Skipping every other element after the first

Question

I have the general idea of how to do this in Java, but I am learning Python and not sure how to do it.

I need to implement a function that returns a list containing every other element of the list, starting with the first element.

Thus far, I have and not sure how to do from here since I am just learning how for-loops in Python are different:

def altElement(a):
    b = []
    for i in a:
        b.append(a)

    print b

Answer 1

def altElement(a):
    return a[::2]

Answer 2

Slice notation a[start_index:end_index:step]

return a[::2]

where start_index defaults to 0 and end_index defaults to the len(a).

Answer 3

Alternatively, you could do:

for i in range(0, len(a), 2):
    #do something

The extended slice notation is much more concise, though.

Answer 4

items = range(10)
print items
>>> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
print items[1::2] # every other item after the second; slight variation
>>> [1, 3, 5, 7, 9]
]

Answer 5

b = a[::2]

This uses the extended slice syntax.

Answer 6

There are more ways than one to skin a cat. - Seba Smith

arr = list(range(10)) # Range from 0-9

# List comprehension: Range with conditional
print [arr[index] for index in range(len(arr)) if index % 2 == 0]

# List comprehension: Range with step
print [arr[index] for index in range(0, len(arr), 2)]

# List comprehension: Enumerate with conditional
print [item for index, item in enumerate(arr) if index % 2 == 0]

# List filter: Index in range
print filter(lambda index: index % 2 == 0, range(len(arr)))

# Extended slice
print arr[::2]

Answer 7

Using the for-loop like you have, one way is this:

def altElement(a):
    b = []
    j = False
    for i in a:
        j = not j
        if j:
            b.append(i)

    print b

j just keeps switching between 0 and 1 to keep track of when to append an element to b.