获取“比特”在Python中的浮点数? [英] Get the "bits" of a float in Python?
本文介绍了获取“比特”在Python中的浮点数?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在寻找Java的 Float.floatToBits
的Python等价物。
答案Alex Martelli在这个问题上给出的内容非常简单 - 你可以把它简化为:
>>> import struct
>>>
>>>
>>> def floatToBits(f):
... s = struct.pack('> f',f)
... return struct.unpack('> l',s)[0]
...
...
>>> floatToBits(173.3125)
1127043072
>>>十六进制(_)
'0x432d5000'
一旦你把它作为一个整数执行你需要的任何其他操作。
您可以将操作顺序颠倒过来:
>>> def bitsToFloat(b):
... s = struct.pack('> l',b)
... return struct.unpack('> f',s)[0]
>>> bitsToFloat(0x432d5000)
173.3125
I am looking for the Python equivalent of Java's Float.floatToBits
.
I found this Python: obtain & manipulate (as integers) bit patterns of floats but does anyone know of a less complicated way?
解决方案
The answer that Alex Martelli gives in that question is really pretty simple -- you can reduce it to:
>>> import struct
>>>
>>>
>>> def floatToBits(f):
... s = struct.pack('>f', f)
... return struct.unpack('>l', s)[0]
...
...
>>> floatToBits(173.3125)
1127043072
>>> hex(_)
'0x432d5000'
Once you have it as an integer, you can perform any other manipulations you need to.
You can reverse the order of operations to round-trip:
>>> def bitsToFloat(b):
... s = struct.pack('>l', b)
... return struct.unpack('>f', s)[0]
>>> bitsToFloat(0x432d5000)
173.3125
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