How sorted lambda with two keys?

Question

i'm uses this

ThreadList.append([''.join(map(str,[Thread])), date, rrf[5]])

SendThreadList = ''

ThreadList = sorted(ThreadList, key=lambda Entry: Entry[1], reverse=True)
ThreadList = sorted(ThreadList, key=lambda Entry: Entry[2], reverse=True)

sorted show only the second sorted, how to do sorted with two entrys in key ??

What are you trying to achieve with ''.join(map(str,[Thread]))? Applying map, I get ''.join([str(Thread)]) which of course is just str(Thread) — Eric
– Eric, Commented Jan 6, 2013 at 21:21

Jon Clements · Accepted Answer · 2013-01-06 21:09:05Z

6

If you want to sort by Entry[1], then Entry[2] then you make the lambda return both as a tuple...:

ThreadList = sorted(ThreadList, key=lambda Entry: (Entry[1], Entry[2]), reverse=True)

What's sometimes more readable (and potentially slightly quicker) is to make use of the operator module and use key=operator.itemgetter(1, 2)

answered Jan 6, 2013 at 21:09

Jon Clements

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4 Comments

Jon Clements Over a year ago

@GeorgeSet. Sorry - I don't understand - could you try re-phrasing please?

George Set. Over a year ago

ThreadList = sorted(ThreadList, key=lambda Entry: (Entry[1], Entry[2]), reverse=True)

Jon Clements Over a year ago

@GeorgeSet. What I don't get is "check the Entry[2] is 1" ?

Jon Clements Over a year ago

@GeorgeSet. Yes, but it may be better as a new question with some context about what you're trying to achieve.

Eric · Accepted Answer · 2013-01-06 21:20:28Z

1

While it is the wrong way to do it, your example will work if you swap the order:

ThreadList = sorted(ThreadList, key=lambda Entry: Entry[2], reverse=True)
ThreadList = sorted(ThreadList, key=lambda Entry: Entry[1], reverse=True)

Python sorts are stable - if two entries tie on [1], they will keep their order from the first sort.

answered Jan 6, 2013 at 21:20

Eric

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