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@shuivin
Last active Jan 4, 2021
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Python字典按值排序、包含字典的列表按字典值排序的方法
"""
Python字典按值排序、包含字典的列表按字典值排序的方法
官方文檔參考http://www.python.org/dev/peps/pep-0265/
"""
#-*- encoding=utf-8 -*-
import operator
#按字典值排序(默认为升序)
x = {1:2, 3:4, 4:3, 2:1, 0:0}
sorted_x = sorted(x.iteritems(), key=operator.itemgetter(1))
print sorted_x
#[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
#如果要降序排序,可以指定reverse=True
sorted_x = sorted(x.iteritems(), key=operator.itemgetter(1), reverse=True)
print sorted_x
#[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
#或者直接使用list的reverse方法将sorted_x顺序反转
#sorted_x.reverse()
#取代方法是,用lambda表达式
sorted_x = sorted(x.iteritems(), key=lambda x : x[1])
print sorted_x
#[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
sorted_x = sorted(x.iteritems(), key=lambda x : x[1], reverse=True)
print sorted_x
#[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
#包含字典dict的列表list的排序方法与dict的排序类似,如下:
x = [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]
sorted_x = sorted(x, key=operator.itemgetter('name'))
print sorted_x
#[{'age': 10, 'name': 'Bart'}, {'age': 39, 'name': 'Homer'}]
sorted_x = sorted(x, key=operator.itemgetter('name'), reverse=True)
print sorted_x
#[{'age': 39, 'name': 'Homer'}, {'age': 10, 'name': 'Bart'}]
sorted_x = sorted(x, key=lambda x : x['name'])
print sorted_x
#[{'age': 10, 'name': 'Bart'}, {'age': 39, 'name': 'Homer'}]
sorted_x = sorted(x, key=lambda x : x['name'], reverse=True)
print sorted_x
#[{'age': 39, 'name': 'Homer'}, {'age': 10, 'name': 'Bart'}]
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