Python Miminum Value In Dictionary Of Lists

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 question d

Solution 1:

You can use min() with a key= argument, and specify a key function that compares the way you want.

d = {'a': ['1'], 'b': ['1', '2'], 'c': ['2'], 'd':['1']}

min_value = min(d.values())
min_list = [key for key, value in d.items() if value == min_value]
max_len = len(max(d.values(), key=len))
long_list = [key for key, value in d.items() if len(value) == max_len]

print(min_list)
print(long_list)

Notes:

0) Don't use dict as a variable name; that's the name of the class for dictionary, and if you use it as a variable name you "shadow" it. I just used d for the name here.

1) min_value was easy; no need to use a key= function.

2) max_len uses a key= function, len(), to find the longest value.

Solution 2:

Your problem is that your lists contain strings ('2'), and not integers (2). Leave out the quotes, or use the following:

min_value = min(min(map(int, v) for v in dct.values()))
min_keys = [k for k,v in d.items() if min_value in map(int, v)]

Similarily, to calculate the keys with the max length:

max_length = max(map(len, dct.values()))
maxlen_keys = [k for k,v in d.items() if max_length == len(v)]

Also, it's a bad idea to use dict as a variable name, as doing so overshadows the built-in dict.

Solution 3:

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]))

Solution 4:

I've found such a simple solution:

min_len = len(min(d.values(), key=(lambda value: len(value))))  # 1
min_keys = [key for i, key in enumerate(d) if len(d[key]) == min_len]  # ['e', 'f', 'g']