Python对值列表进行排序，然后按值排序

Question

I have a dictionary, with a name as a key and a list containing scores as the value.

I'm trying to sort the lists in the values and then sort the whole dictionary by the values - making it so it's sorted by the highest score.

Here's the dictionary:

{'Dave_Wright': [10], 'Clara_Johnson': [9, 7], 'Amy_Kukri': [9, 8, 9], 'Eden_Jia': [10]}

I'm trying to use code I'v found from other places of stackoverflow, but I've not found a question specific to this yet.

print(d2)
d2 = sorted(d2.items(), key = itemgetter(1))
for key in d2:
    print(key, d2[key])

I'm probably making a mountain out of a molehill with this, but I am genuinely stumped.

Answer 1

Just itemgetter(1) will not do the sorting. You need to write your own key for this:

def get_return(key_value):
    return list(sorted(key_value[1])), key_value[0]

d2 = sorted(d2.items(), key=get_return, reverse=True)

for key, value in d2:
    print(key, value)

Here is a helpful tutorial for sorting dictionaries by keys/values that was recommended by spectre-d

Answer 2

This will print the dictionary in the order you've specified (assuming that you meant highest-score first):

d = {'Dave_Wright': [10], 'Clara_Johnson': [9, 7], 'Amy_Kukri': [9, 8, 9], 'Eden_Jia': [10]}

print ('\n'.join(
    '{}, {}'.format(key, sorted(value, reverse=True))
    for key, value in sorted(d.items(), key=lambda x:x[1], reverse=True)))

Result:

Dave_Wright, [10]
Eden_Jia, [10]
Amy_Kukri, [9, 9, 8]
Clara_Johnson, [9, 7]

Answer 3

Dictionaries aren't usually sortable; though you can sort the key-value pairs in the dictionary.

for name, scores in sorted(d2.items(), key=lambda x: max(x[1]), reverse=True):
    print name, sorted(scores)  # reverse=True?

Answer 4

for l in d.values():
    l.sort(reverse=True) # this will sort the values in place(mutate the dict)
result = sorted(d.items(), key=lambda x: x[1], reverse=True)
Out[518]:
[('Eden_Jia', [10]),
 ('Dave_Wright', [10]),
 ('Amy_Kukri', [9, 9, 8]),
 ('Clara_Johnson', [9, 7])]

Is this the output you expected?