## Is your ice cream float bigger than mine

Posted May 27, 2013 at 07:46 AM | categories: math | tags:

Updated May 28, 2013 at 08:59 AM

Float numbers (i.e. the ones with decimals) cannot be perfectly represented in a computer. This can lead to some artifacts when you have to compare float numbers that on paper should be the same, but in silico are not. Let us look at some examples. In this example, we do some simple math that should result in an answer of 1, and then see if the answer is “equal” to one.

print 3.0 * (1.0/3.0) print 1.0 == 3.0 * (1.0/3.0)

1.0 True

Everything looks fine. Now, consider this example.

print 49.0 * (1.0/49.0) print 1.0 == 49.0 * (1.0/49.0)

1.0 False

The first line looks like everything is find, but the equality fails!

1.0 False

You can see here why the equality statement fails. We will print the two numbers to sixteen decimal places.

print '{0:1.16f}'.format(49.0 * (1.0/49.0) ) print '{0:1.16f}'.format(1.0) print 1 - 49.0 * (1.0/49.0)

0.9999999999999999 1.0000000000000000 1.11022302463e-16

The two numbers actually are not equal to each other because of float math. They are *very, very* close to each other, but not the same.

This leads to the idea of asking if two numbers are equal to each other within some tolerance. The question of what tolerance to use requires thought. Should it be an absolute tolerance? a relative tolerance? How large should the tolerance be? We will use the distance between 1 and the nearest floating point number (this is `eps`

in Matlab). `numpy`

can tell us this number with the `np.spacing`

command.

Below, we implement a comparison function from 10.1107/S010876730302186X that allows comparisons with tolerance.

# Implemented from Acta Crystallographica A60, 1-6 (2003). doi:10.1107/S010876730302186X import numpy as np print np.spacing(1) def feq(x, y, epsilon): 'x == y' return not((x < (y - epsilon)) or (y < (x - epsilon))) print feq(1.0, 49.0 * (1.0/49.0), np.spacing(1))

2.22044604925e-16 True

For completeness, here are the other float comparison operators from that paper. We also show a few examples.

import numpy as np def flt(x, y, epsilon): 'x < y' return x < (y - epsilon) def fgt(x, y, epsilon): 'x > y' return y < (x - epsilon) def fle(x, y, epsilon): 'x <= y' return not(y < (x - epsilon)) def fge(x, y, epsilon): 'x >= y' return not(x < (y - epsilon)) print fge(1.0, 49.0 * (1.0/49.0), np.spacing(1)) print fle(1.0, 49.0 * (1.0/49.0), np.spacing(1)) print fgt(1.0 + np.spacing(1), 49.0 * (1.0/49.0), np.spacing(1)) print flt(1.0 - 2 * np.spacing(1), 49.0 * (1.0/49.0), np.spacing(1))

True True True True

As you can see, float comparisons can be tricky. You have to give a lot of thought to how to make the comparisons, and the functions shown above are not the only way to do it. You need to build in testing to make sure your comparisons are doing what you want.

Copyright (C) 2013 by John Kitchin. See the License for information about copying.