How can I capitalize the first letter of each word in a string?

Question

s = \'the brown fox\'

...do something here...

s should be:

\'The Brown Fox\'

What\'s the easiest

Answer 1

Just because this sort of thing is fun for me, here are two more solutions.

Split into words, initial-cap each word from the split groups, and rejoin. This will change the white space separating the words into a single white space, no matter what it was.

s = 'the brown fox'
lst = [word[0].upper() + word[1:] for word in s.split()]
s = " ".join(lst)

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EDIT: I don't remember what I was thinking back when I wrote the above code, but there is no need to build an explicit list; we can use a generator expression to do it in lazy fashion. So here is a better solution:

s = 'the brown fox'
s = ' '.join(word[0].upper() + word[1:] for word in s.split())

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Use a regular expression to match the beginning of the string, or white space separating words, plus a single non-whitespace character; use parentheses to mark "match groups". Write a function that takes a match object, and returns the white space match group unchanged and the non-whitespace character match group in upper case. Then use re.sub() to replace the patterns. This one does not have the punctuation problems of the first solution, nor does it redo the white space like my first solution. This one produces the best result.

import re
s = 'the brown fox'

def repl_func(m):
    """process regular expression match groups for word upper-casing problem"""
    return m.group(1) + m.group(2).upper()

s = re.sub("(^|\s)(\S)", repl_func, s)


>>> re.sub("(^|\s)(\S)", repl_func, s)
"They're Bill's Friends From The UK"

I'm glad I researched this answer. I had no idea that re.sub() could take a function! You can do nontrivial processing inside re.sub() to produce the final result!