method to remove zero digits from fraction in decimal

[Skaperen]

when a decimal number is a whole (integral, integer-like) number, it gets converted to string without the ".0". but if i take the result of calculation, such as conversion from Fahrenheit to Celsius, 98.6 gives me 37.0. but when the fraction part is zero, i want to make it like whole. basically i want to have a whole value if the fraction is zero. what decimal method can do this. i looked through the ones in the library manual but i could not find one. must i implement one (i'm sure i can, but this seem silly if there already is one).

[bowlofred]

when a decimal number is a whole (integral, integer-like) number, it gets converted to string without the ".0"

I would say the string method of a int doesn't have the decimal, while a float does.

>>> print(int(2))
2
>>> print(float(2))
2.0

You could always convert to an int if appropriate.

if x == int(x):
    x = int(x)

[rob101]

Not too sure if I fully understand, but here goes:

number = 37.0

print(f"number with the decimal place:    {number}")
print(f"number without the decimal place: {number:.0f}")

Output:
number with the decimal place:    37.0
number without the decimal place: 37

[ibreeden]

But if I understand Skaperen right, he only wants to convert a float to an int if the decimal part equals zero.

[rob101]

But if I understand Skaperen right, he only wants to convert a float to an int if the decimal part equals zero.

So, more like this, you think?

number = float(input("number: "))

if number % 1 == 0:
    print(f"{int(number)} does not have a decimal place")
else:
    print(f"{number} has a decimal place")

[deanhystad]

I don't think you can automatically print a float with no fraction component and have it look like an int. You can specify zero precision, but that prints 1.2 as "1". I cannot find or thing of a way to do this without testing if the value has no fraction component and use different formats based on the result.

def min_float_str(value:float):
    int_val = int(value)
    return str(int_val) if value == int_val else str(value)

print(min_float_str(1.2), min_float_str(1.0))

Output:
1.2 1

[Skaperen]

Not too sure if I fully understand, but here goes:

number = 37.0

print(f"number with the decimal place:    {number}")
print(f"number without the decimal place: {number:.0f}")

Output:
number with the decimal place:    37.0
number without the decimal place: 37

that's float. use decimal, like decimal.Decimal(111)/decimal.Decimal(3). and don't add on your own formatting.

i guess i didn't make it clear enough that i am referring to type decimal.Decimal.

[Skaperen]

I don't think you can automatically print a float with no fraction component and have it look like an int.

i meant decimal.Decimal("37") vs decimal.Decimal("37.0").

what i want is a way to have decimal.Decimal("37.0") print as "37" while decimal.Decimal("37.1") prints as "37.1". since decimal.Decimal("37") prints as intended, i was looking for a method like my_decimal_number.remove_fraction_if_zero()

[Skaperen]

But if I understand Skaperen right, he only wants to convert a float to an int if the decimal part equals zero.

almost.

i want this to be in the decimal.Decimal type which may or may not store the fraction, which can be zero if stored.

sorry, my first post just said "decimal" instead of "decimal.Decimal".

[deanhystad]

So you want decimal to resize to a minimum precision instead of remembering the precision it inherited from a math operation.

from decimal import Decimal as D

a = D("3")
b = D("2")
c = b / a
d = c * a
print(a, b, c, d)

Output:
3 2 0.6666666666666666666666666667 2.000000000000000000000000000

d should be 2 instead of 2.000000000000000000000000000

One way to make a new decimal with all trailing zeros removed.

from decimal import *
from decimal import Decimal as D

def rstrip(value:D)->D:
    """I'm a happy little function :D with a tragic ending D:"""
    value_str = str(value)
    if "." in value_str:
        return D(value_str.rstrip(".0"))
    return value

a = D("3")
b = D("2")
c = rstrip(b / a)
d = rstrip(c * a)
print(a, b, c, d, D(1000))

Output:
3 2 0.6666666666666666666666666667 2 1000