using list operator “in” with floating point values

Question

I have a list with floats, each number with 3 decimals (eg. 474.259). If I verify the number in the list like this:

if 474.259 in list_sample:
    print \"so         


        

Answer 1

You can use numpy.isclose() instead of Python's in.

import numpy as np
other_list = np.array([474.251001, 123.456])
number = other_list[0]
number = round(number, 3)
if number == 474.251:
    print "number == 474.251"
if number in other_list:
    print "number in other_list"
if any(np.isclose(number, other_list, rtol=1e-7)):
    print 'any(np.isclose(number, other_list, rtol=1e-7))'

Output:

number == 474.251
any(np.isclose(number, other_list, rtol=1e-7))

Answer 2

Comparing floating point numbers for exact equality usually won't do what you want. This is because floating point numbers in computers have a representation (storage format) which is inherently inaccurate for many real numbers.

I suggest reading about it here: http://floating-point-gui.de/ and doing something like a "fuzzy compare" using an "epsilon" tolerance value to consider the numbers equal so long as they differ by less than x% or whatever.

Answer 3

You could also following an approach, where you compare the values based on an arbitrary precision.

For example, convert all your floats like this:

def internalPrecision(number):
    precision = 1000
    return int(round(number * precision)) 

If you do this, both operators == and in should work.