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

Asked By: seiryuu10

||

Source

Answer 1

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).

Answered By: Darius Bacon

Answer 2

b = a[::2]

This uses the extended slice syntax.

Answered By: John Zwinck

Answer 3

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

Answered By: Muhammad Alkarouri

Answer 4

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.

Answered By: bchurchill

Answer 5

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]
]

Answered By: jdi

Answer 6

Alternatively, you could do:

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

The extended slice notation is much more concise, though.

Answered By: Joel Cornett

Answer 7

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]

Answered By: Mr. Polywhirl

Answer 8

def skip_elements(elements):
  new_list = []
  for index,element in enumerate(elements):
    if index == 0:
      new_list.append(element)
    elif (index % 2) == 0: 
      new_list.append(element)
  return new_list

Also can use for loop + enumerate.
elif (index % 2) == 0: ## Checking if number is even, not odd cause indexing starts from zero not 1.

Answered By: Ievgen Chuchukalo

Answer 9

def skip_elements(elements):
    # Initialize variables
    i = 0
    new_list=elements[::2]
    return new_list

# Should be ['a', 'c', 'e', 'g']:    
print(skip_elements(["a", "b", "c", "d", "e", "f", "g"]))
# Should be ['Orange', 'Strawberry', 'Peach']:
print(skip_elements(['Orange', 'Pineapple', 'Strawberry', 'Kiwi', 'Peach'])) 
# Should be []:
print(skip_elements([]))

Answered By: Jyotsna Jha

Answer 10

Or you can do it like this!

    def skip_elements(elements):
        # Initialize variables
        new_list = []
        i = 0

        # Iterate through the list
        for words in elements:

            # Does this element belong in the resulting list?
            if i <= len(elements):
                # Add this element to the resulting list
                new_list.append(elements[i])
            # Increment i
                i += 2

        return new_list

    print(skip_elements(["a", "b", "c", "d", "e", "f", "g"])) # Should be ['a', 'c', 'e', 'g']
    print(skip_elements(['Orange', 'Pineapple', 'Strawberry', 'Kiwi', 'Peach'])) # Should be ['Orange', 'Strawberry', 'Peach']
    print(skip_elements([])) # Should be []

Answered By: Rifat Hossain

Answer 11

def skip_elements(elements):
   new_list = [ ]
   i = 0
   for element in elements:
       if i%2==0:
         c=elements[i]
         new_list.append(c)
       i+=1
  return new_list

Answered By: Gaya3

Answer 12

def skip_elements(elements):
    # code goes here
    new_list=[]
    for index,alpha in enumerate(elements):
        if index%2==0:
            new_list.append(alpha)
    return new_list


print(skip_elements(["a", "b", "c", "d", "e", "f", "g"]))  
print(skip_elements(['Orange', 'Pineapple', 'Strawberry', 'Kiwi', 'Peach']))

Answered By: TE A 24

Answer 13

def skip_elements(elements):
    # Initialize variables
    new_list = []
    i = 0

    # Iterate through the list
    for words in elements:
        # Does this element belong in the resulting list?
        if i <= len(elements):
            # Add this element to the resulting list
            new_list.insert(i,elements[i])
        # Increment i
        i += 2

    return new_list

Answered By: ree_d_wan

Answer 14

# Initialize variables
new_list = []
i = 0
# Iterate through the list
for i in range(len(elements)):
    # Does this element belong in the resulting list?
    if i%2==0:
        # Add this element to the resulting list
        new_list.append(elements[i])
    # Increment i
    i +=2

return new_list

Answered By: Mohammad Asif

Answer 15

def skip_elements(elements):
# Initialize variables
new_list = []
i = 0

# Iterate through the list
for i in range(0,len(elements),2):
    # Does this element belong in the resulting list?
    if len(elements) > 0:
        # Add this element to the resulting list
        new_list.append(elements[i])        
return new_list

print(skip_elements(["a", "b", "c", "d", "e", "f", "g"])) # Should be ['a', 'c', 'e', 'g']
print(skip_elements(['Orange', 'Pineapple', 'Strawberry', 'Kiwi', 'Peach'])) # Should be ['Orange', 'Strawberry', 'Peach']
print(skip_elements([])) # Should be []

Answered By: Mayank Raj Vaid

Skipping every other element after the first

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