Adding additional logic to a lambda function to get the minimum value from a python dictionary

Question

I have a dictionary with values mapping some object to an integer; e.g. {node: 1, node: 2}.

I want to get the key that corresponds to the minimum value in this dictionary, which I know I can do with:

min_key = min(my_dict, key=lambda k: my_dict[k])

However, I'd also like to add the constraint that the key is not contained within some other set. Here's the pseudo-code that I wished worked:

min_key = min(my_dict, key=lambda k: my_dict[k] where k not in my_set)

Is there a way to write this as a one-liner in python as part of a lambda, or do I now have to explicitly loop through the dictionary and add the logic within the loop like this?

min_key, min_value = None, float('inf')
for k,v in my_dict.items():
    if v < min_value and k not in my_set:
        min_key = k
return min_key

That's not a valid dictionary, you can't have duplicate keys. — Barmar
– Barmar, Commented Dec 10, 2021 at 1:23
A standard trick is to return math.inf for the values you want to ignore when taking the minimum. Likewise use -math.inf for finding the maximum. So your example would be key=lambda k: math.inf if k in my_set else my_dict[k] — Frank Yellin
– Frank Yellin, Commented Dec 10, 2021 at 1:27

Barmar · Accepted Answer · 2021-12-10 21:53:59Z

5

Replace my_dict with a dictionary comprehension that returns the filtered dictionary.

min_key = min({k:v for k, v in my_dict.items() if k not in my_set}, 
                key = lambda k: my_dict[k])

edited Dec 10, 2021 at 21:53

answered Dec 10, 2021 at 1:24

Barmar

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3 Comments

Kelly Bundy Over a year ago

No point building a dict with values, as min is going to ignore them.

Barmar Over a year ago

You're right, we just need a list of keys.

Barmar Over a year ago

Since you posted that answer, I won't bother editing mine.

Kelly Bundy · Accepted Answer · 2021-12-10 21:51:19Z

1

Just take the minimum over the filtered keys instead of all keys:

min_key = min(my_dict.keys() - my_set, key=my_dict.get)

(Note I also replaced your key function, no need to write your own.)

edited Dec 10, 2021 at 21:51

answered Dec 10, 2021 at 21:46

Kelly Bundy

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Comments

score 1 · Accepted Answer · 2021-12-10 22:01:25Z

1

It's similar to @Barmar's answer but you can also use set.difference between my_dict and my_set to filter the relevant dictionary:

out = min(set(my_dict).difference(my_set), key = lambda k: my_dict[k])

edited Dec 10, 2021 at 22:01

answered Dec 10, 2021 at 1:43

user7864386

2 Comments

Kelly Bundy Over a year ago

Putting extra effort into the values that min is going to ignore anyway seems wasteful.

user7864386 Over a year ago

@KellyBundy you're right, that's an excellent point. Thanks.

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