How to convert string like '001100' to numpy.array([0,0,1,1,0,0]) quickly?

Question

I have a string consists of 0 and 1, like '00101'. And I want to convert it to numpy array numpy.array([0,0,1,0,1].

I am using for loop like:

import numpy as np
X = np.zeros((1,5),int)
S = '00101'
for i in xrange(5):
    X[0][i] = int(S[i])

But since I have many strings and the length of each string is 1024, this way is very slow. Is there any better way to do this?

Answer 1

map should be a bit faster than a list comp:

import  numpy as np

arr = np.array(map(int,'00101'))

Some timings show it is on a string of 1024 chars:

In [12]: timeit np.array([int(c) for c in s])
1000 loops, best of 3: 422 µs per loop

In [13]: timeit np.array(map(int,s))
1000 loops, best of 3: 389 µs per loop

Just calling list in s and using dtype=int is faster:

In [20]: timeit np.array(list(s), dtype=int)
1000 loops, best of 3: 329 µs per loop

Using fromiter and passing dtype=int is faster again:

In [21]: timeit  np.fromiter(s,dtype=int)
1000 loops, best of 3: 289 µs per loop

Borrowing from this answer, using fromstring and uint8 as the dtype is the fastest:

In [54]: timeit  np.fromstring(s, 'int8') - 48
100000 loops, best of 3: 4.54 µs per loop

Even rebinding the name and changing the dtype is still by far the fastest:

In [71]: %%timeit
   ....: arr = np.fromstring(s, 'int8') - 48
   ....: arr = arr.astype(int)
   ....: 
100000 loops, best of 3: 6.23 µs per loop

Even considerably faster than Ashwini's join:

In [76]: timeit  np.fromstring(' '.join(s), sep=' ', dtype=int)
10000 loops, best of 3: 62.6 µs per loop

As @Unutbu commented out,np.fromstring(s, 'int8') - 48 is not limited to ones and zeros but will work for all strings composed of ASCII digits.