python how to sort a list by the values ​​of a dictionary within the list

1. Sort by dictionary value (default is ascending order)

x = {1:2, 3:4, 4 :3, 2:1, 0:0}
1. sorted_x = sorted(x.iteritems(), key=operator.itemgetter(1)) print sorted_x
Output result: [(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
If you want to sort in descending order, you can specify reverse=True

2. sorted_x = sorted(x.iteritems(), key=operator.itemgetter(1), reverse=True) print sorted_x
Output result: [(3, 4), (4, 3), (1, 2), (2 , 1), (0, 0)]

2. Or directly use the reverse method of list to reverse the order of sorted_x

sorted_x.reverse()

3. The more common method is to use lambda expression

sorted_x = sorted(x.iteritems(), key=lambda x : x[1] ) print sorted_x

Output result: [(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
orted_x = sorted( x.iteritems(), key=lambda x : x[1], reverse=True) print sorted_x
Output result: [(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]

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