python float类型使用方法

今天我们来看看python 中float 方法

1: 获取值的最简化结果

def
as_integer_ratio(self):
# real signatureunknown; restored from __doc__

"""

float.as_integer_ratio() -> (int,int)



Return a pair of integers, whoseratio is exactly equal to the original

float and with a positivedenominator.

Raise OverflowError on infinities anda ValueError on NaNs.



>>>(10.0).as_integer_ratio() #### 这里给出了使用方法，返回了元组类型

(10, 1)

>>> (0.0).as_integer_ratio()

(0, 1)

>>>(-.25).as_integer_ratio()

(-1, 4)

"""


2:复数取共轭

a = 1.1
b = 10.5
c = 2 + 1.3J                      ### 这个例子才能体现出来

print("a",a.conjugate())
print("b",b.conjugate())
print("c",c.conjugate())

a 1.1
b 10.5
c (2-1.3j)

3:十六进制转为浮点数

def fromhex(self, string): # real signature unknown; restored from __doc__
    """
    float.fromhex(string) -> float
    
    Create a floating-point number from a hexadecimal string.
    >>> float.fromhex('0x1.ffffp10')    ### 使用方法
    2047.984375
    >>> float.fromhex('-0x1p-1074')
    -5e-324
    """
    return 0.0

4:将浮点数转换为十六进制

def hex(self): # real signature unknown; restored from __doc__
    """
    float.hex() -> string
    
    Return a hexadecimal representation of a floating-point number.
    >>> (-0.1).hex()                  #### 使用方法
    '-0x1.999999999999ap-4'
    >>> 3.14159.hex()
    '0x1.921f9f01b866ep+1'
    """
    return ""

5:判断是不是整数，返回bool 值

def is_integer(self, *args, **kwargs): # real signature unknown
    """ Return True if the float is an integer. """
    pass

测试：

a = 1.1
b = 10.5
c = 2 + 1.3J
d = 5.0

print("a",a.is_integer())
print("b",b.is_integer())
print("d",d.is_integer())

########         返回 结果 bool类型
a False
b False
d True