Python: sort list of nested lists according to two criteria

Question

In Python, I have a nested list like the following:

[ [[x1,y1,z1], [['0.9', 4], [0.8, 3], [0.5, 10], [0.1, 11]], 
  [[x2,y2,z2], [['1.0', 8], [0.8, 3], [0.2, 1], [0.1, 8]]
...]

So each element is in the form:

 [[3-tuple], [[val1, occurrences_of_val1], [val2, occurrences_of_val2],...]]

The second nested list is already sorted by the first item (val1 > val2 > val3 ...), but I want also the 3-tuples [x,y,z] to appear sorted in descending order according to two criteria:

value of val1 (highest first)
in case of same val1, highest occurrences_of_val1 (possibly applied to val2 if the two values above are the same)

How do I do this? Probably with itemgetter, but I'm not sure in this case.

Could you clarify the link between (x,y,z) and [[val_1, nb_1],[val_2,nb_2],...]]` ? — Pierre GM
– Pierre GM, Commented Aug 20, 2012 at 14:27
@RickyRobinson: x[1][0] gives you both items [val1, occurrences_of_val1], which will sort the way you requested. — Steven Rumbalski
– Steven Rumbalski, Commented Aug 20, 2012 at 14:35
@mgilson: That was not originally specified, but this would do it: yourList.sort(key=lambda x: x[1], reverse=True) — Steven Rumbalski
– Steven Rumbalski, Commented Aug 20, 2012 at 14:39
@RickyRobinson: Then don't narrow down to x[1][0], just leave the key as x[1]. — Steven Rumbalski
– Steven Rumbalski, Commented Aug 20, 2012 at 14:41

Steven Rumbalski · Accepted Answer · 2012-08-20 14:39:49Z

7

yourList.sort(key=lambda x: x[1], reverse=True)

answered Aug 20, 2012 at 14:39

Steven Rumbalski

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8 Comments

mgilson Over a year ago

I was just about to suggest you post this as an answer.

DSM Over a year ago

Do we know what's going on with the fact val1 is a string but none of the others are?

Steven Rumbalski Over a year ago

@DSM: I noticed that. It will cause issues if the values are not between '0.0' and '9.9'. There is some datatype strangeness. Also, I think the OP might be better off with a list of objects.

Ricky Robinson Over a year ago

I just realised val1 is indeed a string, but it seems that values (they are included in [0,1]) get sorted in the expected way.

Ricky Robinson Over a year ago

It's ok then, val_i is in [0,1] for each i

|

kiriloff · Accepted Answer · 2013-05-08 08:12:12Z

3

I tested this

X=[ [[4,5,6], [[3.0, 4], [0.8, 3], [0.5, 10], [0.1, 11]]],
  [[2,1,3], [[2.0, 8], [0.8, 3], [0.2, 1], [0.1, 8]]]]
>>> X.sort(key=lambda x: x[1])
>>> X
[[[2, 1, 3], [[2.0, 8], [0.80000000000000004, 3], [0.20000000000000001, 1], [0.10000000000000001, 8]]], [[4, 5, 6], [[3.0, 4], [0.80000000000000004, 3], [0.5, 10], [0.10000000000000001, 11]]]]

Most importantly, X.sort(key=lambda x: x[1]) is sorting by second element of X, that is by this elements [[val1, occurrences_of_val1], [val2, occurrences_of_val2],...]. Implicitely, it is sorting by val1, then in case of equality by occurrences_of_val1 ...

answered May 8, 2013 at 8:12

kiriloff

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1 Comment

BVBC Over a year ago

Can you explain how this line works: X.sort(key=lambda x: x[1])? It works but I don't quite understand it. Isn't it we need to pass in a function as an argument to the list.sort([fun])?

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