Python dict基于dict中的两个值排序

Question

I have a dictionairy that is build as

>> dict = {'foo':[20,15],'bar':[10,5],'is':[35,3],'a':[20,10],'word':[50,1]}

I want to find the key that has the highest of list value [0] and the lowest of value [1] .. (or an approximation of it) but its giving me a total brainfreeze

So for this example the desired result would be

>> 'word':[50,1]

It has been suggested I should more clearly define my paramaters: right now I'm looking to print the top 10 highest results from the [0] value as long as the second value remains below 5

Thank you for taking the time to read the question

Answer 1

You can use max function with a proper key function :

>>> max(dict.items(),key=lambda x: (x[1][0],-x[1][1]))
('word', [50, 1])

Note that in this case the priority of x[1][0] (max value) is more than second one,So for some dictionaries like following :

>>> dict = { 'foo': [35,5], 'word': [60, 25]}

It will returns :

('word', [60, 25])

You can also get items based on the difference of the values (which seems more close to what you want):

>>> dict = { 'foo': [70,5], 'word': [68,1]}
>>> max(dict.items(),key=lambda x: (x[1][0]-x[1][1]))
('word', [68, 1])

Answer 2

Try with below code. I am not sure whether it satisfies all your scenarios

    dict = {'a': [20, 10], 'word': [50, 1], 'is': [35, 3], 'foo': [20, 15], 'bar': [10, 5]}

    value = max(dict.values())
    b = value[1]
    for each in dict.values():
        if value[0] == each[0] and each[1] < b:
             value = each

    print (dict.keys()[dict.values().index(value)],value)