Python: Sorting a list using another list as an order

Question

I'm curious if there is a way to optimize in a situation I'm facing currently.

I have a list of strings representing categories to group and order data by:

['first', 'third', 'second']

This corresponds to a list of dicts containing objects of those categories which need to be sorted according to them:

[{'color':'yellow', 'section':'third'},{'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]

The data list should be sorted via the order given in the first set, in this case resulting in:

[{'color':'red', 'section':'first'},{'color':'yellow', 'section':'third'},{'color': 'blue', 'section':'second'}]

My current solution:

sortedList = []
for section in orderList:
  for item in dataList:
    if item['section'] == section: sortedList.append(item)

Is there a cleaner way I can be sorting this?

There may be more attributes, but they are all unique keys referencing a single string. — DivinusVox
– DivinusVox, Commented Mar 28, 2013 at 9:23

Volatility · Accepted Answer · 2013-03-28 09:18:29Z

3

You can use the built in sorted function.

>>> lst = ['first', 'third', 'second']
>>> dcts = [{'color':'yellow', 'section':'third'}, {'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]
>>> sorted(dcts, key=lambda dct: lst.index(dct['section']))
[{'section': 'first', 'color': 'red'}, {'section': 'third', 'color': 'yellow'}, {'section': 'second', 'color': 'blue'}]

answered Mar 28, 2013 at 9:18

Volatility

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Comments

jamylak · Accepted Answer · 2013-03-28 09:26:46Z

3

>>> dicts = [{'color':'yellow', 'section':'third'},{'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]
>>> L = ['first', 'third', 'second']
>>> order = dict(zip(L, range(len(L)))) # Dictionary for O(1) lookup
>>> sorted(dicts, key=lambda d: order[d['section']])
[{'color': 'red', 'section': 'first'}, {'color': 'yellow', 'section': 'third'}, {'color': 'blue', 'section': 'second'}]

This method will be O(N) instead of O(N log N) for the sort:

>>> sorted_sections = ['first', 'third', 'second']
>>> dicts = [{'color':'yellow', 'section':'third'},{'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]
>>> dict_by_section = {d['section']:d for d in dicts}
>>> [dict_by_section[section] for section in sorted_sections]
[{'color': 'red', 'section': 'first'}, {'color': 'yellow', 'section': 'third'}, {'color': 'blue', 'section': 'second'}]

edited Mar 28, 2013 at 9:26

answered Mar 28, 2013 at 9:19

jamylak

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5 Comments

DSM Over a year ago

The .index approaches work fine for small sizes but using a dict will scale better.

georg Over a year ago

There are two problems with this "optimization". First, building a helper dict is O(n), second, it requires twice as much memory.

jamylak Over a year ago

@thg435 I don't see those as problems. Also it won't require twice as much memory since it's only storing references to each dictionary.

jamylak Over a year ago

@DivinusVox Yeah but that's why I asked if they would exist or not and you said they were all unique keys. Anyway it's all good

DivinusVox Over a year ago

@jamylak Aha, I misread your original comment. My bad. Thanks for the help.

NPE · Accepted Answer · 2013-03-28 09:18:09Z

2

You could just use sorted() with key:

In [6]: o = ['first', 'third', 'second']

In [7]: l = [{'color':'yellow', 'section':'third'},{'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]

In [8]: sorted(l, key=lambda x:o.index(x['section']))
Out[8]: 
[{'color': 'red', 'section': 'first'},
 {'color': 'yellow', 'section': 'third'},
 {'color': 'blue', 'section': 'second'}]

This does a linear search on o. If o can be large, @jamylak's solution should be preferred.

answered Mar 28, 2013 at 9:18

NPE

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Comments

Serdalis · Accepted Answer · 2013-03-28 09:18:32Z

2

here is a more optimised version for you:

sort_key = lambda x: ks.index(x['section'])

print(sorted(dicts, key=sort_key))

answered Mar 28, 2013 at 9:18

Serdalis

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Comments

Wei Li · Accepted Answer · 2013-03-28 09:26:41Z

0

orderList = ['first', 'third', 'second']
dataList = [{'color':'yellow', 'section':'third'},{'color':'red', 'section':'first'}, {'color': 'blue', 'section':'second'}]

orderDict = dict((v,offset) for offset, v in enumerate(orderList))

print sorted(dataList, key=lambda d: orderDict[d['section']])

answered Mar 28, 2013 at 9:26

Wei Li

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