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This article mainly introduces "how to sort Python dictionary". In daily operation, I believe many people have doubts about how to sort Python dictionary. Xiaobian consulted all kinds of materials and sorted out simple and easy operation methods. I hope to help you answer the doubts about "how to sort Python dictionary"! Next, please follow the small series to learn together!

A dictionary is a data structure in Python, where each dictionary element consists of a key-value pair. The key and value of the dictionary are organized in the shape of a set, respectively, for quick query. Collections are stored like a tree, so searching is very fast. We can obtain iterable objects for key sets and value sets separately through the dictionary keys method and values method, as follows:

x = {'x':20,'a':12,'b':5} print(x.keys()) print(x.values())

Executing this code will output the following:

dict_keys(['x', 'a', 'b']) dict_values([20, 12, 5])

PS: dict_keys and dict_values are Python's two inner classes, both of which organize data in a tree structure. The problem is that the collection is unordered (because it is stored in a tree structure), but due to certain requirements we expect ordered keys, which leads to the following problems:

Q1: Can the collection be sorted?

Q2: After sorting, can I get pairs of key-values?

Q3: In addition to the key value sorting, can you sort by value?

To answer these questions, go on to the following.

1. Can collections be sorted?

The answer to this question is: No. Since it is called a collection, it must be disordered due to the form in which the data is stored, but we can do a compromise. Since the collection cannot be sorted, we can sort the data in the collection and put it into a sortable data structure (for example, a list), which can solve our problem to some extent. Look at the code below:

x = {'x':20,'a':12,'b':5} keys = sorted(x.keys()) values = sorted(x.values()) print(type(keys)) print(type(values)) print(keys) print(values)

In this code, the sorted function is used, which can be used to sort the sequence, put the sorted result into a list, and finally return this list, so executing this code will output the following content:

['a', 'b', 'x'] [5, 12, 20]

Now that our first problem is solved, we use the sorted function to sort the collections in ascending order and present them as lists. If you want to sort in descending order, you need to set the reverse parameter to True, as follows:

keys = sorted(x.keys(), reverse=True) values = sorted(x.values(), reverse=True)

2. After sorting, can I get pairs of key-values?

Now let's solve the second problem. This article is about dictionaries, so even sorting keys and values is not what we expect. What we expect is to get key-value pairs in dictionaries, and key-value pairs that are already sorted. This problem is also very easy to solve, since you have already obtained the ordered key, then use the key to obtain the corresponding value from the dictionary, so you can form a pair of key-value, the code is as follows:

x = {'x':20,'a':12,'b':5} keys = sorted(x.keys()) for key in keys: print(f"{key}:{x[key]}")

Executing this code will output the following:

x:20 b:5 a:12

Obviously, the output key-value pairs are sorted in ascending order of key.

3. Can you sort by value other than by keys?

The sorting of values is fine, but sorted values are useless. This is because dictionaries cannot retrieve keys in reverse by value. So only ordered values can be obtained through the previous method. So you need a key parameter to the sorted function, which specifies whether to use key or value for sorting. If you are using Python version 3.7 or above. lambda expression can be specified for the key parameter value as follows:

y1 ={k: v for k, v in sorted (x. items (), key = lambda item: item [1])} print (y1)#in descending order of value y2 ={k: v for k, v in sorted (x. items (), key = lambda item: item [1], reverse = True)} print (y2)

This code uses the for-in expression to generate the dictionary, where x is the dictionary to be sorted. items gets key-value pairs. The lambda expression item is the current key-value pair. [1][2][3][4][5][6][7][8][9][10][11][12][13][14][15][16][17][18][19][19][19][10][19][10][10][19][10][10][10][11][10][10][11][10][11][10][11][12][10][11][12][10][11][12][13][14][14][15][15][16][16][17][18][19][10][19][10][10][10][11][10][10][11][10][11

Executing this code will output the following result:

{'b': 5, 'a': 12, 'x': 20} {'x': 20, 'a': 12, 'b': 5}

If you don't want to use a for-in expression, you can also use the dict function as follows: print (dict (sorted (x. items (), key = lambda item: item [1]))) If you don't want to use lambda expressions, or if you need more complex collations, you can customize the collation function as follows:

def dict_val(x): return x[1] sorted_x = sorted(x.items(), key=dict_val)

If you are using Python version 3.6 or below, you can use the following code:

import operator #by value sorted_x = sorted (x. items (), key = operator. itemgetter (1)) print (type (sorted_x))#list print (sorted_x)#so replace with list import operator #by key sorted_x = sorted (x. items (), key = operator. itemgetter (0)) print (sorted_x)

In this code, an ordered list is returned, the list elements are tuples, the first value is key, and the second value is value. Of course, you can reinsert this data into a new dictionary. Executing this code will output the following:

[('b', 5), ('a', 12), ('x', 20)] [('a', 12), ('b', 5), ('x', 20)]

If you still want to search dictionary values by key, you can convert sorted_x into an ordered dictionary using OrderedDict object, code as follows:

import collections sorted_dict = collections.OrderedDict(sorted_x) print(type(sorted_dict)) print(sorted_dict) print(sorted_dict.get('b'))

Executing this code will output the following:

OrderedDict ([('a', 12),('b', 5),('x', 20)]) 5 At this point, the study of "how to sort Python dictionaries" is over, hoping to solve everyone's doubts. Theory and practice can better match to help everyone learn, go and try it! If you want to continue learning more relevant knowledge, please continue to pay attention to the website, Xiaobian will continue to strive to bring more practical articles for everyone!

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