[Type B] Chapter – 9 Class 12 CS – Sumita Arora Assignment | Q/A

Output:
[2, 5, 1, 7, 3, 6, 8, 9, 2, 5, 1, 7, 3, 6, 8, 9]

No, this code will just duplicate and add the list not as the desired purpose in question.

To make the code to work according to the purpose.
SqLst = [2*i for i in NumLst]

SqLst = sorted([2*i for i in NumLst])

The code for above condition
[i for i in ML if(i%2 == 0)]

Output:
[4, 16, 36, 64, 100]

[i for i in ML if(i%2 != 0)]

gen = []
for i in [0, 9, 21, 32]:
  gen.append(i/2)
print(gen)

Output:
(i) False
(ii) True

Solution:

Output:
5 [21, 12] 3
5 [21, 12] 3
[21, 12] 31 2

ages = [11, 14, 15, 17, 13, 18, 25]
  print(ages)
Elig = [x for x in ages if x in range(12, 18)]
print(Elig)

Solution:

Indentation error

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)

Solution:

Output:
[0, 9, 36, 81, 4]
[0, 9, 36, 81, 4]
[0, 9, 36, 81]

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)

Solution:

Output:
[2, 4, 6, 8]

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

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

Output:

(i)
[[9, 6], [4, 5], 10]

(ii)
[[9, 6], [4, 5, 10]]

Solution:

The code will cause error if ch is 2 because then value 100 being absent in Lst1 and Lst2 and its index being asked.

Also, the error will encounter on ch is 3 as Lst2 is empty and underflow case (nothing to pop out) arise.

# corrected code

Lst1 = [23, 34, 12, 77, 34, 26, 28, 93, 48, 69, 73, 23, 19, 88]
Lst2 = []
print("List1 original 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 Lst1.count(100) >0:
    print(Lst1.index(100))
  if Lst2.count(100) >0:
    print(Lst2.index(100))
elif ch == 3:
  print(Lst1.pop())

(i) The if statement is called twice and list comprehension is called in incorrect way
#correct code
[y for y in range(100) if (y % 2 == 0 and y % 5 == 0)]

(ii) The twice if statement
# correct code
(y for y in range(100) if (y % 2 == 0 and y % 5 == 0 ))

#correct code
["good" if i < 3 else "better" for i in range(6)]

No requirement of the colon in list comprehension.

Q15:
(i) The if statement is called twice and list comprehension is called in incorrect way
#correct code
[y for y in range(100) if (y % 2 == 0 and y % 5 == 0)]

(ii) The twice if statement is the error in this case.
# correct code
(y for y in range(100) if (y % 2 == 0 and y % 5 == 0 ))

Q16:
#correct code
["good" if i < 3 else "better" for i in range(6)]

No requirement of the colon in list comprehension.