from math import sqrt as sqrt
    from math import pow as pow
    d1 = pow((a[0] - b[0]),2)
    d2 = pow((a[1] - b[1]),2)
    distance = sqrt(d1 + d2)
    return distance

How can I use this function to compare every ordered pair to every other ordered pair, ultimately finding the two ordered-pairs with the greatest distance between them?

Psuedopsuedocode:

     distance(a,b)
     distance(a,c)
     ...
     distance(e,f)

Any help would be tremendously appreciated.

Answer 1

try:

    from itertools import combinations

except ImportError:

    def combinations(l, n):
        if n != 2: raise Exception('This placeholder only good for n=2')
        for i in range(len(l)):
            for j in range(i+1, len(l)):
                yield l[i], l[j]


coords_list = [(0,0), (3,4), (6,8)]

def distance(p1, p2):
    return ( ( p2[0]-p1[0] ) ** 2 + ( p2[1]-p1[1] )**2 ) ** 0.5

largest_distance, (p1, p2) = max([
     (distance(p1,p2), (p1, p2)) for (p1,p2) in combinations(coords_list, 2)
     ])


print largest_distance, p1, p2

Answer 2

If you don't mind doing distance calculations between two points that are the same twice, the following will find the greatest distance:

max( [distance(a, b) for a in pairs for b in pairs] )

In order to have the a and b pair instead, then do the following:

import operator
max( [((a,b), distance(a, b)) for a in pairs for b in pairs], key=operator.itemgetter(1))

You can combine this with John Feminella's solution to get the (a,b) tuple without doing excess distance comparisons

Answer 3

in python 2.6, you can use itertools.permutations

import itertools
perms = itertools.permutations(pairs, 2)
distances = (distance(*p) for p in perms)

or

import itertools
combs = itertools.combinations(pairs, 2)
distances = (distance(*c) for c in combs)

Answer 4

Try:

max(distance(a, b) for (i, a) in enumerate(pairs) for b in pairs[i+1:])

This avoid identity-comparisons (e.g. distance(x, x), distance(y, y), etc.). It also avoids doing symmetric comparisons, since distance(x, y) == distance(y, x).

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Update: I like Evgeny's solution to use itertools a little better, as it expresses what you're trying to do more succinctly. Both of our solutions do the same thing. (Note: make sure you use combinations, not permutations -- that will be much slower!)

Answer 5

slightly related, you don't have to compute the euclidean distance yourself, there's math.hypot:

In [1]: a = (1, 2)
In [2]: b = (4, 5)
In [3]: hypot(a[0]-b[0], a[1]-b[1])
Out[3]: 4.2426406871192848