How to sort the items in a dictionary according to the size of the values in the dictionary
practical example
The English grades for a class are stored in dictionary form as {'LiLei': 79, 'Jim': 88, 'Lucy': 92, ...}
Student rankings are calculated based on high and low grades.
Note: The key of the dictionary is the student's name and the value of the dictionary is the student's score.
prescription
Using the built-in function sorted
Method 1: Using zip to convert dictionary data into a list of tuples
Method 2: The key argument passed to the sorted function
Code Demo
# Sort the list using sorted
print(sorted([9, 1, 2, 8, 5]))
from random import randint
# Create a randomized table of grades, using dictionary parsing
c = {x: randint(60, 100) for x in 'abcded'}
print(c)
print(sorted(c))
# Method 1:
'''
Direct usesortedSorting the dictionary，It's not sorted by value，Instead, follow the key。
sorted()Passed in are iterable objects，Some kind of conversion for dictionaries，Let it be.sortedSortable structures。
Converting a dictionary to a list of tuples,as if： [(97, 'a'), (69, 'b')]，Then directly compare the first value of the tuple
'''
# Get all the values of the dictionary, or all the values of the dictionary.
# print(())
# print(())
# Use the zip function, two lists will be put together into a list, because the return is an object so use the list () function to convert it
print(list(zip((), ())))
# Then sort the synthesized list of tuples #
print(sorted(zip((), ())))
# Method 2:
# Return the k:v of the dictionary as a list of tuples
print(())
# Set the key argument of sorted to the value of the dictionary
print(sorted((), key=lambda x: x[1]))summarize
The above is a personal experience, I hope it can give you a reference, and I hope you can support me more.