Write a Python program to check whether the n th element exists in a given list

Python: Check whether the n-th element exists in a given list

Last update on October 08 2020 09:22:26 (UTC/GMT +8 hours)

Python | Check for Nth index existence in list

Sometimes, while working with lists, we can have a problem in which we require to insert a particular element at an index. But, before that it is essential to know that particular index is part of list or not. Let’s discuss certain shorthands that can perform this task error free.

Method #1 : Using len()
This task can be performed easily by finding the length of list using len(). We can check if the desired index is smaller than length which would prove it’s existence.




# Python3 code to demonstrate working of
# Check for Nth index existence in list
# Using len()
# initializing list
test_list = [4, 5, 6, 7, 10]
# printing original list
print("The original list is : " + str(test_list))
# initializing N
N = 6
# Check for Nth index existence in list
# Using len()
res = len(test_list) >= N
# printing result
print("Is Nth index available? : " + str(res))
Output : The original list is : [4, 5, 6, 7, 10] Is Nth index available? : False

Method #2 : Using try-except block + IndexError exception
This task can also be solved using the try except block which raises a IndexError exception if we try to access an index not a part of list i.e out of bound.




# Python3 code to demonstrate working of
# Check for Nth index existence in list
# Using try-except block + IndexError exception
# initializing list
test_list = [4, 5, 6, 7, 10]
# printing original list
print("The original list is : " + str(test_list))
# initializing N
N = 6
# Check for Nth index existence in list
# Using try-except block + IndexError exception
try:
val = test_list[N]
res = True
except IndexError:
res = False
# printing result
print("Is Nth index available? : " + str(res))
Output : The original list is : [4, 5, 6, 7, 10] Is Nth index available? : False

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Article Tags :
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Python Programs
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Check if element exists in list in Python

List is an important container in python as if stores elements of all the datatypes as a collection. Knowledge of certain list operations is necessary for day-day programming. This article discusses one of the basic list operations of ways to check the existence of elements in the list.

Method 1: Naive Method

In Naive 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.

Method 2: Using in

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. List need not be sorted to practice this approach of checking.

Code #1: Demonstrating to check the existence of an element in the list using the Naive method and in.

Python3




# Python code to demonstrate
# checking of element existence
# using loops and in
# Initializing list
test_list = [ 1, 6, 3, 5, 3, 4 ]
print("Checking if 4 exists in list ( using loop ) : ")
# Checking if 4 exists in list
# using loop
for i in test_list:
if(i == 4) :
print ("Element Exists")
print("Checking if 4 exists in list ( using in ) : ")
# Checking if 4 exists in list
# using in
if (4 in test_list):
print ("Element Exists")

Output :



Checking if 4 exists in list ( using loop ) : Element Exists Checking if 4 exists in list ( using in ) : Element Exists

Method 3 : Using set() + in

Converting the list into the set and then using in can possibly be more efficient than only using in. But having efficiency for a plus also has certain negatives. One among them is that the order of 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.

Method 4 : Using sort() + bisect_left()

The conventional binary search way of testing element existence, hence list has to be sorted first and hence not preserving 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.

Code #2 : Demonstrating to check existence of element in list using set() + in and sort() + bisect_left().

Python3




# Python code to demonstrate
# checking of element existence
# using set() + in
# using sort() + bisect_left()
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")

Method 5 : Using count()

We can use the in-built python List method, count(), to check if the passed element exists in List. If the passed element exists in the List, 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.

Code #3 : Demonstrating to check the existence of elements in the list using count().

Python3




"""
Python code to demonstrate
checking of element existence
using List count() method
"""
# 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 then 0
if exist_count > 0:
print("Yes, 15 exists in list")
else:
print("No, 15 does not exists in list")

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Article Tags :
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Practice Tags :
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Python Program to check if element exists in list

In this tutorial, we will learn how to check if an element is present in a list or not. List is an ordered set of values enclosed in square brackets [ ]. List stores some values called elements in it, which can be accessed by their particular index. We will be discussing the various approaches by which we can check for an element in the given list.

The program will accept the list and the element which has to check, it should return True if it exists else, it will return False

Input: [6, 22, 7, 3, 0, 11] element= 0

Output: True

Let us look at the different approaches we can follow to execute the task.

(Python Example for Citizen Data Scientist & Business Analyst)

Write a Python program to check whether the n-th element exists in a given list.

Sample Solution

Python Code:

x = [1, 2, 3, 4, 5, 6] xlen = len(x)-1 print(x[xlen])

Sample Output:

6

How to Check if Element Exists in List in Python

The list is a crucial container in Python because it is able to store elements of all types of data as a collection. Knowing specific list operations is essential to be able to program day-to-day. This article will discuss one of the most fundamental list operations, which is to verify the existence of the elements on the list.

Method 1: Naive Method

In the Naive approach, one employs a loop to iterate through the entire list of elements to verify whether the element is a target element. This is the most efficient method of determining the presence of the element on the list.

Method 2: Using in

Python is the most common method of determining whether an element is present in a list. This method returns True if the element is present in the list and False if it is not within the set. The list does not need to be sorted in order to use this method of checking.

Code #1: A demonstration of how you can check whether an element is in the list by using the Naive method as well as in.

Output:

Check if 5 exists in list ( using loop ) : Given element does not exists Check if 5 exists in list ( using in ) : Given element Exists

Method 3 : Using set() + in

Converting the list to the set and then using the in may make more sense than just making use of in. However, efficiency as a plus has some drawbacks. One of these concerns is that order in which the list is displayed isn't maintained. If you choose to make a list and make a new one, you'll need to make use of additional space. Another issue is that set doesn't allow duplicates, and therefore duplicate elements will be eliminated from the list.

Method 4 : Using sort() + bisect_left()

The traditional binary search method to determine the existence of elements means that list must be sorted first, thus not keeping the order of elements. Bisect_left() will return the very first instance of an element found. It works similarly with lower_bound() in C++ STL.

Note: The bisect function will only give the location of the element but not provide details of whether the element is there or not.

Code #2: Demonstrating to check existence of element in list using set() + in and sort() + bisect_left().

Output:

Checking if 5 exists in list (using set() + in): Given element does not exists Checking if 7 exists in list (using sort() + bisect_left()): Given element does not exists

Method 5 : Using count()

We can make use of the built-in Python List method count() to verify whether the element passed is present in the List. If the element that we passed is in the List, then the count() technique will reveal how many instances it is present throughout the list. If it's a positive number, that means that an element exists within the List.

Code #3: Explaining how to verify the presence of elements on the list by counting().

Output:

Checking if 5 exists in list: Yes, 5 exists in list Checking if 46 exists in list: Yes, 46 exists in list

How to check if a list is empty in Python?

In this short tutorial, find how to check if a list is empty in Python. We also look at why you need to do this so that you understand the purpose better.

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