Slicing a Numpy array based on characters of string elements

Use numpy.char.startswith:

arr[np.char.startswith(arr, "a")]

Output:

array(['abc', 'az', 'ac'], dtype='<U3')

Note that this, by default, uses the first index (i.e. 0). Use can use start parameter to act as indexing:

arr[np.char.startswith(arr, "a", 1)]

Output:

array(['bac', 'bar'], dtype='<U3')

You can use simple List Comprehension:

In [3403]: L2 = ['abc', 'bac', 'foo', 'az', 'bar', 'ac']

In [3404]: arr = np.array(L2)  

In [3411]: res = np.array([i for i in arr if i.startswith('a')])

In [3412]: res
Out[3412]: array(['abc', 'az', 'ac'], dtype='<U3')

OR, if you want to use mask:

Use np.char.startswith

In [3415]: mask = np.char.startswith(arr, 'a')
In [3417]: print(arr[mask])
['abc' 'az' 'ac']

In [192]: alist = ['abc', 'bac', 'foo', 'az', 'bar', 'ac']
In [193]: arr = np.array(alist)

The straightforward list comprehension:

In [194]: [a[0]=='a' for a in alist]
Out[194]: [True, False, False, True, False, True]

It also works with the array, but slower (iteration on arrays is slower than on lists):

In [195]: timeit [a[0]=='a' for a in alist]
707 ns ± 5.61 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [196]: timeit [a[0]=='a' for a in arr]
4.88 µs ± 9.44 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

String startswith is also an option:

In [197]: [a.startswith('a') for a in alist]
Out[197]: [True, False, False, True, False, True]
In [198]: timeit [a.startswith('a') for a in alist]
1.14 µs ± 11.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

numpy doesn't have its own string processing tools. It has np.char functions, but they just apply python string methods, without speed improvement:

In [200]: np.char.startswith(arr, 'a')
Out[200]: array([ True, False, False,  True, False,  True])
In [201]: timeit np.char.startswith(arr, 'a')
12.5 µs ± 297 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

You get the best speed if you start and end with a list. Iterating on array, or converting the boolean list back to array takes time.

Rereading your code I see you want to select the items, not just create the mask. Then the list comprehension should be:

In [215]: [a for a in alist if a[0]=='a']
Out[215]: ['abc', 'az', 'ac']
In [216]: timeit [a for a in alist if a[0]=='a']
645 ns ± 3.36 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

====

As for your failed code

[:] does nothing for you, in any of the expressions:

In [213]: arr[:]
Out[213]: array(['abc', 'bac', 'foo', 'az', 'bar', 'ac'], dtype='<U3')

So you are just checking that the first element of arr is 'a'`. You aren't testing the first string of each element.

In [214]: arr[0]=='a'
Out[214]: False

Starting in numpy 1.23.0, you will be able to slice strings in an array. The following will work, without making any copies:

L2 = np.array(['abc', 'bac', 'foo', 'az', 'bar', 'ac'])
mask = L2[:, None].view('U1')[:, 0] == 'a'
print(mask)      # True, False, False, True, False, True
print(arr[mask]) # 'abc', 'az', 'ac'

I'm writing an even simpler method, built on this change (here) that will allow the following:

mask = np.char.slice_(L2, stop=1) == 'a'

Again, fully vectorized and without copying.