Use the in operator:

if "blah" not in somestring: 
    continue

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.

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.

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.

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.

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.

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

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

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

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', ' ', ' ']