Python Searching Tutorial | Searching In Python Tutorial

Searching – All built-in collections in Python implement a way to check element membership using in.. Learn More here.

Searching for an element

All built-in collections in Python implement a way to check element membership using in.

alist = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
5 in alist # True
10 in alist # False


atuple = ('0', '1', '2', '3', '4')
4 in atuple # False
'4' in atuple # True


astring = 'i am a string'
'a' in astring # True
'am' in astring # True
'I' in astring # False
aset = {(10, 10), (20, 20), (30, 30)}
(10, 10) in aset # True
10 in aset # False


dict is a bit special: the normal in only checks the keys. If you want to search in values you need to specify it. The same if you want to search for key-value pairs.

adict = {0: 'a', 1: 'b', 2: 'c', 3: 'd'}
1 in adict # True - implicitly searches in keys
'a' in adict # False
2 in adict.keys() # True - explicitly searches in keys
'a' in adict.values() # True - explicitly searches in values
(0, 'a') in adict.items() # True - explicitly searches key/value pairs
Section 71.2: Searching in custom classes: contains and iter

To allow the use of in for custom classes the class must either provide the magic method contains or, failing that, an iter-method.

Suppose you have a class containing a list of lists:

class ListList:
def init(self, value):
self.value = value

Create a set of all values for fast access

self.setofvalues = set(item for sublist in self.value for item in sublist)
def iter(self):
print('Using iter.')

A generator over all sublist elements

return (item for sublist in self.value for item in sublist)
def contains(self, value):
print('Using contains.')
Just lookup if the value is in the set return value in self.setofvalues
Even without the set you could use the iter method for the contains-check:
return any(item == value for item in iter(self))

Using membership testing is possible using in:

a = ListList([[1,1,1],[0,1,1],[1,5,1]])
10 in a # False

Prints: Using contains.

5 in a # True

Prints: Using contains.

even after deleting the contains method:

del ListList.contains
5 in a # True

Prints: Using iter.

Note: The looping in (as in for i in a) will always use iter even if the class implements a contains method.

Searching: Getting the index for strings: str.index(), str.rindex() and str.find(), str.rfind()

String also have an index method but also more advanced options and the additional str.find. For both of these there is a complementary reversed method.

astring = 'Hello on StackOverflow'
astring.index('o') # 4
astring.rindex('o') # 20
astring.find('o') # 4
astring.rfind('o') # 20

The difference between index/rindex and find/rfind is what happens if the substring is not found in the string:

astring.index(‘q’) # ValueError: substring not found astring.find(‘q’) # -1

All of these methods allow a start and end index:
astring.index('o', 5) # 6
astring.index('o', 6) # 6 - start is inclusive
astring.index('o', 5,7)#6
astring.index('o', 5,6)# - end is not inclusive

ValueError: substring not found

astring.rindex(‘o’, 20) # 20
astring.rindex(‘o’, 19) # 20 – still from left to right

astring.rindex('o', 4, 7) # 6

Searching: Getting the index list and tuples: list.index(), tuple.index()

list and tuple have an index-method to get the position of the element:

alist = [10, 16, 26, 5, 2, 19, 105, 26]
search for 16 in the list alist.index(16) # 1
alist[1] # 16

ValueError: 15 is not in list

But only returns the position of the first found element:

atuple = (10, 16, 26, 5, 2, 19, 105, 26)
atuple.index(26) # 2
atuple[2] # 26
atuple[7] # 26 - is also 26!

Searching key(s) for a value in dict

dict have no builtin method for searching a value or key because dictionaries are unordered. You can create a function that gets the key (or keys) for a specified value:

def getKeysForValue(dictionary, value):
foundkeys = []
for keys in dictionary:
if dictionary[key] == value:
return foundkeys
This could also be written as an equivalent list comprehension:
def getKeysForValueComp(dictionary, value):
return [key for key in dictionary if dictionary[key] == value]

If you only care about one found key:

def getOneKeyForValue(dictionary, value):
return next(key for key in dictionary if dictionary[key] == value)

The first two functions will return a list of all keys that have the specified value:

adict = {'a': 10, 'b': 20, 'c': 10}
getKeysForValue(adict, 10) # ['c', 'a'] - order is random could as well be ['a', 'c']
getKeysForValueComp(adict, 10) # ['c', 'a'] - dito
getKeysForValueComp(adict, 20) # ['b']
getKeysForValueComp(adict, 25) # []

The other one will only return one key:

getOneKeyForValue(adict, 10) # 'c' - depending on the circumstances this could also be 'a'
getOneKeyForValue(adict, 20) # 'b'

and raise a StopIteration-Exception if the value is not in the dict:

getOneKeyForValue(adict, 25)

Searching: Getting the index for sorted sequences:

Sorted sequences allow the use of faster searching algorithms: bisect.bisect_left()1:

import bisect
def index_sorted(sorted_seq, value):
"""Locate the leftmost value exactly equal to x or raise a ValueError"""
i = bisect.bisect_left(sorted_seq, value)
if i != len(sorted_seq) and sorted_seq[i] == value:
return i
raise ValueError
alist = [i for i in range(1, 100000, 3)] # Sorted list from 1 to 100000 with step 3
index_sorted(alist, 97285) # 32428
index_sorted(alist, 4) # 1
index_sorted(alist, 97286)


For very large sorted sequences the speed gain can be quite high. In case for the first search approximately 500 times as fast:

%timeit index_sorted(alist, 97285)
100000 loops, best of 3: 3 µs per loop %timeit alist.index(97285)
1000 loops, best of 3: 1.58 ms per loop

While it’s a bit slower if the element is one of the very first:

%timeit index_sorted(alist, 4)
100000 loops, best of 3: 2.98 µs per loop %timeit alist.index(4)
1000000 loops, best of 3: 580 ns per loop

Searching: Searching nested sequences

Searching in nested sequences like a list of tuple requires an approach like searching the keys for values in dict but needs customized functions.

The index of the outermost sequence if the value was found in the sequence: – Python® Notes for Professionals 360

def outer_index(nested_sequence, value):
return next(index for index, inner in enumerate(nested_sequence)
for item in inner
if item == value)
alist_of_tuples = [(4, 5, 6), (3, 1, 'a'), (7, 0, 4.3)] outer_index(alist_of_tuples, 'a') # 1 outer_index(alist_of_tuples, 4.3) # 2

or the index of the outer and inner sequence:

def outer_inner_index(nested_sequence, value):
return next((oindex, iindex) for oindex, inner in enumerate(nested_sequence)
for iindex, item in enumerate(inner)
if item == value)
outer_inner_index(alist_of_tuples, 'a') # (1, 2)
alist_of_tuples[1][2] # 'a'
outer_inner_index(alist_of_tuples, 7) # (2, 0)
alist_of_tuples[2][0] # 7
In general (not always) using next and a generator expression with conditions to find the first occurrence of the searched value is the most efficient approach.

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