how to convert a float-point number to a fixed-point number with a certain width and precision. Is there any library in python can be used to do this?

Question

I need a function to convert a float number to a fixed-point number, represented in binary so that I can put it to the hardware accelerator which can only accept fix-point number. for example, when I input a float number 5.625, giving data width 8, precision 4(4 bits for mantissa), it will return b'01011010, or h'5a or d'90. Is there any functions like this in python libraries?

Answer 1

That's the same as multiplying by 16 and converting to an int.

>>> int(5.625 * (2 ** 4))
90

The 16 depends on the number of bits in the precision, but it's easily calculated.

If the number has to go in multiple bytes (like 2 or 4), use the struct module. E.g. for 300.625 with 32 data bits, 8 of which are precision:

>>> int(300.625 * (2**8))
76960
>>> struct.pack('I', 76960)  # 'I' is 4-byte unsigned int
b'\xaa,\x01\x00'

Answer 2

You can use fxpmath in python. You can find info about it at:

https://github.com/francof2a/fxpmath

In your example case:

from fxpmath import Fxp

x = Fxp(5.625, signed=False, n_word=8, n_frac=4)
print(x)         # float value
print(x.bin())   # binary representation
print(x.hex())   # hex repr
print(x.val)     # raw val (decimal of binary stored)

Outputs:

5.625
01011010
0x5a
90

Answer 3

several fixed point libraries like numfi can transform floating point number to fixed point and give bin/dec/hex representation

>>> from numfi import numfi
>>> x = numfi(5.625,1,8,4)
>>> x.int # d'90
array([90], dtype=int64)
>>> x.bin # b'01011010
array(['01011010'], dtype='<U8')
>>> x.hex # h'5a
array(['5A'], dtype='<U2')

If you prefer simpler solution, you can do quantization yourself:

num = 5.625
w = 8 # data width
f = 4 # fraction width
quantized = np.round(num*(2**f)).astype(int) #note: overflow is not considered here
dec_repr = np.base_repr(quantized, base=10)
bin_repr = np.binary_repr(quantized, width=8)
hex_repr = np.base_repr(quantized, base=16)

Answer 4

Try to use decimal module.

You may pass it a float formatted as a string with required format and precision:

>>> from decimal import Decimal

>>> '{:5f}'.format(10.0)
'10.000000'

>>> Decimal('{:5f}'.format(10.0))
Decimal('10.000000')

Answer 5

def float2fix(val, width, precision):
    integer = abs(int(val * 2 ** precision))
    if val < 0:
        integer = 2 ** (width-1) - integer
    if val >= 0:
        fix_str = bin(integer).replace('0b', '').rjust(width, '0')
    else:
        fix_str = '1'+bin(integer).replace('0b', '').rjust(width-1, '0')
    return fix_str