Match a whole word in a string using dynamic regex

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Why not use a word boundary?

match_string = r'\b' + word + r'\b'
match_string = r'\b{}\b'.format(word)
match_string = rf'\b{word}\b'          # Python 3.7+ required

If you have a list of words (say, in a words variable) to be matched as a whole word, use

match_string = r'\b(?:{})\b'.format('|'.join(words))
match_string = rf'\b(?:{"|".join(words)})\b'         # Python 3.7+ required

In this case, you will make sure the word is only captured when it is surrounded by non-word characters. Also note that \b matches at the string start and end. So, no use adding 3 alternatives.

Sample code:

import re
strn = "word hereword word, there word"
search = "word"
print re.findall(r"\b" + search + r"\b", strn)

And we found our 3 matches:

['word', 'word', 'word']

NOTE ON “WORD” BOUNDARIES

When the “words” are in fact chunks of any chars you should re.escape them before passing to the regex pattern:

match_string = r'\b{}\b'.format(re.escape(word)) # a single escaped "word" string passed
match_string = r'\b(?:{})\b'.format("|".join(map(re.escape, words))) # words list is escaped
match_string = rf'\b(?:{"|".join(map(re.escape, words))})\b' # Same as above for Python 3.7+

If the words to be matched as whole words may start/end with special characters, \b won’t work, use unambiguous word boundaries:

match_string = r'(?<!\w){}(?!\w)'.format(re.escape(word))
match_string = r'(?<!\w)(?:{})(?!\w)'.format("|".join(map(re.escape, words)))

If the word boundaries are whitespace chars or start/end of string, use whitespace boundaries, (?<!\S)...(?!\S):

match_string = r'(?<!\S){}(?!\S)'.format(word)
match_string = r'(?<!\S)(?:{})(?!\S)'.format("|".join(map(re.escape, words)))

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