# What will be the output of the following Python code snippet a=[1, 4, 3, 5, 2] b=[3, 1, 5, 2, 4] a==b set(a)==set(b)

1.  TrueFalse

2.  FalseFalse

3.  FalseTrue

4.  TrueTrue

4

False
True

Explanation :
No Explanation available for this question

# What will be the output of the following Python code snippet l=[1, 2, 4, 5, 2, 'xy', 4] set(l) l

1.  {1, 2, 4, 5, 2, ‘xy’, 4} [1, 2, 4, 5, 2, ‘xy’, 4]

2.  {1, 2, 4, 5, ‘xy’} [1, 2, 4, 5, 2, ‘xy’, 4]

3.  {1, 5, ‘xy’} [1, 5, ‘xy’]

4.  {1, 2, 4, 5, ‘xy’} [1, 2, 4, 5, ‘xy’]

4
```{1, 2, 4, 5, ‘xy’}
[1, 2, 4, 5, 2, ‘xy’, 4]```

Explanation :
No Explanation available for this question

# What will be the output of the following Python code s1={3, 4} s2={1, 2} s3=set() i=0 j=0 for i in s1: for j in s2: s3.add((i,j)) i+=1 j+=1 print(s3)

1.  {(3, 4), (1, 2)}

2.  Error

3.  {(4, 2), (3, 1), (4, 1), (5, 2)}

4.  {(3, 1), (4, 2)}

4

{(4, 2), (3, 1), (4, 1), (5, 2)}

Explanation :
No Explanation available for this question

# The ____________ function removes the first element of a set and the last element of a list.

1.  remove

2.  pop

4.  dispose

4

pop

Explanation :
No Explanation available for this question

# The difference between the functions discard and remove is that:

1.  Discard removes the last element of the set whereas remove removes the first element of the set

2.  Discard throws an error if the specified element is not present in the set whereas remove does not throw an error in case of absence of the specified element

3.  Remove removes the last element of the set whereas discard removes the first element of the set

4.  Remove throws an error if the specified element is not present in the set whereas discard does not throw an error in case of absence of the specified element

4

Remove throws an error if the specified element is not present in the set whereas discard does not throw an error in case of absence of the specified element

Explanation :
No Explanation available for this question

# What will be the output of the following Python code s1={1, 2, 3} s2={3, 4, 5, 6} s1.difference(s2) s2.difference(s1)

1.  {1, 2}{4, 5, 6}

2.  {1, 2}{1, 2}

3.  {4, 5, 6}{1, 2}

4.  {4, 5, 6}{4, 5, 6}

4

{1, 2}
{4, 5, 6}

Explanation :
No Explanation available for this question

# What will be the output of the following Python code s1={1, 2, 3} s2={4, 5, 6} s1.isdisjoint(s2) s2.isdisjoint(s1)

1.  TrueFalse

2.  False True

3.  TrueTrue

4.  FalseFalse

4

True
True

Explanation :
No Explanation available for this question

# If we have two sets, s1 and s2, and we want to check if all the elements of s1 are present in s2 or not, we can use the function:

1.  s2.issubset(s1)

2.  s2.issuperset(s1)

3.  s1.issuperset(s2)

4.  s1.isset(s2)

4

s2.issuperset(s1)

Explanation :
No Explanation available for this question

# What will be the output of the following Python code s1={1, 2, 3, 8} s2={3, 4, 5, 6} s1|s2 s1.union(s2)

1.  {3}{1, 2, 3, 4, 5, 6, 8}

2.  {1, 2, 4, 5, 6, 8}{1, 2, 4, 5, 6, 8}

3.  {3}{3}

4.  {1, 2, 3, 4, 5, 6, 8}{1, 2, 3, 4, 5, 6, 8}

4

{1, 2, 3, 4, 5, 6, 8}
{1, 2, 3, 4, 5, 6, 8}

Explanation :
No Explanation available for this question

# What will be the output of the following Python code a=set('abc') b=set('def') b.intersection_update(a) a b

1.  set()(‘e’, ‘d’, ‘f’}

2.  {}{}

3.  {‘b’, ‘c’, ‘a’}set()

4.  set()set()

4