Python：用包含子字符串的单词拆分字符串

Question

I have a string text = "Fix me a meeting in 2 days" . I have a list of some words meetingStrings . "meet" is there in meetingStrings . So, I have to split the text by meeting.

Desired Output :

in 2 days

meetingStrings = [
    "appointment",
    "meet",
    "interview"
]
text = "Fix me a meeting in 2 days"
for x in meetingStrings:
    if x in text.lower(): 
        txt = text.split(x, 1)[1]
        print(txt)

This gives Output:

ing in 2 days.

Answer 1

Using re.split() :

import re

meetingStrings = [
    "appointment",
    "meet",
    "interview"
]

text = "Fix me a meeting in 2 days"

print(re.split('|'.join(r'(?:\b\w*'+re.escape(w)+r'\w*\b)' for w in meetingStrings), text, 1)[-1])

Prints:

 in 2 days

Answer 2

With a small change to your code:

meetingStrings = [
    "appointment",
    "meet",
    "interview"
]
text = "Fix me a meeting in 2 days"
for x in meetingStrings:
    if x in text.lower():
        txt = text.split(x, 1)[1]
        print(txt.split(" ", 1)[1]) #<--- Here

Just take your final output, and split at the first occurrence of a space

Answer 3

This expression might also work with an i flag:

(?:meet|interview|appointment)\S*\s+((?:in|after)\s[0-9]+\s+(?:days?|months?|weeks?|years?))

and we can include any desired words that we might want in the non-capturing groups using logical ORs, such as:

(?:in|after|on|from)

---

(?:days?|months?|weeks?|years?|hours?)

---

(?:meet|interview|appointment|session|schedule)

Test

import re

regex = r"(?:meet|interview|appointment)\S*\s+((?:in|after)\s[0-9]+\s+(?:days?|months?|weeks?|years?))"
test_str = "Fix me a meeting in 2 days meetings in 2 months meet in 1 week nomeeting in 2 days meet after 2 days"

print(re.findall(regex, test_str, re.IGNORECASE))

Output

['in 2 days', 'in 2 months', 'in 1 week', 'in 2 days', 'after 2 days']

The expression is explained on the top right panel of this demo if you wish to explore/simplify/modify it.

RegEx Circuit

jex.im visualizes regular expressions:

Answer 4

This is for using a search.
All you need to do is put the text in the middle of a word
then match the word.

The result is in capture group 1.

No whitespace trim

\\b\\w*(?:appointment|meet|interview)\\w*\\b(.*)

https://regex101.com/r/lK4zRz/1

Readable version

 \b 
 \w* 
 (?:
      appointment
   |  meet
   |  interview
 )
 \w* 
 \b 
 ( .* )                        # (1)

---

With whitespace trim

(?m)\\b\\w*(?:appointment|meet|interview)\\w*\\b[^\\S\\r\\n]*(.*?)[^\\S\\r\\n]*$

https://regex101.com/r/v2qAOQ/1

---

Additionally, if you add a .* to the beginning of either regex,
it will always get the last keyword.

Answer 5

Try this:

import re
text = "Fix me a meeting in 2 days"
print(re.split("({})\\w*".format("|".join(meetingStrings)), text)[-1].strip())

Outputs: in 2 days

Answer 6

Without Regex, str.partition -ing:

for x in meetingStrings: 
    pre, _, post = text.lower().partition(x) 
    if post: 
        pre = pre.rpartition(' ')[0] if not pre.endswith(' ') else pre.rstrip() 
        post = post.partition(' ')[-1] if not post.startswith(' ') else post.lstrip() 
        print([pre, post]) 

Example:

In [35]: meetingStrings = [ 
    ...:     "appointment", 
    ...:     "meet", 
    ...:     "interview" 
    ...: ] 
    ...: text = "Fix me a meeting in 2 days" 

    ...: for x in meetingStrings: 
    ...:     pre, _, post = text.lower().partition(x) 
    ...:     if post: 
    ...:         pre = pre.rpartition(' ')[0] if not pre.endswith(' ') else pre.rstrip() 
    ...:         post = post.partition(' ')[-1] if not post.startswith(' ') else post.lstrip() 
    ...:         print([pre, post]) 
    ...:                                                                                                                                                                                                    
['fix me a', 'in 2 days']

Answer 7

Try something like this:

import re

meetingStrings = [
        "appointment",
        "meet",
        "interview"
]
text = "Fix me a meeting in 2 days"

def split_string(text, strings):
    search = re.compile('|'.join(strings))
    start = None
    input = text.split()
    for e, x in enumerate(input):
        if search.search(x):
            if start < e:
                yield ' '.join(input[start:e])
            start = None
        else:
            if start is None:
                start = e
    else:
        if start is not None:
            yield ' '.join(input[start:])

print(' '.join(split_string(text, meetingStrings)))

This might be longer, than other answers, but seems to do exactly what you wanted - split on strings, which contain as substring one of strings passed in.

Answer 8

I have an alternative and much simpler approach, first split all the words in your sentence then chop off the sentence from the location that meetingStrings appear:

l=text.split()
for i in meetingStrings:
    for idx, j in enumerate(l):
        if i in j:
            l=l[idx+1:] 
print(' '.join(l))

Gives:

'in 2 days'

Answer 9

you can just use find() and list slice:

text = "Fix me a meeting in 2 days"
meetingStrings = [
    "appointment",
    "meet",
    "interview"
]


sep = [i for i in meetingStrings if i in text]

idx = text.find(sep[0])
idx_ = text[idx:].find(' ')
print (text[idx+idx_:])

output:

in 2 days