Sort list of dictionaries according to key, value such that the only first three of the key, value pair should come in the given list of dict

Question

Sort the dict by the key 'ordering' if ordering is greater then that dict should appear first and then the dict which contain next greater ordering and the first three greater ordering dicts should be stored in the list.

Input:

my_list = [{u'item_group': u'Bits Special',
         u'item_name': u'Tool Bits-Metric',
         u'ordering': 1},
        {u'item_group': u'Performance Drills 3X',
         u'item_name': u'Drills-Short-Metric',
         u'ordering': 3},
        {u'item_group': u'Hand Taps',
         u'item_name': u'Shank-Metric',
         u'ordering': 4},
        {u'item_group': u'Tool Bits',
         u'item_name': u'T42-Square HSS Tool Bits-BSW',
         u'ordering': 2,}]

Expected o/p:

     op_list = [{u'item_group': u'Hand Taps',u'item_name': u'Shank-Metric',
           u'ordering': 4},{u'item_group': u'Performance Drills 3X',
           u'item_name': u'Drills-Short-Metric',u'ordering': 3}, 
          {u'item_group': u'Tool Bits',u'item_name': u'T42-Square HSS 
           Tool Bits-BSW',u'ordering': 2,}]

I've tried :

 for i in mylist:
       if i['ordering']>mylist['ordering']:
          op_list.append(i)

Answer 1

There are examples showing how to sort a list, giving your own key to use for comparison. You need this revsered (ie descending) and then need to take the top three.

Simplifying what you have, consider

my_list=[{u'ordering':1}, {u'ordering':3}, {u'ordering':4}, {u'ordering':2}]

you can sort this:

>>> sorted(my_list, key=lambda d: d[u'ordering'], reverse=True)
[{'ordering': 4}, {'ordering': 3}, {'ordering': 2}, {'ordering': 1}]

and slice off what you need:

>>> sorted(my_list, key=lambda d: d[u'ordering'], reverse=True)[:3]
[{'ordering': 4}, {'ordering': 3}, {'ordering': 2}]