def get_int_or_float(v):
    number_as_float = float(v)
    number_as_int = int(number_as_float)
    return number_as_int if number_as_float == number_as_int else number_as_float

>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545

如果您不想使用第三方模块，以下可能是最可靠的解决方案：

def string_to_int_or_float(s):
    try:
        f = float(s) # replace s with str(s) if you are not sure that s is a string
    except ValueError:
        print("Provided string '" + s + "' is not interpretable as a literal number.")
        raise
    try:
        i = int(str(f).rstrip('0').rstrip('.'))
    except:
        return f
    return i

它可能不是最快的，但在许多其他解决方案失败的情况下，它可以正确处理字面数字，例如：

>>> string_to_int_or_float('789.')
789
>>> string_to_int_or_float('789.0')
789
>>> string_to_int_or_float('12.3e2')
1230
>>> string_to_int_or_float('12.3e-2')
0.123
>>> string_to_int_or_float('4560e-1')
456
>>> string_to_int_or_float('4560e-2')
45.6

def num(s):
    """num(s)
    num(3),num(3.7)-->3
    num('3')-->3, num('3.7')-->3.7
    num('3,700')-->ValueError
    num('3a'),num('a3'),-->ValueError
    num('3e4') --> 30000.0
    """
    try:
        return int(s)
    except ValueError:
        try:
            return float(s)
        except ValueError:
            raise ValueError('argument is not a string of number')

float(x) if '.' in x else int(x)

Use:

def num(s):
    try:
        for each in s:
            yield int(each)
    except ValueError:
        yield float(each)
a = num(["123.55","345","44"])
print a.next()
print a.next()

这是我所能想到的最具Python风格的方式。