CODEWITHSUNDEEP
pythonnumpyfloating-point
Luca
Luca

Reputation: 841

Is numpy order dependent?

I noticed that the output of numpy.mean() and numpy.var() changes as the ordering of its arguments change.

It has to do only with the precision of floats, or I'm missing something?

import numpy

l1 = [1.0, 0.69, 0.65, 0.7, 0.64, 0.8]
l2 = [1.0, 0.7, 0.69, 0.65, 0.64, 0.8]

assert sorted(l1) == sorted(l2)

print repr(numpy.mean(l1))
print repr(numpy.mean(l2))

print repr(numpy.var(l1))
print repr(numpy.var(l2))

Here the output that I obtain:

0.7466666666666667
0.74666666666666659
0.015522222222222222
0.015522222222222224

Upvotes: 1

Views: 72

Answers (1)

Mike Müller
Mike Müller

Reputation: 85442

NumPy uses (in this case) 64-bit floats. Floating point numbers cannot represent all numbers with unlimited precision. This data type can hold 15 significant digits. This means the last digit in your number is meaningless. Therefore, to compare our numbers use numpy.allclose:

>>> numpy.allclose(numpy.mean(l1), numpy.mean(l2))
True

Upvotes: 3

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