Reputation: 61

Sorting a dictionary which has values as list, and sorting based on the elements inside this list

I have a list

A={'k3': ['b', 3],'k2': ['a', 1],'k1': ['a', 3],'k4': ['c', 2],'k5': ['b', 2]}

I want to sort the above dictionary first by letters 'a','b' and 'c' in an ascending order

and then based on values 3,2,1 in the descending order. So my output should look something like

A={'k1': ['a', 3],'k2': ['a', 1],'k3': ['b', 3],'k5': ['b', 2],'k4': ['c', 2]}

How do I do it?

Upvotes: 1

Answers (6)

Reputation: 7898

Python recently added an OrderedDict

Upvotes: 1

Reputation: 22818

Dictionaries are not sortable. You need may wish to use a list instead.

For example, your input could be represented as a list of tuples:

A=[('k3', ['b', 3]),('k2', ['a', 1]),('k1',['a', 3])...]

And we could sort it like so:

A=sorted(A, key=lambda x: (x[1][0], -x[1][1]))

Upvotes: 0

Reputation: 9091

What about:

sorted(A, key = lambda x: (A[x][0], -A[x][1]))

At least it gives you the order for the keys.

Upvotes: 0

Reputation: 229603

Dictionaries are unordered, so you can't create a sorted dictionary.

If you just want a list of dictionary keys, sorted by the associated values, you could use sorted() like this:

>>> keys = sorted(A, key=lambda k: (A[k][0], -A[k][1]))
>>> keys
['k1', 'k2', 'k3', 'k5', 'k4']

Upvotes: 2

Reputation:

You can not sort dicts - there is no ordering defined on dicts.

Upvotes: 0

Reputation: 20547

AFAIK dictionaries have no concept of "order"

Upvotes: 0