dictionary decomposition and creating a new dictionary with identical values

Question

I have a dictionary myDict

{'1': 5, '2': 13, '3': 23, '4': 17}

I'm using this code, that has served me well, in order to find the key/value in myDict closest to a targetVal

answer = key, value = min(myDict.items(), key=lambda (_, v): abs(v - targetVal))

Assuming targetVal is 14 , answer returns:

('2': 13)

What I need to do now, is deal with identical values in myDict . For example, if myDict was now:

{'1': 5, '2': 13, '3': 23, '4': 13}

I need both those key/value pairs with the value 13 . In instances when the code (above) finds the closest value in myDict , and that value happens to appear more than once, i'd like to create a new dictionary. In this case, answer would return:

{'2': 13, '4': 13}

Is it possible to update the way answer is being found to account for instances when the closest value appears more than once?

Answer 1

Find the minimum value first, then filter your dict .

>>> d = {'1': 5, '2': 13, '3': 23, '4': 13}
>>> target = 13
>>> min_ = min(d.itervalues(), key=lambda v: abs(v - target))
>>> {k:v for k,v in d.iteritems() if v == min_}
{'2': 13, '4': 13}

Answer 2

As you found, min provides only one item which satisfies the minimum condition. You can construct a one-pass solution via a manual loop:

from math import inf

myDict = {'1': 5, '2': 13, '3': 23, '4': 13}
targetVal = 14

res = {}
diff = inf
for k, v in myDict.iteritems():
    current_diff = abs(v - targetVal)
    if current_diff <= diff:
        if current_diff < diff:
            diff = current_diff
            res.clear()
        res.update({k: v})

print(res)

# {'2': 13, '4': 13}