Return more than one value in python function

问题

I was trying to use *args with a for loop in python, however I don't see how to return all the values passed to the function, which should return all the even numbers

def f_even(*args):
    for item in args:
        if item%2 == 0:
            return item

The above code returns only the first value, as I guess after the return it goes out of the function. Indeed, if I use print instead, it works

I'm trying to find a way to return a tuple with all the even numbers when I pass let's say (1,2,3,4,5) to the function

Thank you!

回答1:

In python you can use list comprehension to do this. will make you code more readable and will shrink it too.

def f_even(*args):
   return [elem for elem in args if elem % 2 == 0]

回答2:

You could slightly modify you function and make it a generator by using yield. This way your function wont end after returning first even number but will keep yielding them one by one.

def f_even(*args):
    for item in args:
          if item%2 == 0:
             yield item

for i in f_even(1,2,3,4,5):
    print(i)

Output:

2
4

---

Or if you want to store all yielded values:

even_numbers = list(f_even(1,2,3,4,5))
print(even_numbers) # -> [2, 4]

回答3:

Done, thank you all!!

def f_even(*args): mylist = [] for item in args: if item%2 == 0: mylist.append(item) return mylist

来源：https://stackoverflow.com/questions/53685897/return-more-than-one-value-in-python-function