Round to the nearest 500, Python

Question

I'm looking to find a way to round up to the nearest 500.I've been using:

math.ceil(round(8334.00256 + 250, -3))

Whereby I have a value from a scale in a map I am making in ArcGIS. I have the ability to read and write the scale factor (i.e. 1:8334....basically, you set the thousandth and it defaults to a ratio) If the scale factor isn't a factor of 500, I want to round up to the next 500. The math.ceil will round up any decimal value, and the round(n,-3) will round to the nearest thousandth, but I'm struggling to find a way to round to the nearest 500.

Any suggestions? Thanks, Mike

Solution 1

Scale, round, unscale:

round(x / 500.0) * 500.0

Edit: To round up to the next multiple of 500, use the same logic with math.ceil() instead of round():

math.ceil(x / 500.0) * 500.0

Solution 2

I personally find rounding a but messy. I'd rather use:

(x+250)//500*500

// means integer division.

EDIT: Oh, I missed that you round "up". Then maybe

-(-x//500)*500

Solution 3

Maybe something like this:

round(float(x) / 500) * 500

The "float" conversion is unnecessary if you are using Python 3 or later, or if you run the statement from __future__ import division for sane integer division.

Comments

[Mike]

The issue is, that I need to round UP to the nearest 500, not just round either way. If I plug x = 8001, your logic rounds to 8000. I need 8500....

[Mike]

See comment below Sven's suggestion. Thanks.

[Mike]

Sven, I started working with round(x/500.0 + 0.5)*500 which gives me the result I was looking for. I'm using the math module anyways, so I'm going to use your syntax. Thanks for the help.

[Elias Zamaria]

Sven's idea seems better. I just typed the first reasonable thing I thought of.

[Alex L]

Steven's method is faster: timeit.timeit('math.ceil(x / 500.0) * 500.0', 'import math; x=5200') = 0.539998 vs >>> timeit.timeit('x + (500 - x) % 500', 'x=5200') = 0.23273

[Sven Marnach]

@AlexL: Usually, readability matters more than speed, and I think my approach is more readable. That said, it is also faster (at least on my machine) if you eliminate the attribute look-up for ceil, i.e. if you use from math import ceil.

[PolyGeo]

@StevenRumbalski I think you should write that as an answer (with just a little description added)

[Agent Lu]

@SvenMarnach Hey I was looking through this and needed to round down 500 and I think it would be helpful to add the math.floor() variant. I added an edit for you. Thanks :)