python dictionary sorting in descending order based on values

Question:

I want to sort this dictionary d based on value of sub key key3 in descending order. See below:

``````d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
'124': { 'key1': 6, 'key2': 56, 'key3': 6 },
'125': { 'key1': 7, 'key2': 44, 'key3': 9 },
}
``````

So final dictionary would look like this.

``````d = { '125': { 'key1': 7, 'key2': 44, 'key3': 9 },
'124': { 'key1': 6, 'key2': 56, 'key3': 6 },
'123': { 'key1': 3, 'key2': 11, 'key3': 3 },
}
``````

My approach was to form another dictionary e from d, whose key would be value of key3 and then use reversed(sorted(e)) but since value of key3 can be same, so dictionary e lost some of the keys and their values. makes sense?

How I can accomplish this? This is not a tested code. I am just trying to understand the logic.

Dictionaries do not have any inherent order. Or, rather, their inherent order is “arbitrary but not random”, so it doesn’t do you any good.

In different terms, your `d` and your `e` would be exactly equivalent dictionaries.

What you can do here is to use an `OrderedDict`:

``````from collections import OrderedDict
d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
'124': { 'key1': 6, 'key2': 56, 'key3': 6 },
'125': { 'key1': 7, 'key2': 44, 'key3': 9 },
}
d_ascending = OrderedDict(sorted(d.items(), key=lambda kv: kv[1]['key3']))
d_descending = OrderedDict(sorted(d.items(),
key=lambda kv: kv[1]['key3'], reverse=True))
``````

The original `d` has some arbitrary order. `d_ascending` has the order you thought you had in your original `d`, but didn’t. And `d_descending` has the order you want for your `e`.

If you don’t really need to use `e` as a dictionary, but you just want to be able to iterate over the elements of `d` in a particular order, you can simplify this:

``````for key, value in sorted(d.items(), key=lambda kv: kv[1]['key3'], reverse=True):
do_something_with(key, value)
``````

If you want to maintain a dictionary in sorted order across any changes, instead of an `OrderedDict`, you want some kind of sorted dictionary. There are a number of options available that you can find on PyPI, some implemented on top of trees, others on top of an `OrderedDict` that re-sorts itself as necessary, etc.

Python dicts are not sorted, by definition. You cannot sort one, nor control the order of its elements by how you insert them. You might want to look at collections.OrderDict, which even comes with a little tutorial for almost exactly what you’re trying to do: http://docs.python.org/2/library/collections.html#ordereddict-examples-and-recipes

you can make use of the below code for sorting in descending order and storing to a dictionary:

``````        listname = []
for key, value in sorted(dictionaryName.iteritems(), key=lambda (k,v): (v,k),reverse=True):
diction= {"value":value, "key":key}
listname.append(diction)
``````

List

``````dict = {'Neetu':22,'Shiny':21,'Poonam':23}
print sorted(dict.items())
sv = sorted(dict.values())
print sv
``````

Dictionary

``````d = []
l = len(sv)
while l != 0 :
d.append(sv[l - 1])
l = l - 1
print d`
``````

A short example to sort dictionary is desending order for Python3.

``````a1 = {'a':1, 'b':13, 'd':4, 'c':2, 'e':30}
a1_sorted_keys = sorted(a1, key=a1.get, reverse=True)
for r in a1_sorted_keys:
print(r, a1[r])
``````

Following will be the output

``````e 30
b 13
d 4
c 2
a 1
``````

sort dictionary ‘in_dict’ by value in decreasing order

``````sorted_dict = {r: in_dict[r] for r in sorted(in_dict, key=in_dict.get, reverse=True)}
``````

example above

``````sorted_d = {r: d[r] for r in sorted(d, key=d.get('key3'), reverse=True)}
``````

You can use the operator to sort the dictionary by values in descending order.

``````import operator

d = {"a":1, "b":2, "c":3}
cd = sorted(d.items(),key=operator.itemgetter(1),reverse=True)
``````

The Sorted dictionary will look like,

``````cd = {"c":3, "b":2, "a":1}
``````

Here, operator.itemgetter(1) takes the value of the key which is at the index 1.

Whenever one has a dictionary where the values are integers, the Counter data structure is often a better choice to represent the data than a dictionary.

If you already have a dictionary, a counter can easily be formed by:

``````c = Counter(d['123'])
``````

as an example from your data.

The `most_common` function allows easy access to descending order of the items in the counter

The more complete writeup on the Counter data structure is at https://docs.python.org/2/library/collections.html

You can use dictRysan library. I think that will solve your task.

``````import dictRysan as ry

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 },
'124': { 'key1': 6, 'key2': 56, 'key3': 6 },
'125': { 'key1': 7, 'key2': 44, 'key3': 9 },
}

changed_d=ry.nested_2L_value_sort(d,"key3",True)
print(changed_d)
``````
``````f= {273: 6, 272: 2, 2: 1, 3083: 1, 281: 1, 2490: 5, 366: 5, 1945: 4, 869: 1, 1398: 1, 1920: 5, 3061: 2, 862: 1}

sorted_d = dict( sorted(f.items(), key=operator.itemgetter(1),reverse=True))

print('Dictionary in descending order by value : ',sorted_d)
``````

Here I am adding a generic solution to the problem.

Input: d = {"a":1, "b":2, "c":3}
Output: d= {"c":3, "b":2, "a":1}

``````import operator
dict(sorted(d.items(),key=operator.itemgetter(1),reverse=True))
``````

Above code return a dict with sorted value.

``````sorted(d.items(),key=operator.itemgetter(1),reverse=True)
``````

This line will return [(‘c’, 3), (‘b’, 2), (‘a’, 1)] -> Sorted List of tuples

``````   sql="select id, name, client_name, server_id, idsubdatatable from analytics_data where status=1;"
conn.execute(sql,)
result=conn.fetchall()
for value in result:
data_dict={}
data_dict['id']=value[0]
data_dict['name']=value[1]
data_dict['client_name']=value[2]
data_dict['server_id']=value[3]
data_dict['idsubdatatable']=value[4]
data_result.append(data_dict)
data_result = sorted(data_result, key=lambda i: i['idsubdatatable'],reverse=True)
return (json.dumps({'data': data_result,'status':"SUCCESS"}))
``````

if dictionary value of Key idsubdatatable in database is like 3,5,1,7,2 etc then it will return 7,5,3,2,1

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