Does Python have a string 'contains' substring method?

Question

I'm looking for a string.contains or string.indexof method in Python.

I want to do:

if not somestring.contains("blah"):
   continue

Surpass

Answer 1

Use the in operator:

if "blah" not in somestring: 
    continue

Note: This is case-sensitive.

Insufferable

Note: This is case-sensitive. – Washstand

Answer 2

If it's just a substring search you can use string.find("substring").

You do have to be a little careful with find, index, and in though, as they are substring searches. In other words, this:

s = "This be a string"
if s.find("is") == -1:
    print("No 'is' here!")
else:
    print("Found 'is' in the string.")

It would print Found 'is' in the string. Similarly, if "is" in s: would evaluate to True. This may or may not be what you want.

Swayne

+1 for highlighting the gotchas involved in substring searches. the obvious solution is if ' is ' in s: which will return False as is (probably) expected. – Prefix

@Prefix Obvious it may be, but not entirely correct. What if you have punctuation or it's at the start or end? What about capitalisation? Better would be a case insensitive regex search for \bis\b (word boundaries). – Pareu

Why would this not be what the OP wants – Ruling

@Ruling The problem with substring searches is that, in this example, you're looking for the word is inside "This be a string." That will evaluate to True because of the is in This. This is bad for programs that search for words, like swear filters (for example, a dumb word check for "ass" would also catch "grass"). – Banbury

Answer 3

Does Python have a string contains substring method?

99% of use cases will be covered using the keyword, in, which returns True or False:

'substring' in any_string

For the use case of getting the index, use str.find (which returns -1 on failure, and has optional positional arguments):

start = 0
stop = len(any_string)
any_string.find('substring', start, stop)

or str.index (like find but raises ValueError on failure):

start = 100 
end = 1000
any_string.index('substring', start, end)

Explanation

Use the in comparison operator because

1. the language intends its usage, and
2. other Python programmers will expect you to use it.

>>> 'foo' in '**foo**'
True

The opposite (complement), which the original question asked for, is not in:

>>> 'foo' not in '**foo**' # returns False
False

This is semantically the same as not 'foo' in '**foo**' but it's much more readable and explicitly provided for in the language as a readability improvement.

Avoid using __contains__

The "contains" method implements the behavior for in. This example,

str.__contains__('**foo**', 'foo')

returns True. You could also call this function from the instance of the superstring:

'**foo**'.__contains__('foo')

But don't. Methods that start with underscores are considered semantically non-public. The only reason to use this is when implementing or extending the in and not in functionality (e.g. if subclassing str):

class NoisyString(str):
    def __contains__(self, other):
        print(f'testing if "{other}" in "{self}"')
        return super(NoisyString, self).__contains__(other)

ns = NoisyString('a string with a substring inside')

and now:

>>> 'substring' in ns
testing if "substring" in "a string with a substring inside"
True

Don't use find and index to test for "contains"

Don't use the following string methods to test for "contains":

>>> '**foo**'.index('foo')
2
>>> '**foo**'.find('foo')
2

>>> '**oo**'.find('foo')
-1
>>> '**oo**'.index('foo')

Traceback (most recent call last):
  File "<pyshell#40>", line 1, in <module>
    '**oo**'.index('foo')
ValueError: substring not found

Other languages may have no methods to directly test for substrings, and so you would have to use these types of methods, but with Python, it is much more efficient to use the in comparison operator.

Also, these are not drop-in replacements for in. You may have to handle the exception or -1 cases, and if they return 0 (because they found the substring at the beginning) the boolean interpretation is False instead of True.

If you really mean not any_string.startswith(substring) then say it.

Performance comparisons

We can compare various ways of accomplishing the same goal.

import timeit

def in_(s, other):
    return other in s

def contains(s, other):
    return s.__contains__(other)

def find(s, other):
    return s.find(other) != -1

def index(s, other):
    try:
        s.index(other)
    except ValueError:
        return False
    else:
        return True



perf_dict = {
'in:True': min(timeit.repeat(lambda: in_('superstring', 'str'))),
'in:False': min(timeit.repeat(lambda: in_('superstring', 'not'))),
'__contains__:True': min(timeit.repeat(lambda: contains('superstring', 'str'))),
'__contains__:False': min(timeit.repeat(lambda: contains('superstring', 'not'))),
'find:True': min(timeit.repeat(lambda: find('superstring', 'str'))),
'find:False': min(timeit.repeat(lambda: find('superstring', 'not'))),
'index:True': min(timeit.repeat(lambda: index('superstring', 'str'))),
'index:False': min(timeit.repeat(lambda: index('superstring', 'not'))),
}

And now we see that using in is much faster than the others. Less time to do an equivalent operation is better:

>>> perf_dict
{'in:True': 0.16450627865128808,
 'in:False': 0.1609668098178645,
 '__contains__:True': 0.24355481654697542,
 '__contains__:False': 0.24382793854783813,
 'find:True': 0.3067379407923454,
 'find:False': 0.29860888058124146,
 'index:True': 0.29647137792585454,
 'index:False': 0.5502287584545229}

How can in be faster than __contains__ if in uses __contains__?

This is a fine follow-on question.

Let's disassemble functions with the methods of interest:

>>> from dis import dis
>>> dis(lambda: 'a' in 'b')
  1           0 LOAD_CONST               1 ('a')
              2 LOAD_CONST               2 ('b')
              4 COMPARE_OP               6 (in)
              6 RETURN_VALUE
>>> dis(lambda: 'b'.__contains__('a'))
  1           0 LOAD_CONST               1 ('b')
              2 LOAD_METHOD              0 (__contains__)
              4 LOAD_CONST               2 ('a')
              6 CALL_METHOD              1
              8 RETURN_VALUE

so we see that the .__contains__ method has to be separately looked up and then called from the Python virtual machine - this should adequately explain the difference.

Unremitting

Why should one avoid str.index and str.find? How else would you suggest someone find the index of a substring instead of just whether it exists or not? (or did you mean avoid using them in place of contains - so don't use s.find(ss) != -1 instead of ss in s?) – Sluff

Precisely so, although the intent behind the use of those methods may be better addressed by elegant use of the re module. I have not yet found a use for str.index or str.find myself in any code I have written yet. – Unremitting

Please extend your answer to advice against using str.count as well (string.count(something) != 0). shudder – Soapsuds

How does the operator module version perform? – Sisal

@Sisal it's the same as in_ above - but with a stackframe around it, so it's slower than that: github.com/python/cpython/blob/3.7/Lib/operator.py#L153 – Unremitting

I would just add that __contains__ can also be used for custom (non-string) classes for implementing some cools behaviors. For instance, you could create a Pizza class and create a __contains__ method to allow things like if Ingredient("pepper") in Pizza("margherita") with some internal checking. It could also be used for subclassing lists, or any iterable element (for instance, to check if a specific date is included in a date range). – Frit

Answer 4

if needle in haystack: is the normal use, as @Michael says -- it relies on the in operator, more readable and faster than a method call.

If you truly need a method instead of an operator (e.g. to do some weird key= for a very peculiar sort...?), that would be 'haystack'.__contains__. But since your example is for use in an if, I guess you don't really mean what you say;-). It's not good form (nor readable, nor efficient) to use special methods directly -- they're meant to be used, instead, through the operators and builtins that delegate to them.

Donor

How much faster than a method call? – Intaglio

Answer 5

in Python strings and lists

Here are a few useful examples that speak for themselves concerning the in method:

>>> "foo" in "foobar"
True
>>> "foo" in "Foobar"
False
>>> "foo" in "Foobar".lower()
True
>>> "foo".capitalize() in "Foobar"
True
>>> "foo" in ["bar", "foo", "foobar"]
True
>>> "foo" in ["fo", "o", "foobar"]
False
>>> ["foo" in a for a in ["fo", "o", "foobar"]]
[False, False, True]

Caveat. Lists are iterables, and the in method acts on iterables, not just strings.

If you want to compare strings in a more fuzzy way to measure how "alike" they are, consider using the Levenshtein package

Here's an answer that shows how it works.

Yogurt

Answer 6

If you are happy with "blah" in somestring but want it to be a function/method call, you can probably do this

import operator

if not operator.contains(somestring, "blah"):
    continue

All operators in Python can be more or less found in the operator module including in.

Astronautics

Answer 7

So apparently there is nothing similar for vector-wise comparison. An obvious Python way to do so would be:

names = ['bob', 'john', 'mike']
any(st in 'bob and john' for st in names) 
>> True

any(st in 'mary and jane' for st in names) 
>> False

Duda

That's because there is a bajillion ways of creating a Product from atomic variables. You can stuff them in a tuple, a list (which are forms of Cartesian Products and come with an implied order), or they can be named properties of a class (no a priori order) or dictionary values, or they can be files in a directory, or whatever. Whenever you can uniquely identify (iter or getitem) something in a 'container' or 'context', you can see that 'container' as a sort of vector and define binary ops on it. en.wikipedia.org/wiki/… – Aureaaureate

Worth nothing that in should not be used with lists because it does a linear scan of the elements and is slow compared. Use a set instead, especially if membership tests are to be done repeatedly. – Soapsuds

Answer 8

You can use y.count().

It will return the integer value of the number of times a substring appears in a string.

For example:

string.count("bah")   # gives 0
string.count("Hello") # gives 1

Dantedanton

counting a string is costly when you just want to check if it's there... – Sandpaper

I agree, I had an in depth answer that proposed 3 possible solutions. but this was changed by Jean-Francois Fabre to be what it currently is. Not sure why he would change it so. – Dantedanton

Shifting right is almost certainly not what you want to do here. – Poussette

This is just not the answer to the question. It is by no means an idiomatic way to find out if a string is in another string – Thinskinned

@rsandwick3: I think OP only means ">>" to denote "gives the result", like "->" or "==>". I edited to clarify. – Sever

Answer 9

Here is your answer:

if "insert_char_or_string_here" in "insert_string_to_search_here":
    #DOSTUFF

For checking if it is false:

if not "insert_char_or_string_here" in "insert_string_to_search_here":
    #DOSTUFF

OR:

if "insert_char_or_string_here" not in "insert_string_to_search_here":
    #DOSTUFF

Taliped

PEP 8 prefers "if x not in y" to "if not x in y". – Shamrao

Answer 10

You can use regular expressions to get the occurrences:

>>> import re
>>> print(re.findall(r'( |t)', to_search_in)) # searches for t or space
['t', ' ', 't', ' ', ' ']

Amorphous

It is actually less efficient in terms of time complexity. You are better off using the in operator. But it's a fun solution. In case you insist of using re, re.match is better to use as boolean. – Catalpa