Generate A/B labels for a bunch of hex IDs

[bkowshik]

I want to label a bunch of hex IDs into A/B so that With the code snippet below, I randomly assign a A/B label to hex IDs and the output looks like the map below. Want to know what are other strategies to label hex IDs as A/B.

import random

def split_into_ab(hex_id):
    return random.randint(1, 2) == 1

[dfellis]

If you want it based on the hex_id and therefore the same every time, what about:

def split_into_ab(hex_id):
    return int(hex_id, 16) % 2 == 1

[bkowshik]

Thank you @dfellis

Not sure if this is because of the resolution of the hex IDs or the area I am working with. All the hex IDs in my list, end with an odd number and thus % 2 = 1 for all the hex IDs. Some sample hex IDs from the list of hex IDs that formed the map above are below for reference.

hex_id          ,integer_value
8860145101fffff ,614179795317555199
8860145103fffff ,614179795319652351
8860145105fffff ,614179795321749503
8860145107fffff ,614179795323846655
8860145109fffff ,614179795325943807
886014510bfffff ,614179795328040959
886014510dfffff ,614179795330138111
8860145111fffff ,614179795334332415
8860145113fffff ,614179795336429567

[dfellis]

Oh, right, the bit format is not random enough to work it that way. If you integer divide it by a prime number before doing the modulus that should be enough for your purposes, or you could feed it into a non-cryptographic hash function like xxHash though that is decidedly overkill here.

[dfellis]

Integer division in Python 3 requires the // symbol, so the modified version would be:

def split_into_ab(hex_id):
    return int(hex_id, 16) // 3 % 2

Using 3 here as the prime number since there are only two buckets you're going into, which should keep neighbors from being too regular with each other, I think? It's been a while since I've read up on hashing.

[dfellis]

Just tested in a REPL, I think that version is nicely mixed: