python sorting dictionary by length of values

Question

I have found many threads for sorting by values like here but it doesn’t seem to be working for me…

I have a dictionary of lists that have tuples. Each list has a different amount of tuples. I want to sort the dictionary by how many tuples each list contain.

>>>to_format 
>>>{"one":[(1,3),(1,4)],"two":[(1,2),(1,2),(1,3)],"three":[(1,1)]}
>>>for key in some_sort(to_format):
       print key,
>>>two one three

Is this possible?

Asked By: jwillis0720

||

Source

Answer 1

>>> d = {"one": [(1,3),(1,4)], "two": [(1,2),(1,2),(1,3)], "three": [(1,1)]}
>>> for k in sorted(d, key=lambda k: len(d[k]), reverse=True):
        print k,


two one three

Here is a universal solution that works on Python 2 & Python 3:

>>> print(' '.join(sorted(d, key=lambda k: len(d[k]), reverse=True)))
two one three

Answered By: jamylak

Answer 2

dict= {'a': [9,2,3,4,5], 'b': [1,2,3,4, 5, 6], 'c': [], 'd': [1,2,3,4], 'e': [1,2]}
dict_temp = {'a': 'hello', 'b': 'bye', 'c': '', 'd': 'aa', 'e': 'zz'}

def sort_by_values_len(dict):
    dict_len= {key: len(value) for key, value in dict.items()}
    import operator
    sorted_key_list = sorted(dict_len.items(), key=operator.itemgetter(1), reverse=True)
    sorted_dict = [{item[0]: dict[item [0]]} for item in sorted_key_list]
    return sorted_dict

print (sort_by_values_len(dict))

output:
[{'b': [1, 2, 3, 4, 5, 6]}, {'a': [9, 2, 3, 4, 5]}, {'d': [1, 2, 3, 4]}, {'e': [1, 2]}, {'c': []}]

Answered By: Oleksiy Tsebriy

python sorting dictionary by length of values

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