How to find the first index of any of a set of characters in a string

Question

I\'d like to find the index of the first occurrence of any “special” character in a string, like so:

>>> \"Hello world!\".index([\' \', \'!\'])
5
         


        

Answer 1

You can use enumerate and next with a generator expression, getting the first match or returning None if no character appears in s:

s = "Hello world!"

st = {"!"," "}
ind = next((i for i, ch  in enumerate(s) if ch in st),None)
print(ind)

You can pass any value you want to next as a default return value if there is no match.

If you want to use a function and raise a ValueError:

def first_index(s, characters):
    st = set(characters)
    ind = next((i for i, ch in enumerate(s) if ch in st), None)
    if ind is not None:
        return ind
    raise ValueError

For smaller inputs using a set won't make much if any difference but for large strings it will be a more efficient.

Some timings:

In the string, last character of character set:

In [40]: s = "Hello world!" * 100    
In [41]: string = s    
In [42]: %%timeit
st = {"x","y","!"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 1.71 µs per loop    
In [43]: %%timeit
specials = ['x', 'y', '!']
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 2.64 µs per loop

Not in the string, larger character set:

In [44]: %%timeit
st = {"u","v","w","x","y","z"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 1.49 µs per loop

In [45]: %%timeit
specials = ["u","v","w","x","y","z"]
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 5.48 µs per loop

In string an very first character of character set:

In [47]: %%timeit
specials = ['H', 'y', '!']
min(map(lambda x: (string.index(x) if (x in string) else len(string)), specials))
   ....: 
100000 loops, best of 3: 2.02 µs per loop

In [48]: %%timeit
st = {"H","y","!"}
next((i for i, ch in enumerate(s) if ch in st), None)
   ....: 
1000000 loops, best of 3: 903 ns per loop