Converting binary to float

Question

I have a function which encode float number to binary. I split float number by "." and convert both parts to binary. After i concatenate results. Left part I encode by simple convert bin(left_part). Right part I encode by next algorithm:

    while count:
        val *= 2
        if val >= 1:
            val_bin += "1"
        else:
            val_bin += "0"
        val %= 1
        count -= 1
    return val_bin

I have found this solution on the internet. But I can't found out how to convert back right part to float number in decimal. How i can do it?

I only see the way of full exhaustive search: Example: I have "00110011". We move from right to left. In first step we have "1". That means in the step value may be in interval "1.000 to 1.999". In the next step we have another "1". That means that value in the first step was in interval "1.500 to 1.999". In current step we have the next interval "1.001 to 1.999". Next step we have "0". That means in previous step value was in interval 1.000 to 1.499". etc. And after all iterations we will not have a accurate result. Result always will be in the interval. I may be wrong.

You simply have to understand the logic behind the solution you have found and try to work backwards. It helps if you do it with pen and paper. — Arpan
– Arpan, Commented Jun 4, 2016 at 20:24
To paraphrase @Arpan: What don't you understand about the algorithm that you posted? — Scott Hunter
– Scott Hunter, Commented Jun 4, 2016 at 20:25
@ScottHunter I understand algorithm that i posted. I don't understand how to decode back. I think there will be many possible solutions after i walk backwards through my algorithm and the result will not be accurate enough. — Max
– Max, Commented Jun 4, 2016 at 20:49

John Coleman · Accepted Answer · 2016-06-04 21:32:06Z

2

Here is a version which uses the built-in int function. It is based on the fact that in any base the decimal part of a decimal expansion is the numerator of a fraction where the denominator is a power of the base. The power in question is easily inferred from the length of the decimal part.

#the following code assumes that '.' appears once

def binToFloat(b):
    parts = b.split('.')
    s,t = parts
    x = 0 if len(s) == 0 else int(s,2)
    y = 0 if len(t) == 0 else int(t,2)
    return x + y/2**len(t)

For example:

>>> binToFloat('1101101.1101')
109.8125

The above code assumed Python 3. In Python 2 you would need to rewrite the final line as

return x + float(y)/2**len(t)

edited Jun 4, 2016 at 21:32

answered Jun 4, 2016 at 21:26

John Coleman

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Comments

Keiwan · Accepted Answer · 2016-06-04 21:18:36Z

If you have your number in binary as a string and you only want to interpret it as the part to the right of the decimal point, you can use this function:

def binToFloatAfterPoint(b):
    exp = -1
    result = 0
    for digit in b:
        if digit != "0":
            result += math.pow(int(digit)*2,exp)
        exp -= 1
    return result

However, if you have your whole binary number as one string you can directly convert it into a float by specifying how many decimalPlaces you want using this function:

def binToFloat(b,decimalPlaces):
    exp = len(b) - decimalPlaces - 1
    result = 0
    for digit in b:
        if digit != "0":
            result += math.pow(int(digit)*2,exp)
        exp -= 1
    return result

So say you have

b = "1010"

The output of the first function (and the second with decimalPlaces=4) would be: 0.625 At the same time you could call the second function with other values for decimalPlaces which would give you different floats:

binToFloat(b,0) = 10.0
binToFloat(b,1) = 5.0
binToFloat(b,2) = 2.5
binToFloat(b,3) = 1.25

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