round() returns different result depending on the number of arguments

The round function returns an integer if the second argument is not specified, else the return value has the same type as that of the first argument:

>>> help(round)Help on built-in function round in module builtins:round(number, ndigits=None)    Round a number to a given precision in decimal digits.    The return value is an integer if ndigits is omitted or None. Otherwise    the return value has the same type as the number. ndigits may be negative.

So if the arguments passed are an integer and a zero, the return value will be an integer type:

>>> round(100, 0)100>>> round(100, 1)100

For the sake of completeness:

Negative numbers are used for rounding before the decimal place

>>> round(124638, -2)124600>>> round(15432.346, -2)15400.0

When you specify the number of decimals, even if that number is 0, you are calling the version of the method that returns a float. So it is normal that you get that result.

I think that it is worth to mention, that you can encounter not only round(int) or round(float), but for example: round(class 'numpy.float64'). In that case round will not return integer even when "ndigits is omitted or None." (like Built-in Functions documentation says), but the type of class numpy.float64. I analyzed code example and bumped at round(x) prints "10.0" in python 3.7.3