How can I round a float number to fixed decimal parts using python

Question

I have some prices like 5.35, 10.91, 15.55, which I round with two decimals using

Price = "{:.2f}".format(Price)

But how can I make them to be round based on fixed decimals 0.50 and 0.90 and have

5.50, 10.90, 15.50

Thank you

Do you want every number to have the decimals rounded to the closest value between .50 and .90? So .35 becomes .50 and .82 becomes .90? — Luca Bezerra, Apr 10 '19 at 12:12
You need some custom rounding function as this 0.5, 0.9 is not standard. — smido, Apr 10 '19 at 12:15
Possible duplicate of [Limiting floats to two decimal points](https://stackoverflow.com/questions/455612/limiting-floats-to-two-decimal-points) — Ashish, Apr 10 '19 at 12:25

score 0 · Answer 1 · answered Apr 10 '19 at 12:28

This function should do the trick for the examples that you provided:

def fixed_round(number):
    decimal = number - int(number)
    if (abs(0.5 - decimal)) < (abs(0.9 - decimal)):
        return int(number) + 0.50
    else:
        return int(number) + 0.90

If you want it to round to full numbers, in the case where the decimals are closer to the integer than to 0.5, you will need to adjust for that ;)

ElConrado · Accepted Answer · 2019-04-10T13:19:43.333

0

If you have float number with the fractional part you can use first and second example else with all number (in fact all other type) you can use third and fourth. Num is your fractional part. x is yout 0.5 and y your 0.9

def round(num, x, y) :
    av = (x+y)/2
    if num < av :
        return x
    return y

Or if you want pass numer like 15.98:

def round(num, x, y) :
    av = (x+y)/2
    if num % 1 < av :
        return int(num)+x
    return int(num)+y

You can than call:

round(15.98,0.5,0.9)

Output: 15.9

Or something like:

def round(number):
    x=0.5
    y=0.9
    if type(number) is float: 
        av = (x+y)/2
        if number % 1 < av :
            return int(number)+x
        return int(number)+y
    return number

All of this is rounding down. If you would like round up the construction will be:

def round(number):
    x=0.5
    y=0.9
    if type(number) is float: 
        av = (x+y)/2
        from builtins import round as rd
        if rd(number % 1,2 ) < av :
            return int (number)+x
         return int (number)+y
     return number

edited Apr 10 '19 at 13:19

answered Apr 10 '19 at 12:30

ElConrado

1,477
4
20
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Thank you Elcon :) – Apr 10 '19 at 12:57
Doing it this way 5 rounds to 5.5 instead of 4.9. It may or may not be the expected result though. – Hirabayashi Taro Apr 10 '19 at 13:14
Especially the last two example – ElConrado Apr 10 '19 at 13:17
@Hirabayashi extra profit ;) BUT ElConrado left out the equality if result is already on .50 or .90 – Apr 10 '19 at 13:17

ramazan polat · Answer 3 · 2019-04-10T12:46:30.420

-1

import math


def weird_round(x):
    if round(x % 1, 1) >= 0.9:
        return math.floor(x) + 0.9
    return math.floor(x) + 0.5


prices = [5.35, 10.91, 15.55]

for price in prices:
    text = "{:.2f}".format(weird_round(price))
    print(price, '->', text)

5.35 -> 5.50
10.91 -> 10.90
15.55 -> 15.50

edited Apr 10 '19 at 12:46

answered Apr 10 '19 at 12:22

ramazan polat

7,111
1
48
76

@glibdud now it does. – ramazan polat Apr 10 '19 at 12:46

How can I round a float number to fixed decimal parts using python

3 Answers3