Get quotients and remainder from division of floats

What is the Pythonic approach to getting a quotients and remainder from division of floats? So taking 8 divided by 3 which is 2.66666666666

so I want:

aPythonOperator(8,3) = 2.0, 0.66666666666

but I cannot find a clean way of doing this without the remainder being rounded.
For completeness the Python division operators are:

import math
a = 8.0
b = 3.0
print(f'Division:{a/b}')
print(f'Floored division: {a//b}')
print(f'Modulo: {a%b}')
print(f'Remainder: {math.remainder(a,b)}')  # rounds to int and yields -1 (?)
print(f'Divmod: {divmod(a, b)}')

which gives:

Division:2.6666666666666665
Floored division: 2.0
Modulo: 2.0
Remainder: -1.0

These don't help, but I expected the Divmod function to work. However, Divmod rounds the quotient to an int:

print(f'Divmod: {divmod(a, b)}')
Divmod: (2.0, 2.0)   

I can get the behaviour I want by creating my own version of the Divmod function:

def hackDivmod(a,b):  
    #HACK
    Div = (a/b)
    floredDiv = (a//b) 
    return floredDiv,  Div-floredDiv 

This works as desired:

HackDivmod: (2.0, 0.6666666666666665)

But is there a better way do doing this using built in functions or operators?

I guess you want to use math.modf() combined with native division

>>> import math
>>> math.modf( 8 / 3)
(0.6666666666666665, 2.0)

You could do

def divfract(a, b):
    return a // b, (a % b) / b

This returns the same result as your 'hack' function (which is actually also perfectly fine), including for negative numbers (e.g divfract(-8, 3) == (-3, 0.333...)), whereas math.modf rounds the other way (-2, -0.666...). (There are different ways to calculate a remainder)

You can use the math.modf() to get two separate portions. It is a built in method to split a number into integer and decimal parts. You can use the following custom myDivMod() function is order to make it work as following:

import math
def myDivMod(myParam, myDiv):
    fraction, wholeNum = math.modf( myParam / myDiv)
    return fraction, wholeNum
a,b = myDivMod(8,3)
print(a,b) # output is 0.6666666666666665, 2.0

I hope this solves your issue