# Check if element exists in list in Python

Last Updated : 26 Mar, 2024

The list is an important container in Python as it stores elements of all the data types as a collection. Knowledge of certain list operations is necessary for day-day programming. This article discusses the Fastest way to check if a value exists in a list or not using Python.

Example

`Input: test_list = [1, 6, 3, 5, 3, 4]            3  # Check if 3 exist or not.Output: TrueExplanation: The output is True because the element we are looking is exist in the list.`

### Check if an element exists in the list using the “in” statement

In this method, one easily uses a loop that iterates through all the elements to check the existence of the target element. This is the simplest way to check the existence of the element in the list. Python is the most conventional way to check if an element exists in a list or not. This particular way returns True if an element exists in the list and False if the element does not exist in the list. The list need not be sorted to practice this approach of checking.

Python3 ```lst=[ 1, 6, 3, 5, 3, 4 ] #checking if element 7 is present # in the given list or not i=7 # if element present then return # exist otherwise not exist if i in lst: print("exist") else: print("not exist") ```

Output
```not exist

```

Time Complexity: O(1)
Auxiliary Space: O(n), where n is the total number of elements.

### Find if an element exists in the list using a loop

The given Python code initializes a list named `test_list` with some integer elements. It then iterates through each element in the list using a `for` loop. Inside the loop, it checks if the current element `i` is equal to the value 4 using an `if` statement. If the condition is true, it prints “Element Exists” to the console. The code will output the message if the number 4 is present in the list, and in this case, “Element Exists” will be printed since the number 4 exists in the list `[1, 6, 3, 5, 3, 4]`.

Python3 ```# Initializing list test_list = [1, 6, 3, 5, 3, 4] # Checking if 4 exists in list for i in test_list: if(i == 4): print("Element Exists") ```

Output:

`Element Exists`

Time Complexity: O(n)
Auxiliary Space: O(1)

### Check if an element exists in the list using any() function

It achieves this by utilizing the `any()` function with a generator expression. The generator expression iterates through each element `test_list` and checks if it appears more than once in the list. The result of this check is stored in the variable `result`. Finally, the code prints a message indicating whether there are any duplicate elements, displaying “Does string contain any list element: True” if duplicates exist and “Does string contain any list element: False” if there are no duplicates.

Python3 ```# Initializing list test_list = [1, 6, 3, 5, 3, 4] result = any(item in test_list for item in test_list) print("Does string contain any list element : " +str(bool(result))) ```

Output:

`Does string contain any list element : True`

### Find if an element exists in the list using the count() function

We can use the in-built Python List method, count(), to check if the passed element exists in the List. If the passed element exists in the List, the count() method will show the number of times it occurs in the entire list. If it is a non-zero positive number, it means an element exists in the List. Demonstrating to check the existence of elements in the list using count().

Python3 ```# Initializing list test_list = [10, 15, 20, 7, 46, 2808] print("Checking if 15 exists in list") # number of times element exists in list exist_count = test_list.count(15) # checking if it is more than 0 if exist_count > 0: print("Yes, 15 exists in list") else: print("No, 15 does not exists in list") ```

Output:

`Checking if 15 exists in listYes, 15 exists in list`

### Check if an element exists in the list using sort with bisect_left and set

Converting the list into the set and then using it can possibly be more efficient than only using it. But having efficiency as a plus also has certain negatives. One among them is that the order of the list is not preserved, and if you opt to take a new list for it, you would require to use extra space. Another drawback is that set disallows duplicity and hence duplicate elements would be removed from the original list. In the conventional binary search way of testing element existence, hence list has to be sorted first and hence does not preserve the element ordering. bisect_left() returns the first occurrence of the element to be found and has worked similarly to lower_bound() in C++ STL.

Note: The bisect function will only state the position of where to insert the element but not the details about if the element is present or not.

Demonstrating to check existence of element in list using set() + in and sort() + bisect_left()

Python3 ```from bisect import bisect_left ,bisect # Initializing list test_list_set = [ 1, 6, 3, 5, 3, 4 ] test_list_bisect = [ 1, 6, 3, 5, 3, 4 ] print("Checking if 4 exists in list ( using set() + in) : ") # Checking if 4 exists in list # using set() + in test_list_set = set(test_list_set) if 4 in test_list_set : print ("Element Exists") print("Checking if 4 exists in list ( using sort() + bisect_left() ) : ") # Checking if 4 exists in list # using sort() + bisect_left() test_list_bisect.sort() if bisect_left(test_list_bisect, 4)!=bisect(test_list_bisect, 4): print ("Element Exists") else: print("Element doesnt exist") ```

Output:

`Checking if 4 exists in list ( using set() + in) : Element ExistsChecking if 4 exists in list ( using sort() + bisect_left() ) : Element Exists`

### Check if an element exists in list using find() method

The given Python code checks if the number 15 exists in the list `test_list`. It converts the elements of the list to strings and concatenates them with hyphens. Then, it uses the `find()` method to check if the substring “15” exists in the resulting string. If “15” is found, it prints “Yes, 15 exists in the list”; otherwise, it prints “No, 15 does not exist in the list.”

Python3 ```# Initializing list test_list = [10, 15, 20, 7, 46, 2808] print("Checking if 15 exists in list") x=list(map(str,test_list)) y="-".join(x) if y.find("15") !=-1: print("Yes, 15 exists in list") else: print("No, 15 does not exists in list") ```

Output
```Checking if 15 exists in list
Yes, 15 exists in list

```

### Check if element exists in list using Counter() function

The provided Python code uses the `Counter` class from the `collections` module to calculate the frequency of each element in the `test_list`. It then checks if the frequency of the number 15 is greater than 0. If the frequency is non-zero, it means “15” exists in the list, and the code prints “Yes, 15 exists in the list.” Otherwise, it prints “No, 15 does not exist in the list.” The `Counter` class efficiently counts element occurrences, allowing for a straightforward existence check.

Python3 ```from collections import Counter test_list = [10, 15, 20, 7, 46, 2808] # Calculating frequencies frequency = Counter(test_list) # If the element has frequency greater than 0 # then it exists else it doesn't exist if(frequency[15] > 0): print("Yes, 15 exists in list") else: print("No, 15 does not exists in list") ```

Output
```Yes, 15 exists in list

```

### Find if an an element exists in list using try-except block

One additional approach to check if an element exists in a list is to use the index() method. This method returns the index of the first occurrence of the element in the list or throws a ValueError if the element is not present in the list. To use this method, you can wrap the call to index() in a try-except block to catch the ValueError and return False if it occurs:

Python3 ```def element_exists(lst, element): # Try to get the index of the element in the list try: lst.index(element) # If the element is found, return True return True # If a ValueError is raised, the element is not in the list except ValueError: # Return False in this case return False #Test the function test_list = [1, 6, 3, 5, 3, 4] print(element_exists(test_list, 3)) # prints True print(element_exists(test_list, 7)) # prints False #This code is contributed by Edula Vinay Kumar Reddy ```

Output
```True
False

```

Time complexity: O(n), where n is the length of the list. The index() method iterates through the list to find the element, so the time complexity is linear.
Space complexity: O(1). This approach does not require any additional space.

### Find if an an element exists in list using filter() Function

Step-by-Step Approach

• Define the list my_list and Set element_to_check.
• Use the filter() function to create an iterator (filtered_elements) that contains elements equal to the element_to_check.
• Convert the iterator filtered_elements to a list.
• This step is necessary as the filter() function returns an iterator. The list now contains elements equal to element_to_check.
• Check if the list filtered_list is not empty.
• If the list is not empty, it means the element exists in the original list.
Python ```my_list = [1, 2, 3, 4, 5] element_to_check = 3 # Use filter to create an iterator of elements equal to the target element filtered_elements = filter(lambda x: x == element_to_check, my_list) # Convert the iterator to a list and check if it's not empty if list(filtered_elements): print("Element exists in the list") else: print("Element does not exist in the list") ```

Output
```Element exists in the list
```

Time Complexity: O(n)

Auxiliary Space Complexity: O(n)

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