Sorting a Dictionary by Value Then Key

Sorting a dictionary by value then by key

In [62]: y={100:1, 90:4, 99:3, 92:1, 101:1}
In [63]: sorted(y.items(), key=lambda x: (x[1],x[0]), reverse=True)
Out[63]: [(90, 4), (99, 3), (101, 1), (100, 1), (92, 1)]

The key=lambda x: (x[1],x[0]) tells sorted that for each item x in y.items(), use (x[1],x[0]) as the proxy value to be sorted. Since x is of the form (key,value), (x[1],x[0]) yields (value,key). This causes sorted to sort by value first, then by key for tie-breakers.

reverse=True tells sorted to present the result in descending, rather than ascending order.

See this wiki page for a great tutorial on sorting in Python.

PS. I tried using key=reversed instead, but reversed(x) returns an iterator, which does not compare as needed here.

How do I sort a dictionary by value?

Python 3.7+ or CPython 3.6

Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.

>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

or

>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

Older Python

It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.

For instance,

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))

sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.

And for those wishing to sort on keys instead of values:

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))

In Python3 since unpacking is not allowed we can use

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])

If you want the output as a dict, you can use collections.OrderedDict:

import collections

sorted_dict = collections.OrderedDict(sorted_x)

Sorting a dictionary by value then key

You need to take advantage of the fact that the values are numbers.

>>> [v[0] for v in sorted(d.iteritems(), key=lambda(k, v): (-v, k))]
['peach', 'banana', 'beetroot', 'almond', 'apple']

python sort dictionary items by value and then key

MyDict = {0: {'Score': 80.0, 'studentName': 'dan'},
          1: {'Score': 92.0, 'studentName': 'rob'},
          2: {'Score': 10.0, 'StudentName': 'xyz'}}

This returns the list of key-value pairs in the dictionary, sorted by value from highest to lowest:

sorted(MyDict.items(), key=lambda x: x[1], reverse=True)

For the dictionary sorted by key, use the following:

sorted(MyDict.items(), reverse=True)

The return is a list of tuples because dictionaries themselves can't be sorted.

This can be both printed or sent into further computation.

How to sort a dictionary by value (DESC) then by key (ASC)?

Something like

In [1]: d = {'banana': 3, 'orange': 5, 'apple': 5}

In [2]: sorted(d.items(), key=lambda x: (-x[1], x[0]))
Out[2]: [('apple', 5), ('orange', 5), ('banana', 3)]

Sort a dict according to both key and value at the same time

There's not an easy way to say "this value, ascending, then this one, descending". However, if you negate each of a list of integers, then sort it, then that's the same as sorting it in reverse.

This defines a sorting key which is a tuple:

• The first value is each dict item's value.
• The second value is each dict item's key, but negated.

d = {0: 2, 2: 2, 3: 2, 1: 4, 4: 5}

def sort_key(item):
    key, value = item
    return value, -key

print(sorted(d.items(), key=sort_key))

This outputs:

[(3, 2), (2, 2), (0, 2), (1, 4), (4, 5)]

See? The items are grouped by value, and in the event of a tie, by key in descending order.

How do I sort a list of dictionaries by a value of the dictionary?

The sorted() function takes a key= parameter

newlist = sorted(list_to_be_sorted, key=lambda d: d['name']) 

Alternatively, you can use operator.itemgetter instead of defining the function yourself

from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name')) 

For completeness, add reverse=True to sort in descending order

newlist = sorted(list_to_be_sorted, key=itemgetter('name'), reverse=True)

sort dict by value then if equal by keys

reverse=true applies to both elements of the tuple. That why your alphanumeric output is backwards.

To reverse the order on the first element of the tuple but not on the other you could negate the numbers:

sorted(w.items(),key= lambda a: (-a[1],a[0]))

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If you ever run into problem where you need to sort on multiple tuple elements with varying ASC/DESC order and elements that are not easy to negate like integers you could use the following technique.

Modified multisort from https://docs.python.org/3/howto/sorting.html#sort-stability-and-complex-sorts

Since the sorting is stable the following will work:

def multisort(xs, specs):
     for i, reverse in reversed(specs):
         xs.sort(key=lambda x: x[i], reverse=reverse)
     return xs

items = [('to', 2), ('be', 2), ('or', 1), ('not', 1), ('ae', 2)]
multisort(items, [(0, False), (1, True)])

Output:

[('ae', 2), ('be', 2), ('to', 2), ('not', 1), ('or', 1)]

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