drop trailing zeros from decimal

Question

I have a long list of Decimals and that I have to adjust by factors of 10, 100, 1000,..... 1000000 depending on certain conditions. When I multiply them there is sometimes

Answer 1

Answer from the Decimal FAQ in the documentation:

>>> def remove_exponent(d):
...     return d.quantize(Decimal(1)) if d == d.to_integral() else d.normalize()

>>> remove_exponent(Decimal('5.00'))
Decimal('5')

>>> remove_exponent(Decimal('5.500'))
Decimal('5.5')

>>> remove_exponent(Decimal('5E+3'))
Decimal('5000')

Answer 2

Answer is mentioned in FAQ (https://docs.python.org/2/library/decimal.html#decimal-faq) but does not explain things.

To drop trailing zeros for fraction part you should use normalize:

>>> Decimal('100.2000').normalize()
Decimal('100.2')
>> Decimal('0.2000').normalize()
Decimal('0.2')

But this works different for numbers with leading zeros in sharp part:

>>> Decimal('100.0000').normalize()
Decimal('1E+2')

In this case we should use `to_integral':

>>> Decimal('100.000').to_integral()
Decimal('100')

So we could check if there's a fraction part:

>>> Decimal('100.2000') == Decimal('100.2000').to_integral()
False
>>> Decimal('100.0000') == Decimal('100.0000').to_integral()
True

And use appropriate method then:

def remove_exponent(num):
    return num.to_integral() if num == num.to_integral() else num.normalize()

Try it:

>>> remove_exponent(Decimal('100.2000'))
Decimal('100.2')
>>> remove_exponent(Decimal('100.0000'))
Decimal('100')
>>> remove_exponent(Decimal('0.2000'))
Decimal('0.2')

Now we're done.

Answer 3

There's probably a better way of doing this, but you could use .rstrip('0').rstrip('.') to achieve the result that you want.

Using your numbers as an example:

>>> s = str(Decimal('2.5') * 10)
>>> print s.rstrip('0').rstrip('.') if '.' in s else s
25
>>> s = str(Decimal('2.5678') * 1000)
>>> print s.rstrip('0').rstrip('.') if '.' in s else s
2567.8

And here's the fix for the problem that gerrit pointed out in the comments:

>>> s = str(Decimal('1500'))
>>> print s.rstrip('0').rstrip('.') if '.' in s else s
1500

Answer 4

Use the format specifier %g. It seems remove to trailing zeros.

>>> "%g" % (Decimal('2.5') * 10)
'25'
>>> "%g" % (Decimal('2.5678') * 1000)
'2567.8'

It also works without the Decimal function

>>> "%g" % (2.5 * 10)
'25'
>>> "%g" % (2.5678 * 1000)
'2567.8'

Answer 5

This should work:

'{:f}'.format(decimal.Decimal('2.5') * 10).rstrip('0').rstrip('.')

Answer 6

Slightly modified version of A-IV's answer

NOTE that Decimal('0.99999999999999999999999999995').normalize() will round to Decimal('1')

def trailing(s: str, char="0"):
    return len(s) - len(s.rstrip(char))

def decimal_to_str(value: decimal.Decimal):
    """Convert decimal to str

    * Uses exponential notation when there are more than 4 trailing zeros
    * Handles decimal.InvalidOperation
    """
    # to_integral_value() removes decimals
    if value == value.to_integral_value():
        try:
            value = value.quantize(decimal.Decimal(1))
        except decimal.InvalidOperation:
            pass
        uncast = str(value)
        # use exponential notation if there are more that 4 zeros
        return str(value.normalize()) if trailing(uncast) > 4 else uncast
    else:
        # normalize values with decimal places
        return str(value.normalize())
        # or str(value).rstrip('0') if rounding edgecases are a concern