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While coding a simple command-line tictactoe game to learn Python (here's a link to the entire script on repl.it), I was not happy with my game_over function, since a lot of code is repeated:

Edit: fixed the code by putting the [deck_is_full] if last instead of first

Edit 2: changed the code back to original for clarity

def game_over(deck):
    if deck_is_full(deck):
        return deck_is_full(deck)
    if three_in_line(deck, "X"):
        return three_in_line(deck, "X")
    if three_in_line(deck, "O"):
        return three_in_line(deck, "O")
    return False

Is there a pythonic, repetition-free way to code that? I have found this answer on SO, but it doesn't get rid of the repetition, just gets it into a one-liner.

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2 Answers 2

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def game_over(deck):
    return deck_is_full(deck) or three_in_line(deck, "X") or three_in_line(deck, "O")

Edit: Had a look at your whole code and see that three_in_line returns true strings or the default None. So if you do need False here instead of None, then add it explicitly:

def game_over(deck):
    return deck_is_full(deck) or three_in_line(deck, "X") or three_in_line(deck, "O") or False

Edit 2: Hmm, just a note: If your functions had side effects, my version wouldn't be equivalent to yours, as I only call them once, not twice like you (when they return a true value). But as expected, and as they in my opinion should, they don't have side effects.

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12 Comments

The prople who just copied your post and reposted downvoted this, propped you back up..
note that contrary to what you wrote on a deleted answer, any() does short circuit on the first False condition
@Chris_Rands any does, but the tuple/list-building doesn't.
@Chris_Rands Just removing parentheses wouldn't make it a generator expression. It would make it three separate arguments being passed to any.
@khelwood yes, i got that wrong, would actually require an awkard construct like any(func() for func in (lambda: deck_is_full(deck), lambda: three_in_line(deck, "X"), lambda: three_in_line(deck, "O")))
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Since python 3.8 you can use the walrus operator to assign a variable in the if clause

def game_over(deck):
    if not (result:=deck_is_full(deck)):
        if not (result:=three_in_line(deck, "X")):
            if not (result:=three_in_line(deck, "O")):
                return False
    return result
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5 Comments

Interesting, I didn't know about the walrus operator. By the way, you have fixed an error in my original code (edited since), which if a player won while filling the last remaining space, would declare the game a draw (because the deck becomes full) instead of a victory.
@ArnaudBouffard What do you mean this fixed that? This is equivalent to your original and I even just tested it, it did report a draw in that case.
@superbrain original function checked deck_is_full then three_in_line(deck, "X") then three_in_line(deck, "X"), thus reporting a draw instead of a victory in this case: [O O X] [O X X] [O X O]. Fixed by checking deck_is_full last.
@ArnaudBouffard Yes, but this code checks deck_is_full first.
and reports a draw in that case, you are right @superbrain, I got mixed up

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