Large number to many characters

Question

My goal is to take any size number, and convert it to a string of smaller characters that represent the concatenation of the binary expansion of the number. To illustrate what I am talking about, here is an example using the number 4820:

4820's binary:   0001001011010100   #   16 bits of binary, enough to store the number 4820
Separated:     00010010  11010100   #   Simply split the binary into two char size pieces
Characters:      '\x12'       'Ô'   #   Converted the binary snippets into their respective characters
Together:                 '\x12Ô'   # The concatenation of the characters, this is what I need

I wrote a algorithm in Python to do the job for me, but I'm not sure if there is a easier way to do it, or if my algorithm is as efficient as it can be.

def to_chars(number, length):                             #takes number and expected binary length
    string = bin(number)[2:]                              #built-in binary conversion
    expanded = ('0' * (length - len(string))) + string    #expand the binary
    chars = ''
    for pos in range(int(length / 8)):                    #iterates over chunks of 8 bits
        section = expanded[8 * pos : 8 * (pos + 1)]       #gets the binary snippet
        chars += chr(int(section, 2))                     #converts the binary into a character
    return chars                                          #return the concatenated string

There has to be a lower level, possibly built in way to do the same thing, but faster. In a way, this is just the conversion from one data type to another, but I can't figure out a faster way.

Here are a few benchmarks:

%timeit to_chars(4820, 16)
100000 loops, best of 3: 3.04 µs per loop

%timeit to_chars(4820, 32)
100000 loops, best of 3: 4.6 µs per loop

%timeit to_chars(690655640, 32)
100000 loops, best of 3: 4.59 µs per loop

Answer 1

Well, it's not a huge performance improvement, but at least it's a lot cleaner:

def to_chars2(number, length):
    bin_num = bin(number)[2:].zfill(length)
    return ''.join([chr(int(bin_num[s:s+8], 2)) for s in range(0, len(bin_num), 8)])

Benchmarks:

>>> timeit(lambda: to_chars2(4820, 16))
1.590714931488037
>>> timeit(lambda: to_chars(4820, 16))
1.6366891860961914

Answer 2

Actually, ignore my previous answer. You're looking for the binascii module. It uses C modules for performance.

>>> timeit(lambda: binascii.unhexlify(hex(4820)[2:]))
0.379443883895874

Answer 3

My problem was that I was just not well educated enough about the subject, but after some experimentation, I found the solution. After I tried to save the full range of unicode characters into a file, I ran into a few errors, saying that some of the characters mapped to an undefined symbol, which was out of my control. I decided to try to convert int types into bytes types then, so they can be saved to the file. The solution was extremely simple.

In [21]: x = 4820
In [22]: x.to_bytes(2, byteorder = 'big') # we want it to fit 4820 into 2 bytes, using big endian byte-order
Out[22]: b'\x12\xd4' # This is what I wanted, it just looks a little different

This functionality is exactly what I needed, and I simply overlooked the methods from python's int data type.