CODEWITHSUNDEEP
pythonpython-3.xfloating-pointprecision
Edy Bourne
Edy Bourne

Reputation: 6187

In Python, how to reduce the precision of a float to a given step size?

I have a float that can have arbitrary precision, and a minimum step size that indicates the minimum amount of this number can be increased / decreased by:

num = 3.56891211101
min_step = 0.005

I'd like to have a function that takes this num and the step_size and rounds the num to that given min_step. So in this case the result would be 3.570.

I attempted this:

num = 3.56891211101
min_step = 0.005

def myround(x, base):
    return base * round(x / base)

x = myround(num, min_step)
print(x)
>>> 3.5700000000000003

...it's close, but not quite. I would like the output to be the same as if:

y = 3.570
print(y)
>>> 3.57

What's a simple way of accomplishing this?

I'm on Python 3.8

Upvotes: 0

Views: 1192

Answers (2)

Serge Ballesta
Serge Ballesta

Reputation: 148890

Most Python implementations (including the CPython reference implementation) use IEE 754 floating point numbers. As a result, they are not accurate for decimal values.

The canonic way is to use the decimal module:

from decimal import Decimal, Context

num = 3.56891211101
c = Context(prec=3)
x= c.create_decimal(num)
print(x)

gives as expected

3.57

Upvotes: 1

Edy Bourne
Edy Bourne

Reputation: 6187

I solved it with:

def myround(x, base):
    decimal_places = str(base)[::-1].find('.')
    precise = base * round(x / base)
    return round(precise, decimal_places)

x = myround(num, min_step)
print(x)
>>> 3.57

y = 3.570
print(y)
>>> 3.57

Hope its helpful to others.

Upvotes: 0

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