如何匹配字符串中的第一个单词？

Question

I want to match the word 'St' or 'St.' or 'st' or 'st.' BUT only as the first word of a string. For example 'St. Mary Church Church St.' 'St. Mary Church Church St.' - should find ONLY first 'St.' .

• 'st. Mary Church Church St.' - should find ONLY 'st.'
• 'st Mary Church Church St.' - should find ONLY 'st'

I want to eventually replace the first occurrence with 'Saint'.

Answer 1

Regex sub allows you to define the number of occurence to replace in a string.

ie :

>>> import re
>>> s = "St. Mary Church Church St."
>>> new_s = re.sub(r'^(St.|st.|St|st)\s', r'Saint ', s, 1) # the last argument defines the number of occurrences to be replaced. In this case, it will replace the first occurrence only.
>>> new_s
'Saint Mary Church Church St.'
>>> 

Hope it hepls.

Answer 2

You don't need to use a regex for this, just use the split() method on your string to split it by whitespace. This will return a list of every word in your string:

matches = ["St", "St.", "st", "st."]
name = "St. Mary Church Church St."
words = name.split()   #split the string into words into a list
if words [0] in matches:
    words[0] = "Saint"   #replace the first word in the list (St.) with Saint
new_name = "".join([word + " " for word in words]).strip()   #create the new name from the words, separated by spaces and remove the last whitespace
print(new_name)   #Output: "Saint Mary Church Church St."

Answer 3

Thanks for the question! This is exactly what I was looking for to solve my issue. I wanted to share another regex trick I found while hunting around for this answer. You can simply pass the flag paramater into the sub function. This will allow you to reduce the amount of information you need to pass to the pattern paramater in the tool. This makes the code a little cleaner and reduces the chances of you missing a pattern. Cheers!

import re
s = "St. Mary Church Church St."
new_s = re.sub(r'^(st.|st)\s', r'Saint ', s, 1, flags=re.IGNORECASE) # You can shorten the code from above slightly by ignoring the case
new_s
'Saint Mary Church Church St.'

Answer 4

import re

string = "Some text"

replace = {'St': 'Saint', 'St.': 'Saint', 'st': 'Saint', 'st.': 'Saint'}
replace = dict((re.escape(k), v) for k, v in replace.iteritems())
pattern = re.compile("|".join(replace.keys()))
for text in string.split():
    text = pattern.sub(lambda m: replace[re.escape(m.group(0))], text)

This should work I guess, please check. Source

Answer 5

Try using the regex '^\\S+' to match the first non-space character in your string.

import re 

s = 'st Mary Church Church St.'
m = re.match(r'^\S+', s)
m.group()    # 'st'

s = 'st. Mary Church Church St.'
m = re.match(r'^\S+', s)
m.group()    # 'st.'

Answer 6

Python 3.10 introduced a new Structural Pattern Matching feature (otherwise known as match/case ) which can fit this use-case:

s = "St. Mary Church Church St."

words = s.split()
match words:
    case ["St" | "St." | "st" | "st.", *rest]:
        print("Found st at the start")
        words[0] = "Saint"
    case _:
        print("didn't find st at the start")

print(' '.join(words))

Will give:

Found st at the start
Saint Mary Church Church St.

While using s = "Mary Church Church St." will give:

didn't find st at the start
Mary Church Church St.