Python miminum value in dictionary of lists

Question

Sorry about the question repost...I should have just edited this question in the first place. Flagged the new one for the mods. Sorry for the trouble

Had to re-write the

Answer 1

How about using sorting and lambdas?

#!/usr/bin/env python
d = {'a': ['1'], 'b': ['1', '2'], 'c': ['8', '1'], 'd':['1'], 'e':['1', '2', '3'], 'f': [4, 1]}
sorted_by_sum_d = sorted(d, key=lambda key: sum(list(int(item) for item in d[key])))
sorted_by_length_d = sorted(d, key=lambda key: len(d[key]))
print "Sorted by sum of the items in the list : %s" % sorted_by_sum_d
print "Sorted by length of the items in the list : %s" % sorted_by_length_d

This would output:

Sorted by sum of the items in the list : ['a', 'd', 'b', 'f', 'e', 'c']
Sorted by length of the items in the list : ['a', 'd', 'c', 'b', 'f', 'e']

Be aware I changed the initial 'd' dictionary (just to make sure it was working)

Then, if you want the item with the biggest sum, you get the last element of the sorted_by_sum_d list.

(I'm not too sure this is what you want, though)

Edit:

If you can ensure that the lists are always going to be lists of integers (or numeric types, for that matter, such as long, float...), there's not need to cast strings to integers. The calculation of the sorted_by_sum_d variable can be done simply using:

d = {'a': [1], 'b': [1, 2], 'c': [8, 1], 'd':[1], 'e':[1, 2, 3], 'f': [4, 1]}
sorted_by_sum_d = sorted(d, key=lambda key: sum(d[key]))