Sorting one list to match another in python

Question

Suppose I have these lists:

ids = [4, 3, 7, 8]
objects = [
             {"id": 7, "text": "are"},
             {"id": 3, "text": "how"},
             {"id": 8, "text": "you"},
             {"id": 4, "text": "hello"}
          ]

How can I sort the objects so the order of their ids matches ids? I.e. to get this result:

objects = [
             {"id": 4, "text": "hello"},
             {"id": 3, "text": "how"},
             {"id": 7, "text": "are"},
             {"id": 8, "text": "you"}
          ]

Note that the answers assume 'id' is some unique property among the objects. — zr0gravity7
– zr0gravity7, Commented Jul 16, 2023 at 15:08

Pavel Anossov · Accepted Answer · 2013-04-08 13:55:32Z

11

object_map = {o['id']: o for o in objects}
objects = [object_map[id] for id in ids]

answered Apr 8, 2013 at 13:55

Pavel Anossov

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1 Comment

mgilson Over a year ago

I think I would factor the dict-comp out so you're not doing it over and over again.

NPE · Accepted Answer · 2013-04-08 13:55:52Z

2

In [25]: idmap = dict((id,pos) for pos,id in enumerate(ids))

In [26]: sorted(objects, key=lambda x:idmap[x['id']])
Out[26]: 
[{'id': 4, 'text': 'hello'},
 {'id': 3, 'text': 'how'},
 {'id': 7, 'text': 'are'},
 {'id': 8, 'text': 'you'}]

answered Apr 8, 2013 at 13:55

NPE

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2 Comments

Jon Clements Over a year ago

Strangely almost char for char what I had in my text editor ;)

NPE Over a year ago

@JonClements: We clearly spend far too much time on this site ;)

Yarkee · Accepted Answer · 2013-04-08 13:58:43Z

0

>>> ids = [4,3,7,8]
>>> id_orders = {}
>>> for i,id in enumerate(ids):
...     id_orders[id] = i
... 
>>> id_orders
{8: 3, 3: 1, 4: 0, 7: 2}
>>> 
>>> sorted(objs, key=lambda x: id_orders[x['id']])

answered Apr 8, 2013 at 13:58

Yarkee

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Comments

Melebius · Accepted Answer · 2023-02-20 09:24:09Z

0

Use sorted with a custom key function which simply gets index from ids:

sorted(objects, key=lambda x: ids.index(x['id']))

(Inspired by NPE’s answer.)

answered Feb 20, 2023 at 9:24

Melebius

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2 Comments

Timmmm Over a year ago

Ah yeah this is a much nicer answer.

Timmmm Over a year ago

Actually I take that back, this .index() gives this suboptimal complexity.

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