Sorting a Dictionary with Lists as Values, According to an Element from the List

Sorting a dictionary with lists as values, according to an element from the list

Here is one way to do this:

>>> sorted(myDict.items(), key=lambda e: e[1][2])
[('item2', [8, 2, 3]), ('item1', [7, 1, 9]), ('item3', [9, 3, 11])]

The key argument of the sorted function lets you derive a sorting key for each element of the list.

To iterate over the keys/values in this list, you can use something like:

>>> for key, value in sorted(myDict.items(), key=lambda e: e[1][2]):
...   print key, value
... 
item2 [8, 2, 3]
item1 [7, 1, 9]
item3 [9, 3, 11]

How to sort a dictionary by values which contain lists in python

Sorted in python

Use this code:

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

Output:

[('Key4', ['LONG_A', 9]), ('Key3', ['LONG', 7]), ('Key1', ['LONG', 1])]

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)

Sorting dictionary containing lists

First, you can't sort a dict because it doesn't have order ¹. You can sort it's items however:

sorted(d.items(), key=lambda t: t[1][1])

Should do the trick.

notes

• t[1] # => the "value", t[0] is the "key"
• t[1][1] # => The second element in the value.

You'll get a list of 2-tuples (where the first element is the key, and the second element is the value). Of course, you could always pass this list of 2-tuples directly to collections.OrderedDict if you wanted to construct a dict which has an ordering after the fact ...

^{1More correctly, the order is arbitrary and could change based on key insertions, deletions, python implementation or version ... (which is why it's easier to say that dict's are unordered)}

sorting dictionary with list value (python)

You should not name a Python variable dict. I will call the dict d instead.

$ python
Python 3.8.6 (default, Jan 27 2021, 15:42:20)
[GCC 10.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> d = {"a":[1,2,3,12],"b":[5,6,7,8],"c":[9,10,11,4]}

This yields a sorted list of the dict items:

>>> sorted(list(d.items()), key=lambda x: x[1][3])
[('c', [9, 10, 11, 4]), ('b', [5, 6, 7, 8]), ('a', [1, 2, 3, 12])]

Now we can get the sorted dict:

>>> dict(sorted(list(d.items()), key=lambda x: x[1][3]))
{'c': [9, 10, 11, 4], 'b': [5, 6, 7, 8], 'a': [1, 2, 3, 12]}

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)

Sort list of dictionaries according to the list of values

list_dict.sort(key=lambda x:a_list.index(x['a']))

Sorting dictionary of lists based on first value in list

Did you test what you already had to see whether it still worked? Because it does. When you sort a group of sequences, they are compared "alphabetically" - if the first items of two sequences are equal, their second items are compared, and so on. This means that, in your example, sorted would compare the first element of the value's list. If they are different, the sort order is exactly what it would be if you were comparing the first element only with v[0]. If they are the same, sorting by v[0] would simply ensure an arbitrary order for those elements. Leaving it as (v,k) ensures that if the first elements of the value list are equal, it sorts based on the rest of the elements. If there are two values with identical lists, it sorts based on the key.

>>> import collections
>>> d = {'key1': [2, 3], 'key2': [7, 7], 'key3': [5, 10]}
>>> orderedDict = collections.OrderedDict(sorted(d.iteritems(), key=lambda (k,v):(v,k), reverse=True))
>>> orderedDict
OrderedDict([('key2', [7, 7]), ('key3', [5, 10]), ('key1', [2, 3])])

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