comparison - Python: How to find two equal/closest values between two separate arrays? -

let's have 2 arrays of equal length:

arr1 = (21, 2, 3, 5, 13) arr2 = (10, 4.5, 9, 12, 20) 

which variable arr1 equal / closest variable arr2?

looking @ these 2 lists can closest numbers 4.5 , 5. i've tried implement function returns 2 closest values given 2 lists , kinda works examples above, barely solution because not optimal. , can check function fails when change arrays this:

arr1 = (21, 2, 3, 5, 13) arr2 = (10, 4.5, 9, 12, 18) 

the values function returns 13 , 18

here function:

def get_nearest(arr1, arr2):     lr = [[0, 0, 0]]     x1 in arr1:         x2 in arr2:             r = (x1 / x2 % (x1 + x2))             print x1, x2, r             if r <= 1 , r >= lr[0][2]:                 lr.pop()                 lr.append([x1, x2, r])     return lr 

can come better one?

is speed issue? care ties? if not, simple like

from itertools import product sorted(product(arr1, arr2), key=lambda t: abs(t[0]-t[1]))[0] 

for both

arr1 = (21, 2, 3, 5, 13) arr2 = (10, 4.5, 9, 12, 20) 

and

arr1 = (21, 2, 3, 5, 13) arr2 = (10, 4.5, 9, 12, 18) 

this yields

(5, 4.5) 

explanation:

product(arr1, arr2) = [(a1, a2) (a1, a2) in product(arr1, arr2)] 

yields list of n**2 pairs of numbers:

[(21, 10), (21, 4.5), ..., (13, 12), (13, 20)] 

then sort them absolute difference (|a1 - a2|) using sorted. passing sorted key keyword, tell sorted use sorting criteria lambda t: abs(t[0] - t[1]). pair smallest absolute difference placed in first index of sorted array, can grab tacking [0] on end.

edit:

as suggested piotr in comments, can feed key=func min , max, speeds considerably. try instead:

from itertools import product min(product(arr1, arr2), key=lambda t: abs(t[0]-t[1]))[0]