I have a string like this:
word = 'python'
Based on string.ascii_lowercase, I'd like to create a new array which looks like this:
[15, 24, 19, 7, 14, 13]
My solution for this problem was to do the following:
alphabet = {char: i for i, char in enumerate(string.ascii_lowercase)}
indices = [alphabet[char] for char in word]
print(indices)
Output: [15, 24, 19, 7, 14, 13]
But I'm looking for a more efficient way, without using loops. How can I do that in a vectorized way?
One way is to build an array from the string using np.fromiter and specifying a 'U1' dtype, cast to integer, and subtract the starting position of the alphabet in the unicode table, 97 or we can just use ord('a'): as suggested by Antoine Dubuis:
import numpy as np
word = 'python'
np.fromiter(word, dtype='U1').view(np.uint32) - ord('a')
array([15, 24, 19, 7, 14, 13])
We can use np.frombuffer for a pretty efficient one -
import numpy as np
np.frombuffer(word.encode(), dtype=np.uint8)-97
Timings on a 1M long string :
In [23]: import string
In [24]: p = string.ascii_lowercase
In [25]: word = ''.join([p[i] for i in np.random.randint(0,len(p), 1000000)])
In [26]: %timeit np.frombuffer(word.encode(), dtype=np.uint8)-97
136 µs ± 1.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
# @yatu's soln with np.fromiter
In [27]: %timeit np.fromiter(word, dtype='U1').view(np.uint32) - ord('a')
24.8 ms ± 423 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)