Question

I need to sort this dict with two sequential criteria: “rank” descending, “fruit” ascending (in case of same ranking result).

The dict is structured in this way

'object':['fruit',rank]

For example I’ve:

myDict = {'item2': ['bananas', 3], 'item3': ['cocumbers', 11], 'item1': ['pinapples', 3], 'item4': ['pears', 3]}

My aim is to obtain this:

{'item3': ['cocumbers', 11], 'item2': ['bananas', 3],'item4': ['pears', 3], 'item1': ['pinapples', 3]}

with

sorted(myDict.items(), key=lambda (k, v): v[1], reverse=True)

I obtain only the correct sorting for rank but with no sorting order for the objects with the same ranking:

[('item3', ['cocumbers', 11]), ('item2', ['bananas', 3]), ('item1', ['pinapples', 3]), ('item4', ['pears', 3])]

How solve this in python 2.7?

Thank’s

Answer 1

sorted(myDict.items(), key=lambda (k, v): (-v[1],v[0]))

should do the trick. Since rank is an integer, we can easily make it sort descending by making the key function just return the negative value rather than needing the reverse=True keywords to sorted. In the case of a tie, we fall back on the natural ordering of the strings.

This works because sequences in python sort lexicographically (tuples are a sequence type) — Python compares elements from each sequence in order until one of the elements is non-equal which gives the ordering of your pair of sequences.

In the case where you can’t easily replace reverse=True with a key function, you’ll need to sort twice:

l1 = sorted(myDict.items(),key=lambda (k,v):v[0])
l1.sort(key=lambda (k,v):v[1],reverse=True)

This relies on the documented fact that a python sort must be stable. In other words, the relative order of elements that are equal is the same before and after sorting.