list = [1, 2, 3, 4] if (list) % 2 == 0 : print(list[ ]) anyone help?

Answer 1

if y=10, else x = 77

Answer 2

Jay in your code I guess you were trying to print the list of all even numbers from 1 to 4. If so then use list comprehension;

list = [a for a in range(1,5) if a%2==0]
print(list)

It should give you what you want.

Answer 3

No it's not fine

Answer 4

list = [1, 2, 3, 4]

if ...........(list) % 2 == 0:

print(list[......])
help please

Answer 5

list = [1, 2, 3, 4]

if
(list) % 2 == 0
:

print(list[
]) how to solve

Answer 6

list = [1, 2, 3, 4]
if len (list) %2 == 0:
print (list[0])

Answer 7

Fill in the blanks to print the first element of the list, if it contains even number of elements.
list = [1, 2, 3, 4]

if
len
(list) % 2 == 0
:

print(list[
0
])

This is the correct answer .
reason : Since the question is asking to print the first element of list if the list is even so we use len(list) in the if statement .

references : Yasin Kokdogan ✌

Answer 8

list = [1, 2, 3, 4]

if
len
(list) % 2 == 0
:

print(list[
0
])