如何使用python将浮点数转换为具有预定义位数的固定点 [英] How to use python to convert a float number to fixed point with predefined number of bits

问题描述

我有numpy格式的浮点数32个数字(比方说正数).我想将它们转换为具有预定义位数的定点数，以降低精度.

I have float 32 numbers (let's say positive numbers) in numpy format. I want to convert them to fixed point numbers with predefined number of bits to reduce precision.

例如，使用函数num2fixpt，数字3.1415926在matlab中变为3.25. 该命令是num2fixpt(3.1415926，sfix(5)，2 ^(1 + 2-5)，'Nearest'，'on')，它表示整数部分为3位，小数部分为2位.

For example, number 3.1415926 becomes 3.25 in matlab by using function num2fixpt. The command is num2fixpt(3.1415926,sfix(5),2^(1 + 2-5), 'Nearest','on') which says 3 bits for integer part, 2 bits for fractional part.

我可以使用Python做同样的事情

Can I do the same thing using Python

推荐答案

如果您了解IEEE浮点符号的工作原理，则可以这样做.基本上，您将需要转换为python LONG，执行按位运算符，然后隐式返回.例如:

You can do it if you understand how IEEE floating point notation works. Basically you'll need to convert to a python LONG, do bitwise operators, then covert back. For example:

import time,struct,math
long2bits = lambda L: ("".join([str(int(1 << i & L > 0)) for i in range(64)]))[::-1]
double2long = lambda d: struct.unpack("Q",struct.pack("d",d))[0]
double2bits = lambda d: long2bits(double2long(d))
long2double = lambda L: struct.unpack('d',struct.pack('Q',L))[0]
bits2double = lambda b: long2double(bits2long(b))
bits2long=lambda z:sum([bool(z[i] == '1')*2**(len(z)-i-1) for i in range(len(z))[::-1]])

>>> pi = 3.1415926
>>> double2bits(pi)
'0100000000001001001000011111101101001101000100101101100001001010'
>>> bits2long('1111111111111111000000000000000000000000000000000000000000000000')
18446462598732840960L
>>> double2long(pi)
4614256656431372362
>>> long2double(double2long(pi) & 18446462598732840960L)
3.125
>>>

def rshift(x,n=1):
    while n > 0:
        x = 9223372036854775808L | (x >> 1)
        n -= 1
    return x

>>> L = bits2long('1'*12 + '0'*52)
>>> L
18442240474082181120L
>>> long2double(rshift(L,0) & double2long(pi))
2.0
>>> long2double(rshift(L,1) & double2long(pi))
3.0
>>> long2double(rshift(L,4) & double2long(pi))
3.125
>>> long2double(rshift(L,7) & double2long(pi))
3.140625

这只会截断位数，而不会舍入它们. rshift函数是必需的，因为python的右移运算符用零填充最左边的空白位.请在此处中查看IEEE浮点的说明.

This will only truncate the number of bits though, not round them. The rshift function is necessary because python's right-shift operator fills the empty leftmost bit with a zero. See a discription of IEEE floating point here.

这篇关于如何使用python将浮点数转换为具有预定义位数的固定点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！