7 ways to check if an element is in a list in Python
Sometime or other while coding in Python, we surely require to assure ourselves that the value we are searching exists in a list or not in order to continue our code.
Lists being an important Data structure in Python are used in almost every field where python is used.
Thus checking for a value being an important use case, If the list is large, we need to be very careful to search the elements inside of a list to avoid performance issues.
let’s start exploring about the methods to check for a value inside a list and which method is the fastest among them.
Overview of Lists
List contains a group of elements of a different data type separated by commas, and enclosed inside square brackets [ ].
Example of a list:
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
Methods to check if an element exists in a Python List
After having a brief idea of what a list is, now let’s start to discuss about the different methods to check if an element is in a list or not.
Method 1: Using the for loop
The novice method of solving this problem that can be easily adopted by any beginner is by using the for loop.
The mighty for loop helps us to iterate through the whole list and check if our element is present inside the list or not.
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Bombay"
for i in sample_list:
if (i==required_word):
print("Value Found")
Thus we can easily achieve our output , but is this the fastest and most optimized way of doing it ?
Maybe, we will discuss about it in the upcoming sections.
Method 2: Using the in operator
in operator is one of the most valuable operators in Python having a wide range of use case in almost every data structure in python.
It is one of the most useful and liked keyword in python.
in operator returns a bool as result depending upon if the value is exist or not in a list.
How can we check for the presence of a value using the powerful in operator?
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Chennai"
if required_word in sample_list:
print("Value Found")
else:
print("Value Not Found")
As expected , the value we are searching for is not present in our input.
The code gets smaller but, Is this the fastest way ?
Maybe. Let’s try out the next method.
Method 3: Using the not in operator
In the previous section we discussed about the in operator, now we will be using the same operator with a bit of modification in order to improve the runtime performance.
We will use the not in operator to check if the value is present inside of the list.
This operator will return a bool value as well depending upon if the value is not present inside the list or is it.
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Bombay"
if required_word not in sample_list:
print("Value Not Found")
else:
print("Value Found")
The code is very much similar to that of the previous one.
But the runtime performance is a little bit improved.
Wait until we compare all of them !
Method 4: Using sort() and bisect_left()
This method resembles to one of the most conventional methods in programming that is the binary search.
By using this method, we will first sort the list and thus we would not maintain the order of elements and the bisect_left() method returns the first occurrence of an element inside the list.
Let’s try this algo centric approach of solving our problem
from bisect import bisect_left ,bisect
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Madras"
sample_list.sort()
if bisect_left(sample_list, required_word)!=bisect(sample_list, required_word):
print ("Value Found")
else:
print("Value Not Found")
Seems like a complicated code , but is the code is actually complex ?
We will find break down the answer in a few mins.
Method 5: Using lambda function
Instead of searching for an element inside a list, what if we can just filter all elements from inside the list except for the searched element.
But, how can we do that ?
Since , It’s Python that we are working with, inbuilt functions make our task even simpler.
We can use the inbuilt filter() method to achieve this result.
But this method takes in a lambda function as an argument.
This method returns a filter object which we will append inside an empty list and will get all the elements that are left out after filtering.
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Madras"
unique_list = list(filter(lambda word: required_word in word, sample_list))
print(unique_list)
As discussed above, we wrapped the output inside an empty list and here we have the desired result.
But there is a problem.
This method is comparatively slower than all other methods because the filter() constructor is similar to that of a generator function and since, we are converting the result into a list at the end therefore, the runtime performance is degraded.
Method 6: Using count() method
Again since it’s Python, we have a bunch of inbuilt function which can simplify our code and problem quite efficiently.
In this section, we will see the usage of the count() method in order to tackle our problem.
count() method can be used to check if a particular value exists inside a list or not and if yes, it returns a sequence of the occurrences of the element.
If the value is greater than 0, it implies that the searched element exists inside of the list.
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Delhi"
if (sample_list.count(required_word)) > 0:
print("Value Found")
else:
print("Value Not Found")
Optimal solution ? Can be since it uses inbuilt function.
A little more wait !
Method 7: Using the any() method
As we enter the last method of this article, we will have a look over another inbuilt function that can help us to check if a value exist in list or not.
any() method acts like a helper function which would check if the searched element has occurred at least once inside the list and returns a Boolean value accordingly.
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
required_word = "Calcutta"
if (any(word in required_word for word in sample_list)) > 0:
print("Value Found")
else:
print("Value Not Found")
With this method we complete all the methods that can be used to check for a value inside of a list.
In the next section we will discuss the most awaited part, the title itself.
What’s the fastest method to check if a value exists in a list?
Now we will see the fastest method to check if a value exists in a list by comparing their execution speed.
For calculating the time taken by each of the above methods, we will take the help of timeit module.
The timeit module is an inbuilt class in python which has a method called timeit() and the method takes in 4 parameters namely setup, stmt , timer and number.
Let’s then calculate and disclose the suspense:
import timeit
code_starter = """
from bisect import bisect_left ,bisect
required_word = "Bombay"
sample_list = ["Delhi","Bombay","Madras","Kolkata"]
"""
novice_method = """
for i in sample_list:
if (i==required_word):
# print("Value Found")
pass
"""
in_method = """
if required_word in sample_list:
# print("Value Found")
pass
else:
# print("Value Not Found")
pass
"""
notin_method = """
if required_word not in sample_list:
# print("Value Not Found")
pass
else:
# print("Value Found")
pass
"""
sort_method = """
sample_list.sort()
if bisect_left(sample_list, required_word)!=bisect(sample_list, required_word):
# print ("Value Found")
pass
else:
# print("Value Not Found")
pass
"""
lambda_method = """
unique_list = list(filter(lambda word: required_word in word, sample_list))
# print(unique_list)
pass
"""
count_method = """
if (sample_list.count(required_word)) > 0:
# print("Value Found")
pass
else:
# print("Value Not Found")
pass
"""
any_method = """
if (any(word in required_word for word in sample_list)) > 0:
# print("Value Found")
pass
else:
# print("Value Not Found")
pass
"""
print('Time of Code using for loop: ', timeit.timeit(setup = code_starter , stmt= novice_method, number= 10000000))
print('Time of Code using in operator: ', timeit.timeit(setup = code_starter , stmt= in_method, number= 10000000))
print('Time of Code using the not in operator: ', timeit.timeit(setup = code_starter , stmt= notin_method, number= 10000000))
print('Time of Code Using sort + bisect: ', timeit.timeit(setup = code_starter , stmt= sort_method, number= 10000000))
print('Time of Code using lambda function: ', timeit.timeit(setup = code_starter , stmt= lambda_method, number= 10000000))
print('Time of Code using count function: ', timeit.timeit(setup = code_starter , stmt= count_method, number= 10000000))
print('Time of Code using any function: ', timeit.timeit(setup = code_starter , stmt= any_method, number= 10000000))
Output:
Time of Code using for loop: 1.0890936
Time of Code using in operator: 0.29121489999999994
Time of Code using the not in operator: 0.28329629999999995
Time of Code Using sort + bisect: 2.2233091
Time of Code using lambda function: 6.141415800000001
Time of Code using count function: 0.7794974999999997
Time of Code using any function: 8.35378860000128
We have a Winner !
in and Not in operator stands out to be the fastest way to search an element inside a list.
Conclusion
As we near the completion of this article, the conclusion of this would be that, we can use both in and not in operators but it would be a foolish way if we use the lambda function or the any() method since both of them take very long for their completion.
In this article, we first discussed about the various methods of checking for a value inside of a list and then we saw how to determine the fastest method.
I am Dibyanshu Mohanty, Undergraduate Engineering Student at VIT, Vellore with a keen interest in Web and App Development