Natural Binary Code - finding first n numbers

Question

Outputting/finding first n numbers of Natural Binary Code:

import math  
def binary_print(n):
    m = int(math.ceil(math.log(n, 2)))
    for i in range(n):
        b = str(bin(i)[2:])
        print((m - len(b)) * '0' + b)

My question is:

Do you know any other way to do this in Python? Maybe faster? Or shorter (less code)?

First n natural binary numbers. For example binary_print(3) outputs 00 01 10 — Maciej Ziarko, Feb 08 '11 at 18:15
http://stackoverflow.com/questions/4928297/all-permutations-of-a-binary-sequence-x-bits-long/4928350#4928350 might help. — Josh Bleecher Snyder, Feb 08 '11 at 18:20

Scott Griffiths · Answer 1 · 2011-02-08T18:33:09.847

5

Well this is shorter, not sure about faster:

def binary_print(n):
    print '\n'.join('{:0{}b}'.format(x, (n-1).bit_length()) for x in range(n))

Example usage:

>>> binary_print(6)
000
001
010
011
100
101

edited Feb 08 '11 at 18:33

answered Feb 08 '11 at 18:17

Scott Griffiths

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My Python 2.6.6 says AttributeError: 'int' object has no attribute 'bit_length'. Is it a feature od Python3? – Maciej Ziarko Feb 08 '11 at 18:25

Hugh Bothwell · Accepted Answer · 2011-02-08T18:34:06.427

4

def binary_values(n):
    fmt = "{0:0"+str((n-1).bit_length())+"b}"
    for i in range(n):
        print fmt.format(i)

Note: (n-1).bit_length() fixes a fencepost error (otherwise if n is a power of two it prints 1 too many leading zeros).

Might also be able to speed it up a bit more by unrolling a lookup,

def binary_values(n):
    fmt = ("{0:0"+str((n-1).bit_length())+"b}").format
    for i in range(n):
        print fmt(i)

edited Feb 08 '11 at 18:34

answered Feb 08 '11 at 18:27

Hugh Bothwell

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Yes, my answer was out by one - fixed now. Your answer is faster, I was going for the one-liner! – Scott Griffiths Feb 08 '11 at 18:38
1

I like both the answers. I'm learning Python by solving small problems like this one and I learned a lot analysing your solutions. Thanks. – Maciej Ziarko Feb 08 '11 at 18:54

Natural Binary Code - finding first n numbers

2 Answers2