以列表为值对 python 字典进行排序

Question

I have a dictionary with list of values:-

dict_test = {
        'TYPE'          :       [ 'S1', 'S2', 'S3', 'S4', 'S5' ],
        'MARK'          :       [ 8, 11, 5, 34, 2 ]
}

I am trying to sort by key: MARK

I got below approach from one of the previous post:-

>>> dict((k,sorted(v)) for k, v in dict_test.items())
{'TYPE': ['S1', 'S2', 'S3', 'S4', 'S5'], 'MARK': [2, 5, 8, 11, 34]}

But I noticed that this approach sorted the key: MARK values, but it didn't sort the key: TYPE value accordingly.

Eg TYPE: S1 is mapped to MARK: 8 initially. But after sort, TYPE: S1 is mapped to MARK: 2

So what should I do to make sure if I sort key: MARK it will adjust the order of key: TYPE values as well?

Answer 1

Create a sorted list of (MARK, TYPE) pairs. Then update your dictionary:

z = sorted(zip(dict_test['MARK'], dict_test['TYPE']))
dict_test['TYPE'] = [tup[1] for tup in z]
dict_test['MARK'] = [tup[0] for tup in z]

Answer 2

You need an extra function to get sorted index first and then apply it to your values of dict.

For instance:

# Sort MARK 
idxes = sorted(range(len(dict_test['MARK'])), key=lambda i: dict_test['MARK'][i])

dict_test['TYPE'] = [dict_test['TYPE'][i] for i in idxes]
dict_test['MARK'] = [dict_test['MARK'][i] for i in idxes]

Answer 3

Not the most pythonic approach but this should do the trick.

diction = {mark: type for mark, type in zip(dict_test['MARK'], dict_test['TYPE'])}

sorted_dict_test = {'TYPE': [], 'MARK': []}

for i in sorted(diction.items()):
    sorted_dict_test['MARK'].append(i[0])
    sorted_dict_test['TYPE'].append(i[1])

For python 3.7+:

Please note that since the diction dictionary is built using int keys, it is already sorted ascending and we do not need to use sorted(diction.items()) . We can simply loop over the dictionary and build the sorted_dict_test