Rounding errors with floats in Python using Numpy

Question

I'm having an issue that I believe has to do with working with floats and precision but I'm not very well versed in the various intricacies involved. I'm a math person and in my mind I might as well still be just working with decimals on a chalkboard. I'll begin studying up on this, but in the mean time, I'm wondering if there are any general techniques for working with floats that might address the problem I'll outline below.

I have a numpy array of decimals that I would like to round to the nearest .02. I originally accomplished this by dividing every element of the array by .02, rounding the result, then multiplying by .02 again. The actual data is generated by some code that process an input, but this demonstrates the problem:

x = np.array([.45632, .69722, .40692])
xx = np.round(x/.02)*.02

It seems to round everything correctly, as I can check:

xx
array([0.46, 0.7, 0.4])

However, if I inspect the first and second element, I get:

xx[0]
0.46000000000000002
xx[1]
0.70000000000000007

Each element in the array is of type numpy.float64. The problem occurs later because I involve these numbers with comparison operators to select subsets of the data and what happens then is a little unpredictable:

xx[0] == .46
True

But,

xx[1] == .70
False

As I said, I have a work around for this particular application, but I'm wondering if anyone has a way to make my first approach work or if there are techniques for dealing with these types of numbers that are more general that I should be aware of.

You should probably compare floats without the == operator. Instead, compare the number to a range of values, like "xx[0] is within 1e-10 of 0.46?". Or, just don't use floats (save each number as 100x what it actually is). — irrelephant
– irrelephant, Commented Aug 26, 2014 at 4:11
hava a look at isclose or allclose in (numpy)[docs.scipy.org/doc/numpy-dev/reference/generated/…. — Marcin
– Marcin, Commented Aug 26, 2014 at 4:15

Flair · Accepted Answer · 2022-05-28 16:49:56Z

5

Rather than using == to select subsets of data, try using numpy.isclose(). This allows you to specify a relative/absolute tolerance for your comparison (absolute(a - b) <= (atol + rtol * absolute(b)))

edited May 28, 2022 at 16:49

Flair

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answered Aug 26, 2014 at 11:02

Joel Vroom

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2 Comments

user3749714 Over a year ago

I was actually going to answer my own question and suggest a tolerable difference and the < operator, but this is a useful little function to know about.

K.S. Over a year ago

link is broken (can't edit): docs.scipy.org/doc/numpy/reference/generated/numpy.isclose.html

footonwonton · Accepted Answer · 2019-06-18 09:18:58Z

1

Python's formatcould be used for this too.

print format(xx[1], '.100f')

this code return actual value of xx[1]

ie. xx[1] = 0.70000000000000006661338147750939242541790008544921875

You can check these by code shown below

if xx[1] == 0.70000000000000006661338147750939242541790008544921875:
    print 'true'

edited Jun 18, 2019 at 9:18

footonwonton

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answered Aug 26, 2014 at 6:59

Aslam Khan

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