Combining zip and enumerate
Sometimes, we might want to know the indices for zipped lists. In that case, we can use
enumerate() on top of zip(). Yet, it’s essential to understand what each of those functions does. Zip returns a tuple for each element pair of list elements. Enumerate returns a tuple for each element of a list. Therefore the resulting structure will be tuple of tuples:a = [1, 2, 3, 4]
b = [5, 6, 7, 8]
zip(a, b) # [(1, 5), (2, 6), (3, 7), (4, 8)]
enumerate(a) # [(0, 1), (1, 2), (2, 3), (3, 4)]
enumerate(b) # [(0, 5), (1, 6), (2, 7), (3, 8)]
enumerate(zip(a, b)) # [(0, (1, 5)), (1, (2, 6)), (2, (3, 7)), (3, (4, 8))]
# Therefore, we need to loop over the elements with tuples:
for i, (item1, item2) in enumerate(zip(a, b)):
print(f'index: {i} => a[i]={item1} and b[i]={item2}')
# index: 0 => a[i]=1 and b[i]=5
# index: 1 => a[i]=2 and b[i]=6
# index: 2 => a[i]=3 and b[i]=7
# index: 3 => a[i]=4 and b[i]=8Notice how
item1 and item2 are wrapped to form a tuple, in the for loop.Challenge
Given 3 lists each on a separate line, you are asked to create a new list from those which will represent the sum of the elements of the 3 lists, while skipping the values at odd locations (indexing starts at 0).
The input contains 3 lines - each one contains numbers separated by a space.
The program should print a single line - the resulting list.
Input | Output |
1 2 3 4
5 6 7 8
9 10 11 12 | 15 21 |
Can you solve this in a single line 😎?
Constraints
Time limit: 2 seconds
Memory limit: 512 MB
Output limit: 1 MB