Convert string to numpy array

Question

Input:

mystr = "100110"

Desired output numpy array:

mynumpy == np.array([1, 0, 0, 1, 1, 0])

I have tried:

np.fromstring(mystr, dtype=int, sep='')

but the problem is I can't split my string to every digit of it, so numpy takes it as an one number. Any idea how to convert my string to numpy array?

Solution 1

list may help you do that.

import numpy as np

mystr = "100110"
print np.array(list(mystr))
# ['1' '0' '0' '1' '1' '0']

If you want to get numbers instead of string:

print np.array(list(mystr), dtype=int)
# [1 0 0 1 1 0]

Solution 2

You could read them as ASCII characters then subtract 48 (the ASCII value of 0). This should be the fastest way for large strings.

>>> np.fromstring("100110", np.int8) - 48
array([1, 0, 0, 1, 1, 0], dtype=int8)

Alternatively, you could convert the string to a list of integers first:

>>> np.array(map(int, "100110"))
array([1, 0, 0, 1, 1, 0])

Edit: I did some quick timing and the first method is over 100x faster than converting it to a list first.

Solution 3

Adding to above answers, numpy now gives a deprecation warning when you use fromstring
DeprecationWarning: The binary mode of fromstring is deprecated, as it behaves surprisingly on unicode inputs. Use frombuffer instead.
A better option is to use the fromiter. It performs twice as fast. This is what I got in jupyter notebook -

import numpy as np
mystr = "100110"

np.fromiter(mystr, dtype=int)
>> array([1, 0, 0, 1, 1, 0])

# Time comparison
%timeit np.array(list(mystr), dtype=int)
>> 3.5 µs ± 627 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit np.fromstring(mystr, np.int8) - 48
>> 3.52 µs ± 508 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit np.fromiter(mystr, dtype=int)
1.75 µs ± 133 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)