insensitive - string contains python




Does Python have a string 'contains' substring method? (10)

Does Python have a string contains substring method?

Yes, but Python has a comparison operator that you should use instead, because the language intends its usage, and other programmers will expect you to use it. That keyword is in, which is used as a comparison operator:

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

The opposite (complement), which the original question asks 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__, find, and index

As promised, here's the contains method:

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 private. The only reason to use this is when extending the in and not in functionality (e.g. if subclassing str):

class NoisyString(str):
    def __contains__(self, other):
        print('testing if "{0}" in "{1}"'.format(other, 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

Also, avoid the following string methods:

>>> '**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.

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}

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

I want to do:

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

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

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


Another way to find whether a string contains a few characters or not with the Boolean return value (i.e. True or `False):

str1 = "This be a string"
find_this = "tr"
if find_this in str1:
    print find_this, " is been found in ", str1
else:
    print find_this, " is not found in ", str1

Basically, you want to find a substring in a string in python. There are two ways to search for a substring in a string in Python.

Method 1: in operator

You can use the Python's in operator to check for a substring. It's quite simple and intuitive. It will return True if the substring was found in the string else False.

>>> "King" in "King's landing"
True

>>> "Jon Snow" in "King's landing"
False

Method 2: str.find() method

The second method is to use the str.find() method. Here, we call the .find() method on the string in which substring is to found. We pass the substring to the find() method and check its return value. If its value is other than -1, the substring was found in the string, otherwise not. The value returned is the index where substring was found.

>>> some_string = "valar morghulis"

>>> some_string.find("morghulis")
6

>>> some_string.find("dohaeris")
-1

I would recommend you to use the first method as it is more Pythonic and intuitive.


I see there are already answers but I want to add my two cents as well.

In Python there are functions to do this but the most simple (and mostly preferred) method is to use the keyword in :

"test" in "testtext"
True

"abc" in "abcdefg"
True

"abc" in "Abc"
False

"ABC" in "abc"
False

"abc" in "def"
False

"abc" in ["abc", "def", "ghi"]
True

There are some string methods as well:

"xxabcxx".find("abc")
2 #returns the index of the first match

"xxabcxx".find("cde")
-1 #returns -1 if the substring 
#could not be found in the string

# and:

"xxabcxx".index("abc")
2

"xxabcxx".index("cde")
ValueError: substring not found
#raises ValueError...

About performance:

In general in is the fasted method to find a substring...

find is slightly faster than index


Hope I could help!


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.


If you're looking for case-insensitive search for whole words, rather than a substring contained within another word:

import string

s = 'This is my text example'
if 'is' not in (word.lower() 
    for split_char in string.punctuation + string.whitespace 
    for word in s.split(split_char)):
    # do something

In Python there are two simple ways you can achieve this:

The Pythonic way: Using Python's 'in' Keyword-

in takes two "arguments", one on the left(substring) and one on the right, and returns True if the left argument is contained within the rightside argument and if not,it returns False.

example_string = "This is an example string"
substring = "example"
print(substring in example_string)

Output:

True

The non-Pythonic way: Using Python's str.find:

The find method returns the position of the string within the string or -1 if it's not found. But simply check if the position is not -1.

if example_string.find(substring) != -1:
    print('Substring found!')
else:
    print('Substring not found!')

Output:

Substring found!

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 the in operator:

if "blah" not in somestring: 
    continue




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