Sorting a list of list

Question

So I have a list like below:

points = [[0, 0], [5, 3], [0, 5], [0, 2], [1, 3], [5, 3]]

I have been using

points.sort(key=lambda pair: pair[0])

to sort the list. But this only sorts it by first value without looking at the second value.

The result of this code is as below:

[[0, 0], [0, 5], [0, 2], [1, 3], [5, 3], [5, 3]]

However, I want to sort is such that the result is:

[[0, 0], [0, 2], [0, 5], [1, 3], [5, 3], [5, 3]]

How do I do it?

Also, while we are at it, how do i remove the duplicates from this list? The final result should be like:

[[0, 0], [0, 2], [0, 5], [1, 3], [5, 3]]

No sort key at all will sort the pairs lexicographically: `points.sort()` — user2390182, Jan 09 '18 at 06:21

score 5 · Accepted Answer · answered Jan 09 '18 at 06:31

5

Python already sorts iterable containers (tuples, lists, ...) and strings lexicographically.

>>> points = [[0, 0], [5, 3], [0, 5], [0, 2], [1, 3], [5, 3]]
>>> sorted(points)
[[0, 0], [0, 2], [0, 5], [1, 3], [5, 3], [5, 3]]

answered Jan 09 '18 at 06:31

timgeb

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@NutanNepal not at all, see [my first question](https://stackoverflow.com/questions/20145842/python-sorting-by-multiple-criteria). I didn't get lexicographical sorting either. – timgeb Jan 09 '18 at 06:47

RoadRunner · Answer 2 · 2018-01-09T07:22:23.240

You can have a sort key that first sorts by the first element x[0], and if their are ties, sort by the second element x[1]:

>>> points = [[0, 0], [5, 3], [0, 5], [0, 2], [1, 3], [5, 3], [0, 4]]
>>> sorted(points, key = lambda x: (x[0], x[1]))
[[0, 0], [0, 2], [0, 4], [0, 5], [1, 3], [5, 3], [5, 3]]

If you don't want any duplicates in this final list, then you can do this:

points = [[0, 0], [5, 3], [0, 5], [0, 2], [1, 3], [5, 3]]

sorted_lst = sorted(points, key = lambda x: (x[0], x[1]))

seen = set()
no_dups = []
for lst in sorted_lst:
    curr = tuple(lst)
    if curr not in seen:
        no_dups.append(lst)
        seen.add(curr)

print(no_dups)
# [[0, 0], [0, 2], [0, 5], [1, 3], [5, 3]]

Which has a set seen which keeps track of what sublists have been added, and a list no_dups where the lists get added. If an element is not in seen and add it to the final list, and add it to seen, which indicates that the element has already been added.

Also since type list is not a hashable type, you cannot add them directly to a set. In the above code, the lists are converted to tuples and added to seen, which is a hashable type.

Thanks for your answer to remove duplicates. That was helpful. — nnut, Jan 09 '18 at 06:57
@NutanNepal No worries. Be sure to give it an upvote if you found helpful. — RoadRunner, Jan 09 '18 at 06:58
I tried to. But it says I can't upvote when I'm less than 15 rep. I was able to do it just a while ago though. — nnut, Jan 09 '18 at 07:00

score 1 · Answer 3 · answered Jan 09 '18 at 06:27

1

import itertools

points = [[0, 0], [5, 3], [0, 5], [0, 2], [1, 3], [5, 3]]
points.sort()
nList = list(points for points,_ in itertools.groupby(points))
print nList

Result:

[[0, 0], [0, 2], [0, 5], [1, 3], [5, 3]]

answered Jan 09 '18 at 06:27

Rakesh

81,458
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score 0 · Answer 4 · answered Jan 09 '18 at 06:25

0

Try itemgetter

from operator import itemgetter
sorted(points, key=itemgetter(0, 1))

answered Jan 09 '18 at 06:25

Ernest S Kirubakaran

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Sorting a list of list

4 Answers4