# [Numpy-discussion] int to binary

Warren Weckesser warren.weckesser at gmail.com
Mon Apr 29 12:24:11 EDT 2013

```On 4/29/13, josef.pktd at gmail.com <josef.pktd at gmail.com> wrote:
>  Is there a available function to convert an int to binary
> representation as sequence of 0 and 1?
>
>
>  binary_repr produces strings and is not vectorized
>
>>>> np.binary_repr(5)
> '101'
>>>> np.binary_repr(5, width=4)
> '0101'
>>>> np.binary_repr(np.arange(5), width=4)
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
>   File "C:\Python26\lib\site-packages\numpy\core\numeric.py", line
> 1732, in binary_repr
>     if num < 0:
> ValueError: The truth value of an array with more than one element is
> ambiguous. Use a.any() or a.all()
>
> ------------
> That's the best I could come up with in a few minutes:
>
>
>>>> k = 3;  int2bin(np.arange(2**k), k, roll=False)
> array([[ 0.,  0.,  0.],
>        [ 1.,  0.,  0.],
>        [ 0.,  0.,  1.],
>        [ 1.,  0.,  1.],
>        [ 0.,  1.,  0.],
>        [ 1.,  1.,  0.],
>        [ 0.,  1.,  1.],
>        [ 1.,  1.,  1.]])
>>>> k = 3;  int2bin(np.arange(2**k), k, roll=True)
> array([[ 0.,  0.,  0.],
>        [ 0.,  0.,  1.],
>        [ 0.,  1.,  0.],
>        [ 0.,  1.,  1.],
>        [ 1.,  0.,  0.],
>        [ 1.,  0.,  1.],
>        [ 1.,  1.,  0.],
>        [ 1.,  1.,  1.]])
>
> -----------
> def int2bin(x, width, roll=True):
>     x = np.atleast_1d(x)
>     res = np.zeros(x.shape + (width,) )
>     for i in range(width):
>         x, r = divmod(x, 2)
>         res[..., -i] = r
>     if roll:
>         res = np.roll(res, width-1, axis=-1)
>     return res
>

Here one way, in which each value is and'ed (with broadcasting) with
an array of values with a 1 in each consecutive bit.  The comparison `
!= 0` converts the values from powers of 2 to bools, and then
`astype(int)` converts those to 0s and 1s.  You'll probably want to
adjust how reshaping is done to get the result the way you want it.

In [1]: x = array([0, 1, 2, 3, 15, 16])

In [2]: width = 5

In [3]: ((x.reshape(-1,1) & (2**arange(width))) != 0).astype(int)
Out[3]:
array([[0, 0, 0, 0, 0],
[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[1, 1, 0, 0, 0],
[1, 1, 1, 1, 0],
[0, 0, 0, 0, 1]])

Warren

>
> Josef
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>

```