How to sort a list of lists according to the first element of each list?
For example, giving this unsorted list:
[[1,4,7],[3,6,9],[2,59,8]]
The sorted result should be:
[[1,4,7],[2,59,8],[3,6,9]]
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glibdud
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Shubham Sharda
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7Plain and simple use `sorted(d)`? – GIZ Apr 30 '16 at 13:50
3 Answers
75
Use sorted function along with passing anonymous function as value to the key argument. key=lambda x: x[0] will do sorting according to the first element in each sublist.
>>> lis = [[1,4,7],[3,6,9],[2,59,8]]
>>> sorted(lis, key=lambda x: x[0])
[[1, 4, 7], [2, 59, 8], [3, 6, 9]]
Avinash Raj
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4You should avoid using `l` as variable name, see [PEP 8](https://www.python.org/dev/peps/pep-0008/#names-to-avoid). – Delgan Apr 30 '16 at 13:48
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4@NitishKumarPal You can reverse the result of sorting the list.`sorted(lis, key=lambda x: x[0])[::-1]` – rAntonioH Nov 30 '19 at 18:55
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1`sorted(lis, key=lambda x: -x[0])` is ok to sort-reverse numeric elements but will fail for strings, so if needed use @rAntonioH `sorted(lis, key=lambda x: x[0])[::-1]` – jerome Apr 21 '21 at 08:35
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If you're sorting by first element of nested list, you can simply use list.sort() method.
>>> lis = [[1,4,7],[3,6,9],[2,59,8]]
>>> lis.sort()
>>> lis
[[1, 4, 7], [2, 59, 8], [3, 6, 9]]
If you want to do a reverse sort, you can use lis.reverse() after lis.sort()
>>> lis.reverse()
>>> lis
[[3, 6, 9], [2, 59, 8], [1, 4, 7]]
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li = [[1,4,7],[3,6,9],[2,59,8]]
li.sort(key=lambda x: int(x[0]))
This will sort in place changing the original list though. If you want to keep the original list, it is better to use:
sorted(li, key = lambda x: int(x[0]))
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1What's the point of casting to Int if the list is already all ints? – InfiniteLoop Oct 16 '21 at 10:20