python Decimal – checking if integer

Try math.floor(val) == val or val == int(val).

Decimal does have a “hidden” method called _isinteger() that works kind of the like the float’s is_integer() method:

>>> Decimal(1)._isinteger()
True
>>> Decimal(1.1)._isinteger()
Traceback (most recent call last):
  File "C:Program Files (x86)Wing IDE 4.1srcdebugtserver_sandbox.py", line 1, in <module>
    # Used internally for debug sandbox under external interpreter
  File "C:Python26Libdecimal.py", line 649, in __new__
    "First convert the float to a string")
TypeError: Cannot convert float to Decimal.  First convert the float to a string

As you can see, you would have to catch an exception though. Alternatively, you could do the test on the value BEFORE you pass it to Decimal using the float’s method as you mentioned or by using isinstance.

The mathematical solution is to convert your decimal number to integer and then test its equality with your number.

Since Decimal can have an arbitrary precision, you should not convert it to int or float.

Fortunately, the Decimalclass has a to_integral_value which make the conversion for you. You can adopt a solution like this:

def is_integer(d):
    return d == d.to_integral_value()

Example:

from decimal import Decimal

d_int = Decimal(3)
assert is_integer(d_int)

d_float = Decimal(3.1415)
assert not is_integer(d_float)

You can call as_tuple() on a Decimal object to get the sign, the sequence of digits, and the exponent which together define the Decimal value.

If the exponent of a normalized Decimal is non-negative, then your value doesn’t have a fractional component, i.e., it is an integer. So you can check for this very easily:

def is_integer(dec):
    """True if the given Decimal value is an integer, False otherwise."""
    return dec.normalize().as_tuple()[2] >= 0

Try it and see:

from decimal import Decimal


decimals = [
    Decimal('0'),
    Decimal('0.0000'),
    Decimal('1'),
    Decimal('-1'),
    Decimal('1000000'),
    Decimal('0.1'),
    Decimal('-0.0000000009'),
    Decimal('32.4')]

for d in decimals:
    print("Is {} an integer? {}".format(d, is_integer(d)))

Is 0 an integer? True
Is 0.0000 an integer? True
Is 1 an integer? True
Is -1 an integer? True
Is 1000000 an integer? True
Is 0.1 an integer? False
Is -9E-10 an integer? False
Is 32.4 an integer? False

Decimal class has a function to_integral_exact which converts decimal to an integer and also signals if there were any non-zero digits discarded.
https://docs.python.org/3/library/decimal.html#decimal.Decimal.to_integral_exact

Using this information we can implment float’s is_integer for Decimal too:

import decimal
def is_integer(value: decimal.Decimal) -> bool:
   is_it = True
   context = decimal.getcontext()
   context.clear_flags()
   exact = value.to_integral_exact()
   if context.flags[decimal.Inexact]:
     is_it = False
   context.clear_flags()
   return is_it

Using the function above:

# tested on Python 3.9
>>> is_integer(decimal.Decimal("0.23"))
False
>>> is_integer(decimal.Decimal("5.0000"))
True
>>> is_integer(decimal.Decimal("5.0001"))
False

As of Python 3.6, Decimal has a method as_integer_ratio().

as_integer_ratio() returns a (numerator, denominator) tuple. If the denominator is 1, then the value is an integer.

>>> from decimal import Decimal
>>> Decimal("123.456").as_integer_ratio()[1] == 1
False
>>> Decimal("123.000").as_integer_ratio()[1] == 1
True

Building on what was said above, I used:

>>> not 2.5 % 1
False
>>> not 1.0 % 1
True
>>> not 14.000001 % 1
False
>>> not 2.00000000 % 1
True

So you can use the following one liner:

not value % 1

It will provide you with your desired bool.