How To Round Float Down To A Given Precision?

I need a way to round a float to a given number of decimal places, but I want to always round down. For example, instead of >>> round(2.667, 2) 2.67 I would rather have &

Solution 1:

You've got a friend in quantize and ROUND_FLOOR

>>>from decimal import Decimal,ROUND_FLOOR>>>float(Decimal(str(2.667)).quantize(Decimal('.01'), rounding=ROUND_FLOOR))
2.66
>>>float(Decimal(str(-2.667)).quantize(Decimal('.01'), rounding=ROUND_FLOOR))
-2.67

Note that you can use ROUND_DOWN for positive numbers. As interjay mentions in a comment, ROUND_DOWNRounds towards zero and hence may return incorrect values for negative numbers.

>>>from decimal import Decimal,ROUND_DOWN>>>Decimal(str(2.667)).quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('2.66')
>>>float(Decimal(str(2.667)).quantize(Decimal('.01'), rounding=ROUND_DOWN))
2.66

Solution 2:

Something like this should work for whatever number of digits you want to do:

>>>import math>>>defround_down(num,digits):
        factor = 10.0 ** digits
        return math.floor(num * factor) / factor

>>>round_down(2.667,2)
2.66

Solution 3:

You can use math.floor to "round down" to the nearest whole number. So to round to the 3rd decimal place, you can try math.floor(1000*number) / 1000.

In general, to "round down" a number num to precision n, you can try:

from math import floor

def round_down(num, n):
    multiplier =pow(10,n)
    returnfloor(num * multiplier) / multiplier

Solution 4:

You can also play around this using strings

defround_down(num, prec):
    ifisinstance(num, float):
        s = str(num)
        returnfloat(s[:s.find('.') + prec + 1])
    else:
        raise ValueError

round_down(2.6667, 2)
# 2.66

Patch the code with more checks like precision is not negative among others.