Is there another way to cut the numbers after the 10th decimal without callig round()?

user_15271555 0
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I wanted to create a function that whatever the number received as input, it would cut-off (not round, which could take some decimal higher) every number after the 10th one. It normally works fine, but if I call the function with a number like this:

1234567.123456789123456789

It returns: 1234567.1234567892

And not 1234567.1234567891, as expected.

def cut_to_10(number):
    number = str(number)
    j = 0
    if "." in number:
        for i in number:
            j += 1
            if i == ".":
                before_dot = number[:j - 1]
                after_dot = number[j:]
                if len(after_dot) > 10:
                    after_dot = after_dot[:10]

                return float(before_dot + "." + after_dot)

    if "." not in number:
        return round(float(number), 10)

EDIT 1:

Also tried this, and the same unexpected result returned.

import numpy as np

def cut_to_10(number):
    return np.floor(number * 1e10) / 1e10

EDIT 2:

Considering the amazing explanation given by ShadowRanger, but, considering also that I need more a ceiling precision than a floor one, and given the fact that I still had to work with float, I just decided to change

return float(before_dot + "." + after_dot)

to

return float(before_dot + "." + after_dot) - 2e-10

and it worked fine. 1e-10 didn't work.

python-3.x python
2 Answers

1234567.123456789123456789 is not an actual float. You can type it, but it becomes 1234567.1234567892, the closest representable float value; the rounding is occurring before you actually call round.

>>> 1234567.123456789123456789
1234567.1234567892

If you need that much precision, you need to move to decimal.Decimal, which will work just fine:

>>> decimal.Decimal('1234567.123456789123456789')
Decimal('1234567.123456789123456789')

>>> round(_, 10)
Decimal('1234567.1234567891')

Just don't convert back to float, because 1234567.1234567891 doesn't exist (if it did, that's what 1234567.123456789123456789 would have produced), and you'll be back where you started with 1234567.1234567892.

user_3646960

See the other answer for the correct solution. Leaving this one up just because it seems like a more straightforward approach to doing the truncation once you work past the floating point limitation.

import numpy as np
    
def cut_to_10(number):
    return np.floor(number * 1e10) / 1e10
user_61146510
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