How to iterate through a list of coordinates and calculate distance between them in Python

Question

I have list that has 20 coordinates (x and y coordinates). I can calculate the distance between any two coordinates, but I have a hard time writing an algorithm that will iterate through the list and calculate the distance between the first node and every other node. for example,

ListOfCoordinates = [(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)]

In this case I need a for loop that will interate the list and calculate the distance between the first coordinate and the second coordinates, distance between first coordinate and third coordinate, etc. I am in need of an algorithm to help me out, then I will transform it into a python code. Thanks

Thanks for ll the feedback. It's been helpful.

Answer 1

Whenever you need something combinatorics-oriented ("I need first and second, then first and third, then...") chances are the itertools module has what you need.

from math import hypot

def distance(p1,p2):
    """Euclidean distance between two points."""
    x1,y1 = p1
    x2,y2 = p2
    return hypot(x2 - x1, y2 - y1)

from itertools import combinations

list_of_coords = [(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)]

[distance(*combo) for combo in combinations(list_of_coords,2)]
Out[29]: 
[2.8284271247461903,
 5.656854249492381,
 8.48528137423857,
 11.313708498984761,
 14.142135623730951,
 2.8284271247461903,
 5.656854249492381,
 8.48528137423857,
 11.313708498984761,
 2.8284271247461903,
 5.656854249492381,
 8.48528137423857,
 2.8284271247461903,
 5.656854249492381,
 2.8284271247461903]

edit: Your question is a bit confusing. Just in case you only want the first point compared against the other points:

from itertools import repeat

pts = [(1,2), (3,4), (5,6), (7,8), (9,10), (11,12)]

[distance(*pair) for pair in zip(repeat(pts[0]),pts[1:])]
Out[32]: 
[2.8284271247461903,
 5.656854249492381,
 8.48528137423857,
 11.313708498984761,
 14.142135623730951]

But usually in this type of problem you care about all the combinations so I'll leave the first answer up there.