If all elements match requirement not using "if all"

If you look at the docs: https://docs.python.org/3/library/functions.html#all

all(iterable) .
Return True if all elements of the iterable are true (or if the iterable is empty). Equivalent to:

def all(iterable): for element in iterable: if not element: return False return True 

So if all(x % 2 == 1 for x in li): roughly translates to

def are_all_odds(num_list): #Flag to keep track if we encounter an even number flag = True for num in num_list: #Once we encounter a even number, we break the for loop if num % 2 != 1: flag = False break #Return the flag return flag 

We can test this function by doing

print(are_all_odds([1, 2, 3, 4])) #False print(are_all_odds([1, 3, 5])) #True 

Also just a suggestion, list is a python builtin keyword, so don't use it in variables :)

Yes, it is possible.

The Python code you wrote is very idiomatic, keep up that good work.

To see how to do it differently, you can look at programming languages that are less advanced, such as C. That is a very basic programming language which lacks the features for this if all statement. Searching for "c all elements array true" should give you the code you are looking for. For such a simple piece of code, it's easy to translate the code back into Python.