Check if a word is in a string in Python

Split and match full words

For cases where full word matches are essential, avoid partial matches:

words = phrase.split()  # Splits "phrase" into its atomic word particles!
found = word in words   # Checks if the full "word" is present among the atomic particles! 
print(found)  # True only if 'exact' is an individual word in 'phrase'
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Boundary-aware regex match

To ensure matches respect word boundaries:

import re
pattern = re.compile(r'\b{}\b'.format(word))  # Boundaries defined, no crossing over!
found = bool(pattern.search(phrase))          
print(found)  # True if and only if 'exact' is a completely independent entity in 'phrase'
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Advanced string search techniques

Precise word matching with regex

For fine-tuned full word searches, engage regex boundaries:

import re
found = bool(re.search(r'\b{}\b'.format(re.escape(word)), phrase))  # Boundaries, we respect!
print(found)  # Prints: True. Bingo! 'exact' spotted in 'phrase'
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Ignore case while matching

Don't discriminate based on case; match regardless of case with modified regex patterns:

found = bool(re.search(r'\b{}\b'.format(re.escape(word)), phrase, re.IGNORECASE)) # All are equal!
print(found)  # True for all 'exact' forms - 'Exact', eXacT, EXACT etc.
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Compiled regex: Faster is better

Boost your search speed with precompiled regex patterns for repeated searches:

pattern = re.compile(r'\b{}\b'.format(re.escape(word)), re.IGNORECASE)
found = bool(pattern.search(phrase))  # That was fast, wasn't it?
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Context matters!

Just know, in checks for occurence anywhere in the string:

# Possibly incorrect match
password = 'supersecret'
found = 'secret' in password  # It's not a secret anymore! Returns True
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Dynamic patterns with f-strings

With Python 3.6+, use f-strings for efficient pattern formation:

pattern = re.compile(rf'\b{word}\b')  # 'word' right inside the f-string; how cool is that?
found = bool(pattern.search(phrase))
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Python's lesser-known methods

Ever met str.find()? Another way to seek words:

index = phrase.find(word)
found = index >= 0
print(found)  # True if 'exact' found playing hide and seek in 'phrase'
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But, remember str.find() can be slow compared to in or regex when checking presence.

Particularities of string methods

str.count() might also be useful, with a pinch of salt:

count = phrase.count(word)
print(count > 0)  # True if 'exact' shows up in the 'phrase' party at least once
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Mind you, just like in, str.count() can count partial matches like 'exactly'.

When the "exact" isn't the only goal

At times, a partial match is just what the doctor ordered:

if 'act' in phrase:
    print('Partial match found!')  # 'act' found moonlighting within 'exact'
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Use this with caution, keeping an eye out for unintended broader matches.