I'm learning python and in an exercise I need to write a function that takes an arbitrary number of arguments and returns a list containing only those arguments that are even.
My code is wrong I know: (But what is wrong with this code ?)
def myfunc(*args): for n in args: if n%2 == 0: return list(args) myfunc(1,2,3,4,5,6,7,8,9,10) 8 Answers
Do a list-comprehension which picks elements from args that matches our selection criteria:
def myfunc(*args): return [n for n in args if n%2 == 0] print(myfunc(1,2,3,4,5,6,7,8,9,10)) # [2, 4, 6, 8, 10] This also could be helpful, however, the previous comment looks more advanced:
def myfunc(*args): lista = [] for i in list(args): if not i % 2: lista.append(i) return lista Pick evens
def myfunc(*args): abc = [] for n in args: if n%2==0: abc.append(n) return abc def myfunc(*args): mylist = [] for x in list(args): if x % 2 == 0: mylist.remove(x) return mylist def myfunc(*args): even=[] for n in args: if n %2==0: even.append(n) else: pass return even myfunc(1,2,3,4,8,9) def myfunc(*args): #solution 1 # Create an empty list mylist = [] for number in args: if number %2 == 0: mylist.append(number) else: pass # return the list return mylist #solution 2 # Uses a list comprehension that includes the logic to find all evens and the list comprehension returns a list of those values # return [n for n in args if n%2 == 0] You need to create an empty list to contain the even numbers. Also convert your arguments to a list. Then append the even numbers to the newly created list.
def myfunc(*args): new_list = [] for num in list(args): if num % 2 == 0: new_list.append(num) else: pass return new_list def myfunc(*args): x=[] for i in list(args): if i%2==0: x.append(i) return x 
