Case-insensitive String Startswith In Python

Answer :

You could use a regular expression as follows:

In [33]: bool(re.match('he', 'Hello', re.I)) Out[33]: True   In [34]: bool(re.match('el', 'Hello', re.I)) Out[34]: False  

On a 2000-character string this is about 20x times faster than lower():

In [38]: s = 'A' * 2000  In [39]: %timeit s.lower().startswith('he') 10000 loops, best of 3: 41.3 us per loop  In [40]: %timeit bool(re.match('el', s, re.I)) 100000 loops, best of 3: 2.06 us per loop 

If you are matching the same prefix repeatedly, pre-compiling the regex can make a large difference:

In [41]: p = re.compile('he', re.I)  In [42]: %timeit p.match(s) 1000000 loops, best of 3: 351 ns per loop 

For short prefixes, slicing the prefix out of the string before converting it to lowercase could be even faster:

In [43]: %timeit s[:2].lower() == 'he' 1000000 loops, best of 3: 287 ns per loop 

Relative timings of these approaches will of course depend on the length of the prefix. On my machine the breakeven point seems to be about six characters, which is when the pre-compiled regex becomes the fastest method.

In my experiments, checking every character separately could be even faster:

In [44]: %timeit (s[0] == 'h' or s[0] == 'H') and (s[1] == 'e' or s[1] == 'E') 1000000 loops, best of 3: 189 ns per loop 

However, this method only works for prefixes that are known when you're writing the code, and doesn't lend itself to longer prefixes.

How about this:

prefix = 'he' if myVeryLongStr[:len(prefix)].lower() == prefix.lower() 

Another simple solution is to pass a tuple to startswith() for all the cases needed to match e.g. .startswith(('case1', 'case2', ..)).

For example:

>>> 'Hello'.startswith(('He', 'HE')) True >>> 'HEllo'.startswith(('He', 'HE')) True >>>