How to select the dictionary that minimizes a key along a set of dictionaries in python-CodePudding

I have two(n) Python dictionaries dict_1 = {"obj_value": 1, "other_key": 10} and dict_2 = {"obj_value": 0, "other_key": 10} and I need a rutine that returns the dictionary that minimizes the obj_value key, in this case, dict_2

Any ideas about this problem without using conditionals (if)? I would want to extend this to more than two dictionaries.

CodePudding user response：

You can use min

min(dict1, dict2,.., key=lambda x : x.get("obj_value"))

Also if there are list of dictionaries then l_o_d = [{dict}] Then You can use

min(*l_o_d, key=lambda x : x.get("obj_value"))

CodePudding user response：

For the case of two dictionaries as you have it, a conditional is good enough:

chosen_dict = dict_1 if dict_1["object_value"] < dict_2["object_value"] else dict_2

For the general case, it depends a bit on the data structure you are using. Let's assume a simple extension with a list of dictionaries, like:

all_dictionaries = [
    {"obj_value": 1, "other_key": 10},
    {"obj_value": 0, "other_key": 10},
    {"obj_value": 2, "other_key": 10},
    ...
    {"obj_value": 999, "other_key": 10},
]

In this case you can simply sort the list by the key, like so:

chosen_dict = sorted(all_dictionaries, key=lambda x: x["obj_value"])[0]

This sorts the list by the key attribute, with this picking out the value for the obj_value in each dictionary, then we simply take the 0th element, which is a dictionary with the lowest value for the object_value key.

You can do something similar with min:

chosen_dict = min(all_dictionaries, key=lambda x: x["obj_value"])