How to check if a string contains a substring in Python

1The in operatorRecommended

substring in string is a plain boolean test — no imports, no return code to interpret. It reads like English, so use it directly in an if rather than comparing something to -1.

text = "CompileBytes runs your code in the browser"

if "code" in text:
    print("found it")

print("browser" in text)
print("Python" in text)
print("Code" in text)

Output: found it, True, False, False. That last line is the gotcha — in is case-sensitive, so "Code" misses the lowercase code in the text. Section 3 fixes that. Use not in for the negative check.

2find() and index() — when you need the position

Both return the index of the first occurrence. They differ only in how they report a miss: .find() returns -1, while .index() raises ValueError. Pick .find() when "not there" is a normal outcome and .index() when it means your data is wrong and you want the crash.

text = "CompileBytes runs your code in the browser"

print(text.find("code"))
print(text.find("Python"))

print(text.index("code"))
try:
    text.index("Python")
except ValueError as err:
    print("index raised ValueError:", err)

Output: 23, -1, 23, then index raised ValueError: substring not found. Never write if text.find("x"): — a match at position 0 is falsy, so that silently misses hits at the start. Compare against -1, or just use in. Both methods also take start and end offsets, and .rfind() / .rindex() search from the right.

3Case-insensitive matching

There is no ignorecase flag on in — you normalise both sides yourself. .lower() is enough for ASCII, but .casefold() is the more correct choice: it is an aggressive, matching-oriented fold that handles cases .lower() leaves alone, such as German ß → ss.

text = "CompileBytes Runs Your Code"

print("runs" in text)
print("runs" in text.lower())

# .casefold() is more aggressive than .lower() outside ASCII
title = "Große Straße"
print("GROSSE".lower() in title.lower())
print("GROSSE".casefold() in title.casefold())

Output: False, True, False, True. The third line is .lower() failing for real: "Große".lower() is still große, so grosse isn't in it — .casefold() turns it into grosse and the match succeeds. Fold once outside the loop if you're testing many needles against the same text.

4Patterns with re.search(), edges with startswith()

in matches literal text — "c.de" in text is False even though the regex c.de matches code. When you want a shape (a date, a number, an id), use re.search(), which returns a match object or None. When you only care about the start or the end, the dedicated .startswith() / .endswith() methods say so more precisely — and both accept a tuple of candidates.

import re

log = "ERROR 2026-07-23 disk usage at 91%"

# a pattern, not a fixed substring
print(bool(re.search(r"\d{4}-\d{2}-\d{2}", log)))
print(re.search(r"\d+%", log).group())

# anchored checks — no regex needed
print(log.startswith("ERROR"))
print(log.endswith(("%", "!")))

Output: True, 91%, True, True. Use re.search(), not re.match() — match() only anchors at the start. And if your needle comes from user input, wrap it in re.escape() or characters like . and * will be read as pattern syntax.

5Which should you use?

6Common variation: check many substrings at once

Don't chain or across a dozen checks. Put the needles in a list and feed a generator to any() — it short-circuits on the first hit — or to all() when every term must appear.

text = "the deploy failed with a timeout on the database"
keywords = ["timeout", "panic", "segfault"]

print(any(word in text for word in keywords))
print(all(word in text for word in keywords))

# which ones actually matched?
print([word for word in keywords if word in text])

# first match, or None
print(next((word for word in keywords if word in text), None))

Output: True, False, ['timeout'], timeout. The comprehension gives you every match and next(..., None) gives you the first one without an IndexError on an empty result.