Question 1

Question 5Which of the following is the correct expansion of list_1 = [expr(i) for i in list_0 if func(i)] ?(a)list_1 = []
for i in list_0:
    if func(i):
        list_1.append(i)
(b)for i in list_0:
    if func(i):
        list_1.append(expr(i))
(c)list_1 = []
for i in list_0:
    if func(i):
        list_1.append(expr(i))
(d) none of these

Accepted Answer

list_1 = []
for i in list_0:
    if func(i):
        list_1.append(expr(i))
Reason —for i in list_0 — This part iterates over each element i in the list list_0.if func(i) — This part applies the function func() to each element i and checks if the result is True. Only elements for which func(i) returns True will be included in the resulting list.expr(i) — This part applies some expression or function call expr() to each selected element i.Therefore, the correct expansion would be:list_1 = []
for i in list_0:
    if func(i):
        list_1.append(expr(i))

Question 2

Question 16How are lists internally stored ? How are 2D lists internally stored ?

Accepted Answer

Lists in Python are stored similarly to strings in memory. However, lists store references to their elements at each index, which can vary in size. This means that each index of a list holds a reference to the actual object stored elsewhere in memory. Each of the individual items in the list occupies its own memory location.When it comes to 2D lists, also known as lists of lists, they follow a similar principle. Internally, a 2D list is represented as an array of pointers to inner lists. Each element of the outer list is a reference to an inner list object, which in turn is another array of pointers to the elements it contains.

Question 3

Question 1Create a list SqLst that stores the doubles of elements of another list NumLst. Following code is trying to achieve this. Will this code work as desired ? What will be stored in SqLst after following code ?NumLst = [2, 5, 1, 7, 3, 6, 8, 9]
SqLst = NumLst * 2

Accepted Answer

No, the above code will not work as desired.[2, 5, 1, 7, 3, 6, 8, 9, 2, 5, 1, 7, 3, 6, 8, 9] will be stored in SqLst after the above code is executed. The multiplication operator * with a list NumLst, duplicates the elements of the list NumLst by 2 times.ExplanationThe correct code is:NumLst = [2, 5, 1, 7, 3, 6, 8, 9]
SqLst = []
for num in NumLst:
    SqLst.append(num * 2)
print(SqLst)Output[4, 10, 2, 14, 6, 12, 16, 18]

Question 4

Question 2Change the above code so that it works as stated in previous question.

Accepted Answer

NumLst = [2, 5, 1, 7, 3, 6, 8, 9]
SqLst = []
for num in NumLst:
    SqLst.append(num * 2)
print(SqLst)Output[4, 10, 2, 14, 6, 12, 16, 18]
ExplanationNumLst = [2, 5, 1, 7, 3, 6, 8, 9] — This line creates a list called NumLst containing the given numbers [2, 5, 1, 7, 3, 6, 8, 9].SqLst = [] — This line initializes an empty list called SqLst.for num in NumLst: — This line starts a for loop that iterates over each element num in the NumLst list.SqLst.append(num * 2) — Inside the loop, each num is multiplied by 2, and the result is appended to the SqLst list.print(SqLst) — This line prints the resulting list SqLst.

Question 5

Question 3Modify your previous code so that SqLst stores the doubled numbers in ascending order.

Accepted Answer

NumLst = [2, 5, 1, 7, 3, 6, 8, 9]
SqLst = []
for num in NumLst:
    SqLst.append(num * 2)

n = len(SqLst)
for i in range(n):
    for j in range(0, n-i-1):
        if SqLst[j] > SqLst[j+1]:
            SqLst[j], SqLst[j+1] = SqLst[j+1], SqLst[j]

print(SqLst)Output[2, 4, 6, 10, 12, 14, 16, 18]

Question 6

Question 4Consider a list ML = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]. Write code using a list comprehension that takes the list ML and makes a new list that has only the even elements of this list in it.

Accepted Answer

ML = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
A = [i for i in ML if i % 2 == 0]
print(A)Output[4, 16, 36, 64, 100]
ExplanationML = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] — This line initializes a list ML.A = [i for i in ML if i % 2 == 0] — This line uses a list comprehension to create a new list A. It iterates over each element i in the list ML, and only includes i in A if i % 2 == 0 is true. This condition checks if the element i is even.print(A) — This line prints the list A, which contains only the even numbers from the list ML.

Question 7

Question 6Write equivalent for loop for the following list comprehension :gen = (i/2 for i in [0, 9, 21, 32])
print(gen)

Accepted Answer

In the above code, Python would raise an error because round brackets are used in list comprehension. List comprehensions work with square brackets only.ExplanationThe corrected code is:gen = [i/2 for i in [0, 9, 21, 32]]
print(gen)The equivalent for loop for the above list comprehension is:gen = []
for i in [0, 9, 21, 32]:
    gen.append(i/2)
print(gen)

Question 8

Question 7Predict the output of following code if the input is :(i) 12, 3, 4, 5, 7, 12, 8, 23, 12(ii) 8, 9, 2, 3, 7, 8Code :s = eval(input("Enter a list : "))
n = len(s)
t = s[1:n-1]
print(s[0] == s[n-1] and
t.count(s[0]) == 0)

Accepted Answer

OutputEnter a list : [12, 3, 4, 5, 7, 12, 8, 23, 12]
False
Enter a list : [8, 9, 2, 3, 7, 8]
True
Explanations = eval(input("Enter a list : ")) — This line prompts the user to enter a list.n = len(s) — This line finds the length of the list s.t = s[1:n-1] — This line slices the list s starting from index 1 up to index n-1, which effectively removes the first and last elements from the list s and stores it in list t.print(s[0] == s[n-1] and t.count(s[0]) == 0) — The condition s[0] == s[n-1] and t.count(s[0]) == 0 checks if the first and the last element of the list s are same [s[0] == s[n-1]] and that element does not appear anywhere else in list s apart from the first and last position [t.count(s[0]) == 0]. For case (i), 12 is the first and the last element of list but as it also occurs at the 5th index hence the output is False. For case (ii), 8 is the first and the last element of list and it doesn't appear anywhere else hence the output is True.

Question 9

Question 8Predict the output :def h_t(NLst):                
    from_back = NLst.pop()            
    from_front = NLst.pop(0)
    NLst.append(from_front)
    NLst.insert(0, from_back)
NLst1 = [[21, 12], 31]
NLst3 = NLst1.copy()
NLst2 = NLst1
NLst2[-1] = 5
NLst2.insert(1, 6)
h_t(NLst1)
print(NLst1[0], NLst1[-1], len(NLst1)) 
print(NLst2[0], NLst2[-1], len(NLst2)) 
print(NLst3[0], NLst3[-1], len(NLst3))

Accepted Answer

Output5 [21, 12] 3
5 [21, 12] 3
[21, 12] 31 2
Explanationdef h_t(NLst): — This line defines a function named h_t that takes a single argument NLst.from_back = NLst.pop() — This line removes and returns the last element from the list NLst, storing it in the variable from_back.from_front = NLst.pop(0) — This line removes and returns the first element from the list NLst, storing it in the variable from_front.NLst.append(from_front) — This line appends the value stored in from_front to the end of the list NLst.NLst.insert(0, from_back) — This line inserts the value stored in from_back at the beginning of the list NLst.NLst1 = [[21, 12], 31] — This line initializes NLst1 as a nested list.NLst3 = NLst1.copy() — This line creates a shallow copy of NLst1 and assigns it to NLst3.NLst2 = NLst1 — This line assigns the reference of NLst1 to NLst2, meaning both NLst1 and NLst2 point to the same list object.NLst2[-1] = 5 — This line modifies the last element of NLst2 to 5. Since NLst1 and NLst2 reference the same list object, this change also affects NLst1.NLst2.insert(1, 6) — This line inserts 6 at index 1 in NLst2. Again, since NLst1 and NLst2 reference the same list object, this change also affects NLst1.h_t(NLst1) — This line calls the function h_t with NLst1 as an argument.print(NLst1[0], NLst1[-1], len(NLst1)) — This line prints the first element, last element, and length of NLst1.print(NLst2[0], NLst2[-1], len(NLst2)) — This line prints the first element, last element, and length of NLst2.print(NLst3[0], NLst3[-1], len(NLst3)) — This line prints the first element, last element, and length of NLst3.

Question 10

Question 9Predict the output :ages = [11, 14, 15, 17, 13, 18, 25]
    print(ages)
Elig = [x for x in ages if x in range(14, 18)] 
print(Elig)

Accepted Answer

The above code will raise an error due to an indentation error in line 2.ExplanationThe correct code is:ages = [11, 14, 15, 17, 13, 18, 25]
print(ages)
Elig = [x for x in ages if x in range(14, 18)] 
print(Elig)Output[11, 14, 15, 17, 13, 18, 25]
[14, 15, 17]
ages = [11, 14, 15, 17, 13, 18, 25] — This line initializes a list ages.print(ages) — This line prints the list ages.Elig = [x for x in ages if x in range(14, 18)] — This line uses list comprehension to create new list Elig, by taking the values between 14 to 17 from ages list.print(Elig) — This line prints the list Elig.

Question 11

Question 10Predict the output :L1= [x ** 2 for x in range(10) if x % 3 == 0]
L2 = L1
L1.append(len(L1))       
print(L1)
print(L2)      
L2.remove(len(L2) - 1) 
print(L1)

Accepted Answer

Output[0, 9, 36, 81, 4]
[0, 9, 36, 81, 4]
[0, 9, 36, 81]
ExplanationL1 = [x ** 2 for x in range(10) if x % 3 == 0] — This line creates a list L1 containing the squares of multiples of 3 between 0 to 9.L2 = L1 — This line assigns list L1 to list L2, meaning both L1 and L2 reference the same list in memory (shallow copy).L1.append(len(L1)) — This line appends the length of L1 to L1. So, the length of L1 is 4, and it appends 4 to L1.print(L1) — This prints the modified L1 list.print(L2) — This prints L2, which is pointing to the same list as L1.L2.remove(len(L2) - 1) — This line removes the last element of L2 since len(L2) - 1 gives the last index of L2.print(L1) — This prints L1 again. Since both L1 and L2 reference the same list, modifying L2 also affects L1. Therefore, L1 also reflects the change made to L2.

Question 12

Question 11Predict the output :def even(n):
    return n % 2 == 0
list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9]
ev = [n for n in list1 if n % 2 == 0]
evp = [n for n in list1 if even(n)]
print(evp)

Accepted Answer

Output[2, 4, 6, 8]
Explanationdef even(n) — This line defines a function named even that takes a parameter n.return n % 2 == 0 — This line returns True if the input number n is even (i.e., when n % 2 equals 0), and False otherwise.list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9] — This line initializes a list named list1 with integers from 1 to 9.ev = [n for n in list1 if n % 2 == 0] — This line creates a new list ev using a list comprehension. It iterates over each element n in list1 and includes n in ev if it's even (i.e., if n % 2 == 0).evp = [n for n in list1 if even(n)] — This line creates a new list evp using a list comprehension. It iterates over each element n in list1 and includes n in evp if the function even(n) returns True for that n. Effectively, this also creates a new list from the even numbers of list1. Therefore, both ev and evp will contain same values.print(evp) — This line prints the list evp.

Question 13

Question 12(i)Predict the output.b = [[9, 6], [4, 5], [7, 7]]
x = b[:2] 
x.append(10)
print(x)

Accepted Answer

Output[[9, 6], [4, 5], 10]
Explanationb = [[9, 6], [4, 5], [7, 7]] — This line initializes a list b containing three sublists, each containing two elements.x = b[:2] — This line creates a new list x by slicing the list b from index 0 to index 1. So, x will contain the first two sublists of b. At this point, x will be [[9, 6], [4, 5]].x.append(10) — This line appends the integer 10 to the end of the list x. x now becomes [[9, 6], [4, 5], 10].print(x) — This line prints the list x.

Question 14

Question 12(ii)Predict the output.b = [[9, 6], [4, 5], [7, 7]]
x = b[:2] 
x[1].append(10)
print(x)

Accepted Answer

Output[[9, 6], [4, 5, 10]]
Explanationb = [[9, 6], [4, 5], [7, 7]] — This line initializes a list b containing three sublists, each containing two elements.x = b[:2] — This line creates a new list x by slicing the list b from index 0 to index 1. So, x will contain the first two sublists of b. At this point, x will be [[9, 6], [4, 5]].x[1].append(10) — This line accesses the second sublist of x, which is [4, 5], and appends 10 at its end. Now x becomes [[9, 6], [4, 5, 10]]print(x) — This line prints the list x.

Question 15

Question 13Find the Error. Consider the following code, which runs correctly at times but gives error at other times. Find the error and its reason.Lst1 = [23, 34, 12, 77, 34, 26, 28, 93, 48, 69, 73, 23, 19, 88]
Lst2 = []
print("List1 originally is: ", Lst1) 
ch = int(input("Enter 1/2/3 and \
predict which operation was performed?")) 
if ch == 1:
   Lst1.append(100)
   Lst2.append(100)
elif ch == 2:
   print(Lst1.index(100)) 
   print(Lst2.index(100)) 
elif ch == 3:
   print(Lst1.pop())
   print(Lst2.pop())

Accepted Answer

Lst1 = [23, 34, 12, 77, 34, 26, 28, 93, 48, 69, 73, 23, 19, 88]
Lst2 = []
print("List1 originally is: ", Lst1) 
ch = int(input("Enter 1/2/3 and \
predict which operation was performed?")) 
if ch == 1:
   Lst1.append(100)
   Lst2.append(100)
elif ch == 2:
   print(Lst1.index(100)) # Error 1
   print(Lst2.index(100)) # Error 2
elif ch == 3:
   print(Lst1.pop())
   print(Lst2.pop()) # Error 3When the user selects option 1 (ch == 1), the code works correctly, i.e., it appends 100 to both Lst1 and Lst2 at the end. However, errors occur when the user selects option 2 (ch == 2) or option 3 (ch == 3). The errors are as follows:Error 1 — Attempting to find the index of 100 in the list Lst1 using Lst1.index(100), an error occurs because 100 is not present in Lst1.Error 2 — Attempting to find the index of 100 in the list Lst2 using Lst2.index(100), an error occurs because 100 is not present in Lst2.Error 3 — Attempting to remove an item from an empty list Lst2 using Lst2.pop(), an error occurs because there are no items to remove.

Question 16

Question 14Suggest the correction for the error(s) in previous question's code.

Accepted Answer

For the case of finding the index of 100 in Lst1 and Lst2, we can use conditional checks to verify if 100 is present in each list before attempting to find its index. If 100 is not found, we print an error message.For the case of popping elements from Lst2, we use conditional checks to verify if the list is empty before attempting to pop elements. If a list is empty, we print an error message indicating that popping is not possible.The corrected code is :Lst1 = [23, 34, 12, 77, 34, 26, 28, 93, 48, 69, 73, 23, 19, 88]
Lst2 = []
print("List1 originally is: ", Lst1) 
ch = int(input("Enter 1/2/3 and predict which operation was performed? ")) 
if ch == 1:
    Lst1.append(100)
    Lst2.append(100)
elif ch == 2:
    if 100 in Lst1:
        print("Index of 100 in Lst1:", Lst1.index(100))
    else:
        print("Error: 100 is not in Lst1")
    if 100 in Lst2:
        print("Index of 100 in Lst2:", Lst2.index(100))
    else:
        print("Error: 100 is not in Lst2")
elif ch == 3:
    print(Lst1.pop())
    if Lst2:
        Lst2.pop()
    else:
        print("Error: Cannot pop from an empty list (Lst2)")

Question 17

Question 15(ii)Find the error. Consider the following code and predict the error(s):(y for y in range(100) if y % 2 == 0 and if y % 5 == 0)

Accepted Answer

In the code, round brackets are used for list comprehensions, but list comprehensions work with square brackets only.The syntax error arises from the use of two if statements within the list comprehension. To resolve this error, we should use a single if statement with both conditions combined using the and operator.

Question 18

Question 16Find the error in the following list comprehension :["good" if i < 3: else: "better" for i in range(6)]

Accepted Answer

The code contains a syntax error due to the placement of the colon (:). There should not be colon in list comprehension.

Source: This question is from Data Structures - I : Linear Lists, Computer Science &mdash; Class 12, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Data Structures - I : Linear Lists: Question, Error, Comprehension, Good, Else, Better. These are fundamental topics in Computer Science that students are expected to master as part of the CBSE Class 12 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Computer Science papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberWrite programs with proper indentation and comments.
Trace through your code with sample inputs to verify correctness.
Explain the logic behind each step of your solution.
Familiarise yourself with common library functions and methods.
Related Questions from Data Structures - I : Linear ListsQ1: Question 1 Create a list SqLst that stores the doubles of elements of another list NumLst....
Q10: Question 10 Predict the output : L1 = [ x ** 2 for x in range ( 10 ) if x % 3 == 0 ] L2 = L1 L1 ....
Q11: Question 11 Predict the output : def even ( n ): return n % 2 == 0 list1 = [ 1 , 2 , 3 , 4 , 5 , 6...
Q12: Question 12(i) Predict the output. b = [[ 9 , 6 ], [ 4 , 5 ], [ 7 , 7 ]] x = b [: 2 ] x . append (...
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Practice more questions from Data Structures - I : Linear Lists &mdash; Computer Science, Class 12 CBSE

Question 19

Question 17Suggest corrections for the errors in both the previous questions.

Accepted Answer

(i)y for y in range(100) if y % 2 == 0 and if y % 5 == 0The list comprehension should be enclosed in square brackets, and we should use a single if statement with both conditions combined using the and operator.The corrected code is:[y for y in range(100) if y % 2 == 0 and y % 5 == 0](ii)(y for y in range(100) if y % 2 == 0 and if y % 5 == 0)The list comprehension should be enclosed in square brackets, and we should use a single if statement with both conditions combined using the and operator.The corrected code is:[y for y in range(100) if y % 2 == 0 and y % 5 == 0](iii)["good" if i < 3: else: "better" for i in range(6)]The list comprehension does not include colon.The corrected code is:["good" if i < 3 else "better" for i in range(6)]

Data Structures - I : Linear Lists

Multiple Choice Questions

Type A Short Answer Questionsconceptual Questions

Type B Application Based Questions