Replace First and Last Word of String in the Most Pythonic Way

Question

I am looking for the most pythonic way to replace the first and last word of a string (doing it on a letter basis won't work for various reasons). To demonstrate what I'm trying to do, here is an example.

a = "this is the demonstration sentence."

I'd like the result of my python function to be:

b = "This is the demonstration Sentence."

The tricky part of it is that there might be spaces on the front or the end of the string. I need those to be preserved.

Here's what I mean:

a = " this is a demonstration sentence. "

The result would need to be:

b = " This is a demonstration Sentence. "

Would also be interested in opinions on whether a regex would do this job better than python's inbuilt methods, or vice versa.

Answer 1

import re
a = " this is a demonstration sentence. "
print(re.sub(r'''(?x)      # VERBOSE mode
             (             # 
              ^            # start of string
              \s*          # zero-or-more whitespaces 
              \w           # followed by an alphanumeric character
              )        
             |             # OR
             (
             \w            # an alphanumeric character
             \S*           # zero-or-more non-space characters
             \s*           # zero-or-more whitespaces
             $             # end of string
             )
             ''',
             lambda m: m.group().title(),
             a))

yields

 This is a demonstration Sentence. 

Answer 2

Does this work for you:

In [9]: a = "this is the demonstration sentence."

In [10]: left, _, right = a.strip().partition(' ')

In [11]: mid, _, right = right.rpartition(' ')

In [12]: Left = left.title()

In [13]: Right = right.title()

In [14]: a = a.replace(left, Left, 1).replace(right, Right, 1)

In [15]: a
Out[15]: 'This is the demonstration Sentence.'

Answer 3

Here's a regex solution:

def cap(m):
    return m.group(0).title()

re.sub(r'(?:^\s*\w+)|(?:[^\s]+\s*$)',cap," this is a demonstration sentence. ")
' This is a demonstration Sentence. '

Sorry, that's the best I can do ...

Regex breakdown:

(?:^\s*\w+)    #match (optional) whitespace and then 1 word at the beginning of the string
|              #regex "or"
(?:[^\s]+\s*$) #match a string of non-whitespace characters followed by (optional) whitespace and the end of the line.

Answer 4

Similar to inspectorG4dget, but using .rsplit() giving it the maxsplit argument, and .capitalize() instead.

Note: .split() also accepts an optional maxsplit argument, to split from the left.

>>> a = " this is a demonstration sentence. "
>>> part_one, part_two = a.rsplit(" ", 1)
>>> " ".join([part_one.capitalize(), part_two.capitalize()])
'This is the demonstration Sentence.'

.rsplit() splits the text from the right, where the maxsplit argument tells it how many splits to perform. The value 1 will give you one " split " from the right.

>>> a.rsplit(" ", 1)
['this is the demonstration', 'sentence.']

Answer 5

sentence = " this is a demonstration sentence. "
sentence = sentence.split(' ')  # Split the string where a space occurs

for word in sentence:
    if word:  # If the list item is not whitespace
        sentence[sentence.index(word)] = word.title()
        break  # now that the first word's been replaced, we're done

# get the last word by traversing the sentence backwards
for word in sentence[::-1]:
    if word:
        sentence[sentence.index(word)] = word.title()
        break

final_sentence = ' '.join(sentence)