Filtering python dict by abs minimal value, returning many-CodePudding

I would like to filter a python dictionary to include only the minimum absolute values, returning a new dict with all key,value pairs which have the same min(abs()) match. I've coded up the following solution:

def return_closest_vals(dc):
    # min_val = dc[min(dc, key=dc.get)]

    min_val = 1e3
    keys_matching_min_val = {}
    for d, v in dc.items():
        if abs(v) < min_val:
            min_val = abs(v)
            keys_matching_min_val = {d:v} # updating min_val means keys matchin are outdated. Reset with new val
        elif abs(v) == min_val:
            keys_matching_min_val[d] = v

    return keys_matching_min_val

if __name__ == "__main__":

    test_dict = {"a":1, "b":2, "c":-1, "d":2, "e":1}
    expected_outcome = {"a":1, "c":-1, "e":1}
    print(return_closest_vals(test_dict))

but I don't think it's very elegant. Is there any nicer way to do this?

CodePudding user response：

Solution that generates the correct output however it loops over the provided dictionary twice. If one pass is your requirement then your original solution is best.

def return_closest_vals(dc):
    if not dc: return {}

    min_val = min(abs(v) for v in dc.values())
    return {k:v for k,v in dc.items() if abs(v) == min_val}

CodePudding user response：

I think you want something like

new_dict = {k: v for k,v in old_dict.items() if abs(v) == abs(min([*old_dict.values()]))}