Quickest way to get dictionary key that is closest to 1 in Python

Question

Hi I wanted to ask if this is the quickest and taking the least resources way to get value closest to 1 in dictionary or is there better, more effective way of doing this.

import operator

dct = {"a": 0.1, "b": 0.2, "c": 0.7, "d": 1, "e": 0.5}

sorted_orders = sorted(dct.items(), key=operator.itemgetter(1))
sorted_orders = str(sorted_orders.pop()[:1])
a = len(sorted_orders) - 3
sorted_orders = sorted_orders[2:a]
print sorted_orders

Desired output is contents of key closest to 1 here: d

Answer 1

This is how I would do it:

closest = sorted(dct.values(), key=lambda x: abs(1-x))[0]

I sort the items with a key that gives the distance between the number and 1 . ( abs(1-x1) ). The first item, therefore, is the value that is closest to 1 . You could use min() as in the answer by Gábor Erdős, but if you want to know the whole order, use this.

Answer 2

This should do it

min(list(dct.values()), key=lambda x: abs(x - 1))

Notes:

1. This method works for lists as well

2. This is not the most efficient way. A faster approach would be to use bisect