Python : Best way to sort a python dictionary based on values ; values are lists

Question

I have a python dictionary

d = {
    "Random" : [1,2,3], "Stupid" :  [1], "Gross" : [1,1,1,1,1], "Ugly" : [2,1,1,1]
    }

The above list should be sorted based on the count of the list values.

I tried this : sorted(d, key=d.get, reverse=True)

But I don't seem to get the right result.

falsetru · Accepted Answer · 2013-07-19 07:26:06Z

4

Do you mean length?

>>> sorted(d, key=lambda k: -len(d[k]))
['Gross', 'Ugly', 'Random', 'Stupid']
>>> sorted(d, key=lambda k: len(d[k]), reverse=True)
['Gross', 'Ugly', 'Random', 'Stupid']

answered Jul 19, 2013 at 7:26

falsetru

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3 Comments

Deepankar Bajpeyi Over a year ago

This works for me. Thanks, I need to get used to lambda . Sigh Btw why not use count instead of len /.

Pawel Miech Over a year ago

count counts the occurences of an item in a list, length returns the number of all items in a list

falsetru Over a year ago

@DeepankarBajpeyi, [1,1,1,3].count(1) gives 3. len([1,1,1,3]) gives 4. If you want 3, use count.

Steve Barnes · Accepted Answer · 2013-07-19 07:44:33Z

1

Since you are starting from a dictionary presumably you need to end with one, rather than a list so:

>>> import collections
>>> d = {
...     "Random" : [1,2,3], "Stupid" :  [1], "Gross" : [1,1,1,1,1], "Ugly" : [2,1,1,1]
...     }
>>> sd = collections.OrderedDict(sorted(d.items(), key=lambda t: -len(t[1])))
>>> sd
OrderedDict([('Gross', [1, 1, 1, 1, 1]), ('Ugly', [2, 1, 1, 1]), ('Random', [1, 2, 3]), ('Stupid', [1])])
>>>

answered Jul 19, 2013 at 7:44

Steve Barnes

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