在 python 中向下舍入到小数点后两位

Question

I need to round down and it should be two decimal places.
Tried the following,

a = 28.266
print round(a, 2)

28.27

But the expected value is 28.26 only.

Answer 1

Seems like you need the floor :

import math
math.floor(a * 100)/100.0

# 28.26

Answer 2

It seems you want truncation, not rounding.

A simple way would be to combine floor division // and regular division / :

>>> a = 28.266
>>> a // 0.01 / 100
28.26

Instead of the regular division you could also multiply (as noted in the comments by cmc ):

>>> a // 0.01 * 0.01
28.26

Similarly you could create a function to round down to other more/less decimals. But because floats are inexact numbers, this can lead to inaccuracies.

def round_down(value, decimals):
    factor = 1 / (10 ** decimals)
    return (value // factor) * factor

print(round_down(28.266, 2))
# 28.26

But as said it's not exactly exact:

for i in range(0, 8):
    print(i, round_down(12.33333, i))

0 12.0
1 12.3
2 12.33
3 12.333
4 12.333300000000001 # weird, but almost correct
5 12.33332           # wrong
6 12.33333
7 12.33333

There are other (more precise) approaches though:

A solution using the fraction module

A fraction can represent a decimal number much more exact than a float . Then one can use the "multiply, then floor, then divide" approach mentioned by Psidom but with significantly higher precision:

import fractions
import math

a = 28.266

def round_down(value, decimals):
    factor = 10 ** decimals
    f = fractions.Fraction(value)
    return fractions.Fraction(math.floor(f * factor),  factor)

print(round_down(28.266, 2))
# 1413/50  <- that's 28.26

And using the test I did with the floats:

for i in range(0, 8):
    print(i, round_down(12.33333, i))

0 12
1 123/10
2 1233/100
3 12333/1000
4 123333/10000
5 1233333/100000
6 1233333/100000
7 1233333/100000

However creating a Fraction will not magically fix an inexact float , so typically one should create the Fraction from a string or a "numerator-denominator pair" instead of from float.

A solution using the decimal module

You could also use the decimal module, which offers a variety of rounding modes, including rounding down.

For this demonstration I'm using a context manager to avoid changing the decimal rounding mode globally:

import decimal

def round_down(value, decimals):
    with decimal.localcontext() as ctx:
        d = decimal.Decimal(value)
        ctx.rounding = decimal.ROUND_DOWN
        return round(d, decimals)

print(round_down(28.266, 2))  # 28.26

Which gives more sensible results for the rounding:

for i in range(0, 8):
    print(i, round_down(12.33333, i))

0 12
1 12.3
2 12.33
3 12.333
4 12.3333
5 12.33333
6 12.333330
7 12.3333300

As with Fraction a Decimal should be created from a string to avoid the intermediate inexact float. But different from Fraction the Decimal have limited precision, so for values with lots of significant figures it will also become inexact.

However "rounding down" is just one of the available options. The list of available rounding modes is extensive:

Rounding modes

decimal.ROUND_CEILING Round towards Infinity.

decimal.ROUND_DOWN Round towards zero.

decimal.ROUND_FLOOR Round towards -Infinity.

decimal.ROUND_HALF_DOWN Round to nearest with ties going towards zero.

decimal.ROUND_HALF_EVEN Round to nearest with ties going to nearest even integer.

decimal.ROUND_HALF_UP Round to nearest with ties going away from zero.

decimal.ROUND_UP Round away from zero.

decimal.ROUND_05UP Round away from zero if last digit after rounding towards zero would have been 0 or 5; otherwise round towards zero.

Answer 3

With Python 3.9, you can use quantize()

from decimal import *    
>>> Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32')

I tested the above answer and it worked. But I find this 1 liner is quite simple. So I also write my answer here.

Source: https://docs.python.org/3/library/decimal.html

Answer 4

simply try this:

import math
a = 28.266
print((math.floor(a * 100)) / 100.0)

Output:

28.26

Answer 5

here's a simple function that isn't affected by float precision errors

def truncate_float(n, places):
    return int(n * (10 ** places)) / 10 ** places

Tests:

>>> truncate_float(28.266, 3)
28.266
>>> truncate_float(28.266, 2)
28.26
>>> truncate_float(28.266, 1)
28.2

Answer 6

There's an even simpler way of doing this generically, by subtracting a small quantity before rounding, like so:

a = 28.269
digits = 2
print(round(a - 0.5/10**digits, digits))

This is based on the intuition that one way of rounding a float to the nearest integer is by adding 0.5, then truncating. The above solution does the opposite by subtracting a half of the minimal 'tick' that the desired precision allows, then rounding.

Answer 7

Simple function which you can use in your codes. This function you can also use for integer floor numbers.

import math
def floorDecimal(number, decimal):
    return math.floor(number * pow(10, decimal))/pow(10, decimal)

Example of using:

number = 256.789
newNumber = floorDecimal(number, 2) # newNumber is 256.78
newNumber = floorDecimal(number, -2) # newNumber is 200

Answer 8

Here is the function I use with f for the float and d for the number of decimals

from math import floor

def floor_decimal(f, d):
    n = 10 ** d
    return floor(f * n) / n