python列表字典排序

Posted by yan on February 18, 2019

按字典值排序(默认为升序)

x = {1:2, 3:4, 4:3, 2:1, 0:0}

  1. 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
  2. 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()

更为常用的方法是,用lambda表达式

  1. sorted_x = sorted(x.iteritems(), key=lambda x : x[1])
    print sorted_x ——-»» [(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
  2. 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}]

  1. sorted_x = sorted(x, key=operator.itemgetter(‘name’))
    print sorted_x ——-»» [{‘age’: 10, ‘name’: ‘Bart’}, {‘age’: 39, ‘name’: ‘Homer’}]
  2. sorted_x = sorted(x, key=operator.itemgetter(‘name’), reverse=True)
    print sorted_x ——-»» [{‘age’: 39, ‘name’: ‘Homer’}, {‘age’: 10, ‘name’: ‘Bart’}]
  3. sorted_x = sorted(x, key=lambda x : x[‘name’])
    print sorted_x ——-»» [{‘age’: 10, ‘name’: ‘Bart’}, {‘age’: 39, ‘name’: ‘Homer’}]
  4. sorted_x = sorted(x, key=lambda x : x[‘name’], reverse=True)
    print sorted_x ——-»» [{‘age’: 39, ‘name’: ‘Homer’}, {‘age’: 10, ‘name’: ‘Bart’}]