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I was wondering if there was a simple alternative to lambda in my code.

def add_attack(self, attack_name):
        if attack_name in self.known_attacks and attack_name not in self.attacks:
            try:
                assert(len(self.attacks) < 4)
                self.attacks[attack_name] = self.known_attacks.get(attack_name)
                return True
            except:
                #find the min value of self.attacks
                minval = min(self.attacks.keys(), key=(lambda k: self.attacks[k]))
                for keys, values in self.attacks.items():
                    if self.attacks[minval] == values and min(minval, keys) == keys:
                        minval = keys
                del self.attacks[minval]
                self.attacks[attack_name] = self.known_attacks.get(attack_name)
                return True
        else:
            return False

I'm still learning python, and the lambda function is throwing me off since I haven't learned that much about it yet. Instead of using lambda, can someone help me out with another function to replace lambda? Thanks!

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  • The lambda expression is not really that intimidating if you think about what it does: it takes whatever arguments are given to it (usually you will find this argument list in the documentation to the function you are passing the lambda to); it evaluates whatever expression is inside it, and returns that value. In this case, the argument list is the keys of self.attacks, and the value returned as the "key" is that key's value in self.attacks. The min function uses the key to decide which one is the smallest. What about the lambda are you confused by? Commented Nov 28, 2022 at 7:03
  • Note that bare exceptions are frowned upon. In this case better to have: except AssertionError:. Commented Nov 28, 2022 at 7:36

2 Answers 2

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You could define a function for it:

def return_attacks(self,k):
    return self.attacks[k]

And use that function in the key:

minval = min(self.attacks.keys(), key=(self.return_attacks))

I would strongly recommend you get comfortable with lambda functions - and I think it is clear to you now that lambda x : expr(x) is equivalent to func when

def func(x):
    return expr(x)
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1 Comment

Your "return_attacks" function is basically self.attacks.get. You could just use that as the key (although the behavior when the key k doesn't exist would be different (although not relevant in this use-case).
1

A lambda should not scare you! It's just a small anonymous function.

It can take any number of arguments, but can only have one expression.

minval = min(self.attacks.keys(), key=(lambda k: self.attacks[k]))

Here you are getting the result of the expression min() as minval

The min function can take keys, here is more about that. I can see it can be confusing, but this key is not the same thing with a dictionary key. This key is just a way to tell the min function how it should behave.

If we go back to the code:

So the line basically finds the minimum value in self.attacks.keys(), with a lambda function that returns every element in self.attacks[]

If you do not want to use lambda, you can write a function in your class that does exactly the same thing.

def find_min_key(self, my_dict):
    return min(my_dict, key= my_dict.get)

You can use this as:

min_val = self.find_min_key(self.attacks)
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1 Comment

Please don't use dict as a variable name, it will shadow the builtin dict function.

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