Python list is the most widely used data structure, and a good understanding of it is necessary. This Python list exercise aims to help developers learn and practice list operations. All questions are tested on Python 3.
Also Read:
This Python list exercise includes the following: –
The exercise contains 10 questions and solutions provided for each question. You need to solve and practice different list programs, questions, problems, and challenges.
Questions cover the following list topics:
- list operations and manipulations
- list functions
- list slicing
- list comprehension
When you complete each question, you get more familiar with the Python list type. Let us know if you have any alternative solutions in the comment section below.
- Use Online Code Editor to solve exercise questions.
- Read the Complete guide on Python List to solve this exercise.
Table of contents
- Exercise 1: Reverse a list in Python
- Exercise 2: Concatenate two lists index-wise
- Exercise 3: Turn every item of a list into its square
- Exercise 4: Concatenate two lists in the following order
- Exercise 5: Iterate both lists simultaneously
- Exercise 6: Remove empty strings from the list of strings
- Exercise 7: Add new item to list after a specified item
- Exercise 8: Extend nested list by adding the sublist
- Exercise 9: Replace list’s item with new value if found
- Exercise 10: Remove all occurrences of a specific item from a list.
Exercise 1: Reverse a list in Python
Given:
list1 = [100, 200, 300, 400, 500]Expected output:
[500, 400, 300, 200, 100]
Show Hint
Use the list function reverse()
Show Solution
Solution 1: list function reverse()
list1 = [100, 200, 300, 400, 500]
list1.reverse()
print(list1)
Solution 2: Using negative slicing
-1 indicates to start from the last item.
list1 = [100, 200, 300, 400, 500]
list1 = list1[::-1]
print(list1)
Exercise 2: Concatenate two lists index-wise
Write a program to add two lists index-wise. Create a new list that contains the 0th index item from both the list, then the 1st index item, and so on till the last element. any leftover items will get added at the end of the new list.
Given:
list1 = ["M", "na", "i", "Ke"]
list2 = ["y", "me", "s", "lly"]Expected output:
['My', 'name', 'is', 'Kelly']
Show Hint
Use list comprehension with the zip() function
Show Solution
Use the zip() function. This function takes two or more iterables (like list, dict, string), aggregates them in a tuple, and returns it.
list1 = ["M", "na", "i", "Ke"]
list2 = ["y", "me", "s", "lly"]
list3 = [i + j for i, j in zip(list1, list2)]
print(list3)Exercise 3: Turn every item of a list into its square
Given a list of numbers. write a program to turn every item of a list into its square.
Given:
numbers = [1, 2, 3, 4, 5, 6, 7]Expected output:
[1, 4, 9, 16, 25, 36, 49]
Show Hint
Iterate numbers from a list one by one using a for loop and calculate the square of the current number
Show Solution
Solution 1: Using loop and list method
- Create an empty result list
- Iterate a numbers list using a loop
- In each iteration, calculate the square of a current number and add it to the result list using the
append()method.
numbers = [1, 2, 3, 4, 5, 6, 7]
# result list
res = []
for i in numbers:
# calculate square and add to the result list
res.append(i * i)
print(res)
Solution 2: Use list comprehension
numbers = [1, 2, 3, 4, 5, 6, 7]
res = [x * x for x in numbers]
print(res)
Exercise 4: Concatenate two lists in the following order
list1 = ["Hello ", "take "]
list2 = ["Dear", "Sir"]Expected output:
['Hello Dear', 'Hello Sir', 'take Dear', 'take Sir']
Show Hint
Use a list comprehension to iterate two lists using a for loop and concatenate the current item of each list.
Show Solution
list1 = ["Hello ", "take "]
list2 = ["Dear", "Sir"]
res = [x + y for x in list1 for y in list2]
print(res)Exercise 5: Iterate both lists simultaneously
Given a two Python list. Write a program to iterate both lists simultaneously and display items from list1 in original order and items from list2 in reverse order.
Given
list1 = [10, 20, 30, 40]
list2 = [100, 200, 300, 400]Expected output:
10 400 20 300 30 200 40 100
Show Solution
- The
zip()function can take two or more lists, aggregate them in a tuple, and returns it. - Pass the first argument as a
list1and seconds argument as alist2[::-1](reverse list using list slicing) - Iterate the result using a
forloop
list1 = [10, 20, 30, 40]
list2 = [100, 200, 300, 400]
for x, y in zip(list1, list2[::-1]):
print(x, y)Exercise 6: Remove empty strings from the list of strings
list1 = ["Mike", "", "Emma", "Kelly", "", "Brad"]Expected output:
["Mike", "Emma", "Kelly", "Brad"]
Show Hint
Use a filter() function to remove the None / empty type from the list
Show Solution
Use a filter() function to remove None type from the list
list1 = ["Mike", "", "Emma", "Kelly", "", "Brad"]
# remove None from list1 and convert result into list
res = list(filter(None, list1))
print(res)
Exercise 7: Add new item to list after a specified item
Write a program to add item 7000 after 6000 in the following Python List
Given:
list1 = [10, 20, [300, 400, [5000, 6000], 500], 30, 40]Expected output:
[10, 20, [300, 400, [5000, 6000, 7000], 500], 30, 40]
Show Hint
The given list is a nested list. Use indexing to locate the specified item, then use the append() method to add a new item after it.
Show Solution
Use the append() method
list1 = [10, 20, [300, 400, [5000, 6000], 500], 30, 40]
# understand indexing
# list1[0] = 10
# list1[1] = 20
# list1[2] = [300, 400, [5000, 6000], 500]
# list1[2][2] = [5000, 6000]
# list1[2][2][1] = 6000
# solution
list1[2][2].append(7000)
print(list1)
Exercise 8: Extend nested list by adding the sublist
You have given a nested list. Write a program to extend it by adding the sublist ["h", "i", "j"] in such a way that it will look like the following list.
Given List:
list1 = ["a", "b", ["c", ["d", "e", ["f", "g"], "k"], "l"], "m", "n"]
# sub list to add
sub_list = ["h", "i", "j"]Expected Output:
['a', 'b', ['c', ['d', 'e', ['f', 'g', 'h', 'i', 'j'], 'k'], 'l'], 'm', 'n']
Show Hint
The given list is a nested list. Use indexing to locate the specified sublist item, then use the extend() method to add new items after it.
Show Solution
list1 = ["a", "b", ["c", ["d", "e", ["f", "g"], "k"], "l"], "m", "n"]
sub_list = ["h", "i", "j"]
# understand indexing
# list1[2] = ['c', ['d', 'e', ['f', 'g'], 'k'], 'l']
# list1[2][1] = ['d', 'e', ['f', 'g'], 'k']
# list1[2][1][2] = ['f', 'g']
# solution
list1[2][1][2].extend(sub_list)
print(list1)
Exercise 9: Replace list’s item with new value if found
You have given a Python list. Write a program to find value 20 in the list, and if it is present, replace it with 200. Only update the first occurrence of an item.
Given:
list1 = [5, 10, 15, 20, 25, 50, 20]Expected output:
[5, 10, 15, 200, 25, 50, 20]
Show Hint
- Use list method
index(20)to get the index number of a 20 - Next, update the item present at the location using the index number
Show Solution
list1 = [5, 10, 15, 20, 25, 50, 20]
# get the first occurrence index
index = list1.index(20)
# update item present at location
list1[index] = 200
print(list1)Exercise 10: Remove all occurrences of a specific item from a list.
Given a Python list, write a program to remove all occurrences of item 20.
Given:
list1 = [5, 20, 15, 20, 25, 50, 20]Expected output:
[5, 15, 25, 50]
Show Solution
Solution 1: Use the list comprehension
list1 = [5, 20, 15, 20, 25, 50, 20]
# list comprehension
# remove specific items and return a new list
def remove_value(sample_list, val):
return [i for i in sample_list if i != val]
res = remove_value(list1, 20)
print(res)
Solution 2: while loop (slow solution)
list1 = [5, 20, 15, 20, 25, 50, 20]
while 20 in list1:
list1.remove(20)
print(list1)
Exercise number 3
numbers = [1, 2, 3, 4, 5, 6, 7]for index, item in enumerate(numbers):
numbers[index] = item ** 2
print(numbers)
Exercise 1
list1 = [100, 200, 300, 400, 500]increase_position = 0
for ele, value in enumerate(range(0, len(list1))):
last_ele = list1[-1]
list1.remove(list1[-1])
list1.insert(increase_position, last_ele)
increase_position += 1
print(list1)
Exercise 2 – model solution adapted for 2 lists of different lengths
(please indent code as appropriate – tags don’t seem to work on my posts)
I propose this solution based on a functional approach for exercise number three:
‘list’ object is not callable
Question:6
An alternate way of step 1:-
An alternate solution to Exercise 3.
Thanks for this site
exercise 1 in 3 different ways
exercise 6 :
why this wrong?
exercise 3
why make it hard
just do
output = [x*x for x in numbers]its q. 10
Question 3
As a begginer I tried this and it worked also
Sorry bro, but your code does not work.
I changed it to the correct version
Here is it:
ex 10
Exercise 5:
Exercise 4:
exercise 10:
Question 6 with list comprehension
For Question 3 using list comprehension
Thanks, it was a simple and useful exercise 😄️💜️
Hello, For Q3: Turning every item of a list into its square,
Q6
Or you can use list comprehension
it goes like this:
l1 = ["Mike", "", "Emma", "Kelly", "", "Brad"]res = [i for i in l1 if i]
print(res)
Question 2
if l2 length is more then extra element is not added in list
yes try this one
This is a good one.
For Exercise 3: I applied this code, but for some reason on the results, it gives me [1, 16, 9, 4, 25, 36, 49]. Could anyone bother explaining to me why?
The function index() takes as a parameter the value of the item in the list and returns the index. If you add
print()inside the loop you will understand what is going wrong. If you run this:you get the following:
This mean that when your i is 4 your list is
[1, 4, 9, 4, 5, 6, 7]So the
numbers.index(i)returns the position of the second item in the list that is the first 4 right now so [1]. And then it changes is to 16.Exercise 3: Turn every item of a list into its square
Given a list of numbers. write a program to turn every item of a list into its square.
Given:
Question 10
One simplest method.
Question 9
I don’t completely agree with the solution provided, because the question say item 20 if found. what if it’s not found then the solution will throw a “ValueError: Item not in list” Error. I’d go with iterating the list and update the item if found. I would suggest the solution below instead.
Sorry I couldn’t find a way to use the
index()function which is the aim of this exerciseInstead of range() you could use enumerate, to avoid finding the length of the list and an arithmetic operation. Like below,
Correct me if am wrong
Question 1
voooowww! Your articles and exercises are helpful. Thanks alot.
Thank you for the great contents, i have learn a lot.
for problem 10, can we do this:
How to get the output from this list as below :
Hey just wanted to say thanks! These are very helpful for newbies like me 🙂
I am glad it helped you, Jenna.
how can I have your contact
Q10
FOR EXERCISE 8, another different(not so different) approach:
EXERCISE 4:
#my idea was to use nested loops: Each time the i block run, the j block runs 2 times. 2 times because that’s the length of the list and that’s how much we will go through the list.
Another approach for exercise2:
exercise3*** .My bad!
FOR EXERCISE 1:
#my idea was to start counting from the end of the list going down by 1, at the same time,I am adding the value at that position into a new list
exercise6:
ex 10
Is this right for ex 10?
For exercise 5, it is also possible to do it this way:
Thank you for the alternative solution, Leonardo.
For 10
another way:
For Exercise 6 :
For Ex:5 I am not using the Zip function as I have not learned it yet. It a long way, but works
Hi,
for Ex3:
How do I modify this code to get the right o/p?
problem = 3
Hi,
For Ex:2:
this following code id giving me right output. But can somebody tell me the best practice please..
Thanks Hinata 🙂
Thanks khadim. I understood my fault now 🙂
From Naruto Uzumaki
do one thing
Anybody else has trouble not doing things the hard way? Here’s my solution for problem 4. Yours was so much simpler!
Hi there,
I recommend this solution for problem 10 in the list exercise
list1 = [5, 20, 15, 20, 25, 50, 20] while 20 in list1: list1.remove(20)hey dude! how to generate this program??
“Program to generate the sequence: -5,10, -15,20, -25 ——up to n terms”
try this
it won’t give the required list:
enter a number25
5 -5 -15 -25 25
this is how it gives result
ok thanks
hows mine?
Here is my solution in two methods:
method-1:
method-2:
Exercise 10: Given a Python list, remove all occurrence of 20 from the list
Result: [5, 15, 25, 50]
if you have multiple 20s , there will be a logic error
I don’t get a logic error
Code:
Output:
please tell how we decide dimension as in ques 8 list[2][1][2]
Index value for question 8:
I hope this might help u
Problem 1:
Other Approach:- The Below also works
Another solution for Q6
Hi all! this code should remove all negative values from the given list and print a new list with only positive values for some reason it prints an iteration of the given list before the positive list. can I get any hints on how to solve that, please?
you have to add at the end of your def a return statement:
return listOfIntegers
Alternative answer for Q10:
list1 = [5, 20, 15, 20, 25, 50, 20] for x in list1: if x == 20: list1.remove(20) print(list1)Alternative answer for Q10:
One another:
Here’s another
Can anyone please point out the mistake, or is this code correct…..
you must loop over the remove function in order to work
your program is correct.it is running without any error
[5,10,15,300,25,50,20]
Any alternative ans for Q-2
question no.10
Thanks..it really helps:-)
Questions 4:
Any alternate answers for Q-10
Here is an alternate solution for question no.10
Alternative answer to Q9 — Note: would have convert back to a list to satisfy the question requirements.
Thanks Vishal, I am a beginner level python learner and I am doing the exercises , I found these helpful because these cover many methods of data structure and challenges as well. I enjoy solving them.
You’re welcome, Qadeer Rizvi.
sir your article are amazing very helpful for me.keep it up and live long
Thank you, Bostan Khan, for your kind words
list1[2][1][2].extend(subList)i can’t understand the indexing value.please heip me out.
I also had a problem with that. I printed all values in order to understand and it helped me.
Try first print:
list1[0]
then look at
list1[1]
then try
list1[0][0]
then
list1[1][0]
etc.
If you will see it then it will be easier to understand indexing.
can you explain Q4 flow of code not able to understand
its work like nested loop,which mean for each iteration of outer loop the inner loop will be executed completely.keep in mind nested loop and execute it then you guess easily
Good Work Vishal !! Do you run any online classes for Python. I am ready to join in your class
Hey Prashanth, I am not running any classes as of now. I will definitely let you know once I started it.
Hi team,
I’m having trouble understanding the specific append questions where you have to place an item inside a specific list (Q7, Q8). Is there a trick people use to figure out how the indexing associated with the lists work? Playing around, it seems most the index changes result in an error.
Thanks!
Question 7
Question 8
I don’t understand the index of lists. Where they come from? Does someone want to explain?
its like array indexing access ,the point here to be noted each list in the list will be consider an element of the list
Hey Marcin
Just count values in first index, then in second, until you reach the destination index to apply append/extend function.
lets say in Q7.
there are two values in first index before second index initiate so use 2 for that, then again we have 2 values in second index before 3rd and destination index start so use 2 again. Now you can use append function here which is third index.
for question1, can we use the below code?
start = 100 stop = 500 step = 100 stop += step for i in reversed(range(start, stop, step)): print(i, end=" ")which gives the same output as expected, maybe a lengthy code but just want to know whether we can use this method
Thanks in advance
The idea is to use the existing data and you need to supply method. If the data change, for example is has no uniform step interval, then your code fails.
question 5: i think this method is more easy and if this method is not preferred then pls reply
list1 = [10,20,30,40] list2 = [100,200,300,400] l2.reverse() for value in range(len(l1)): print(list1[value],list2[value])Hi, great collection of exercises. For question 7 is there any way to run a loop and extract all the elements while ignoring the sub-lists ?
Hey Noma, Please use the following code
list1 = [10, 20, [300, 400, [5000, 6000], 500], 30, 40] for i in list1: if isinstance(i, list): continue else: print(i)Thanks sir it gave a lot of help to me. Thanks again
Thank you Vishal – This is by far the best website on python for practise,
request to share anylinks or references to read furtheron python topics for begineers
Thank you, Pallavi. Within two weeks you will see new topics for beginners on PYnative
Please tell me the difference between lists and tuples.
lists are mutable whereas tuples are immutable.