If else condition is not working from second iteration onwards in python

I am using regex to find whether a keyword is present in a text or not. If it is present, I am extracting all the text AFTER that keyword.

I have a dictionary where I am storing the text as a value. I am iterating through the dictionary values and am using regex to search whether the keyword is present in the text and if yes then extract all the text AFTER that keyword and update the value for that particular key. The issue is that it is not working for the second key value pair. Below is my code:

x = {'text1' : 'whatever text may or may not come here KEYWORD yada yada yada whatever blah',
'text2' : 'this is also KEYWORD another text that might contain yada yada yada blah'}

for k, v in x.items():

    match1 = re.search(r"regex_pattern", v, flags = re.IGNORECASE)
    match2 = re.search(r"regex_pattern", v, flags = re.IGNORECASE)
    match3 = re.search(r"regex_pattern", v, flags = re.IGNORECASE)
    match4 = re.search(r"regex_pattern", v, flags = re.IGNORECASE)

    if match1:
        print('match1 found for ', k)
        key_text1 = v[match1.end():]
    
    elif match2:
        print('match2 found for ', k)
        key_text2 = v[match2.end():]

    elif match3:
        print('match3 found for ', k)
        key_text3 = v[match3.end():]

    elif match4:
        print('match4 found for ', k)
        key_text4 = v[match4.end():]
    
    else:
        print('None of the above matches! found for ', k)


    if key_text1:
        x[k] = key_text1

    elif key_text2:
        x[k] = key_text2

    elif key_text3:
        x[k] = key_text3

    elif key_text4:
        x[k] = key_text4

    else:
        print('nothing!')

The output should look like the following:

x = {'text1' : ' yada yada yada whatever blah', 'text2' : ' another text that might contain yada yada yada blah'}


But for some reason it is not working for the second iteration. I am getting output like this:

x = {'text1' : ' yada yada yada whatever blah', 'text2' : ' yada yada yada whatever blah'}


In the second iteration,it is taking the first value and setting it as the value for the second key. I suspect the issue is that while iterating for the second key value pair, when it encounters the if condition i.e. if key_text1: it was True from the first iteration and hence the first if condition is satisfied. So even if for second value the first condition is not True, since for the first value the condition was True it is taking the first condition to be True for the second value and proeeding.

How can I resolve this?

Your problem is indeed that key_text1 is still valid on your second pass through the loop. The easiest way to deal with this is to make the assigment back to x[k] in the if/else instead of the key_text* variables:

    if match1:
        print('match1 found for ', k)
        x[k] = v[match1.end():]
    
    elif match2:
        print('match2 found for ', k)
        x[k] = v[match2.end():]

    elif match3:
        print('match3 found for ', k)
        x[k] = v[match3.end():]

    elif match4:
        print('match4 found for ', k)
        x[k] = v[match4.end():]
    
    else:
        print('None of the above matches! found for ', k)

Note I would also put your regex's in a list and iterate over that to further simplify the code. For example:

x = {
    'text1': 'whatever text may or may not come here KEYWORD yada yada yada whatever blah',
    'text2': 'this is also KEYWORD2 another text that might conttin yada yada yada blah',
    'text3': 'some text without a key word' 
}
regexes = [r'\bKEYWORD\s+', r'\bKEYWORD1\s+', r'\bKEYWORD2\s+']


for k, v in x.items():
    for regex in regexes:
        match = re.search(regex, v, flags = re.IGNORECASE)
        if match:
            x[k] = v[match.end():]
            break
    else:
        print('No match found for ', k)

This will output No match found for text3 and x will change to:

{
    'text1': 'yada yada yada whatever blah',
    'text2': 'another text that might conttin yada yada yada blah',
    'text3': 'some text without a key word'
}