按值对字典列表进行排序

Question

I have a list of dictionaries, of the form:

neighbour_list = [{1:4}, {3:5}, {4:9}, {5:2}]

I need to sort the list in order of the dictionary with the largest value. So, for the above code the sorted list would look like:

sorted_list = [{4:9}, {3:5}, {1:4}, {5:2}]

Each dictionary within the list only has one mapping.

Is there an efficient way to do this? Currently I am looping through the list to get the biggest value, then remembering where it was found to return the largest value, but I'm not sure how to extend this to be able to sort the entire list.

Would it just be easier to implement my own dict class?

EDIT: here is my code for returning the dictionary which should come 'first' in an ideally sorted list.

temp = 0
element = 0

for d in list_of_similarities:
    for k in d:
        if (d[k] > temp):
            temp = d[k]
            element = k
            dictionary = d

first = dictionary[element]

Answer 1

You can use an anonymous function as your sorting key to pull out the dict value (not sure if i've done this the most efficient way though:

sorted(neighbour_list, key = lambda x: tuple(x.values()), reverse=True)
[{4: 9}, {3: 5}, {1: 4}, {5: 2}]

Note we need to coerce x.values() to a tuple, since in Python 3, x.values() is of type "dict_values" which is unorderable. I guess the idea is that a dict is more like a set than a list (hence the curly braces), and there's no (well-) ordering on sets; you can't use the usual lexicographic ordering since there's no notion of "first element", "second element", etc.

Answer 2

You could list.sort using the dict values as the key.

neighbour_list.sort(key=lambda x: x.values(), reverse=1)

Considering you only have one value, for python2 you can just call next on itervalues to get the first and only value:

neighbour_list.sort(key=lambda x: next(x.itervalues()), reverse=1)
print(neighbour_list)

For python3, you cannot call next on dict.values , it would have to be:

neighbour_list.sort(key=lambda x: next(iter(x.values())), reverse=1)

And have to call list on dict.values :

neighbour_list.sort(key=lambda x: list(x.values()), reverse=1)