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I have a list of orders (each order is a dict) which looks like this (simplified version) :

[{'name' : XXX,
  'id' : { 'order_id_local' : 'xxx_001'}},
 {'name' : XXX,
  'id' : { 'order_id_local' : 'xxx_002'}},
 {'name' : XXX,
  'id' : {}},
 {'name' : XXX,
  'id' : { 'order_id_local' : 'xxx_002'}},
 {'name' : XXX,
  'id' : { 'order_id_local' : 'xxx_003'}},
  ...]

As you can see there could be duplicate for the key 'order_id_local' but also nothing. What I would like to do is to get the last distinct 3 'order_id_local' in a list, beginning from the last one. Here it will be ['xxx_0003', 'xxx_002', 'xxx_001'].

What i did is :

 id_orders = [x['id']['order_id_local'] for x in order_list if 'order_id_local' in x['id']]  
 id_orders = [x for x in id_orders if x is not None]
 id_orders = list(reversed(sorted(set(id_orders))[-3:]))

It works but when i see this id_orders three times and those nested functions, i'm wondering if there is no a more efficient and pythonic way to do this.

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  • \$\begingroup\$ How large is your actual order_list ? \$\endgroup\$ Jan 21, 2020 at 7:38
  • \$\begingroup\$ around an hundred orders per name, and i get the list for each name \$\endgroup\$
    – TmSmth
    Jan 21, 2020 at 22:44

1 Answer 1

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Use a dict comprehension to keep track of the last index for each order_id_local and skip blank entries:

local_ids = {order['id']['order_id_local']:n for n,order in enumerate(data) if order['id']}

Then reverse sort the dictionary keys by their value and slice off the first 3:

sorted(local_ids.keys(), reverse=True, key=local_ids.get)[:3]

result:

['xxx_003', 'xxx_002', 'xxx_001']
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