[FIXED] dictionary decomposition and creating a new dictionary with identical values

Issue

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?

Solution

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}

Answered By - timgeb