python *args and lists

Question

I have this assignment:

Define a function that takes in an arbitrary number of arguments, and returns a list containing only those arguments that are even. Don't run the function simply provide the definition.

What I've tried:

def myfunc(*args):
    return list(args%2==0)

score 0 · Accepted Answer · answered Feb 10 '20 at 10:53

0

It should be as follows:

def a (*args):
  l = []
  for i in args:
    if i%2 == 0:
      l.append(i)
  return l

answered Feb 10 '20 at 10:53

Puja Neve

50
7

score 0 · Answer 2 · edited Mar 25 '20 at 14:32

0

 def myfunc(*args):
     a=[]
     for num in args:
         if (num%2 == 0):
             a.append(num)
     return a `

edited Mar 25 '20 at 14:32

matthias_h

11,356
9
22
40

answered Mar 25 '20 at 14:01

Ajinkya

1
1

While this may answer the question, it was flagged for review. Answers with no explanation are often considered low-quality. Please provide some commentary in the answer for why this is the correct answer. – Dan Mar 26 '20 at 03:37

score -1 · Answer 3 · answered Feb 10 '20 at 08:58

-1

I think this should work.

def a (*args):
  l = []
  for i in args:
    if i%2 == 0:
      l += [i]
  return l

answered Feb 10 '20 at 08:58

ib4real

84
5

score -1 · Answer 4 · answered Feb 10 '20 at 10:35

Not sure if it is requires to be as arguments like in range(), iba4read's post shuld work just fine. Just in case i'll give you an examplie with a list, note that a ´tuple´ not alwais works as a list.

import random

array = []
for r in range(random.randint(2, 10)):
    array.append(r)


def even_numbers_of(array):
    new_array = []
    for i in array:
        if i % 2 == 0:
            new_array.append(i)
    return new_array


print(even_numbers_of(array))

python *args and lists

4 Answers4