打印带有n位小数的整数或浮点数

Question

在Python中,如果后一种情况要求我将打印输出限制为一定数量的数字,如何打印可能是整数或实数的数字？

长话短说,我们有以下例子:

print("{0:.3f}".format(num)) # I cannot do print("{}".format(num))
                             # because I don't want all the decimals

是否存在"Pythy"方式以确保例如,如果num == 1我打印1而不是1.000(我的意思是除了使用if语句混乱我的代码)

Answer 1

使用Python 3*,您可以使用,round()因为除了舍入floats,当应用于整数时,它将始终返回int:

>>> num = 1.2345
>>> round(num,3)
1.234
>>> num = 1
>>> round(num,3)
1

此行为记录在help(float.__round__):

Help on method_descriptor:

__round__(...)
    Return the Integral closest to x, rounding half toward even.
    When an argument is passed, work like built-in round(x, ndigits).

而且help(int.__round__):

Help on method_descriptor:

__round__(...)
    Rounding an Integral returns itself.
    Rounding with an ndigits argument also returns an integer.

*使用Python 2,round()总是returnsa float.

Answer 2

如果您需要为浮点值保持固定宽度，您可以使用printf 样式的格式，如下所示：

>>> num = 1
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1
>>> num = 1.2345
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.234
>>> num = 1.2
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.200