Computer Science
Answer
- For the case of finding the index of 100 in
Lst1
andLst2
, 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)")
Related Questions
Predict the output.
b = [[9, 6], [4, 5], [7, 7]] x = b[:2] x[1].append(10) print(x)
Find 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())
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
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)