sort 2-D list python

Question

I'm relatively new to programming, and I want to sort a 2-D array (lists as they're called in Python) by the value of all the items in each sub-array. For example:

pop = [[1,5,3],[1,1,1],[7,5,8],[2,5,4]]

The sum of the first element of pop would be 9, because 1 + 5 + 3 = 9. The sum of the second would be 3, because 1 + 1 + 1 = 3, and so on.

I want to rearrange this so the new order would be:

newPop = [pop[1], pop[0], pop[3], pop[2]]

How would I do this?

Note: I don't want to sort the elements each sub-array, but sort according to the sum of all the numbers in each sub-array.

arshajii · Answer 1 · 2013-08-28T00:57:03.963

4

You can use sorted():

>>> pop = [[1,5,3],[1,1,1],[7,5,8],[2,5,4]]
>>> newPop = sorted(pop, key=sum)
>>> newPop
[[1, 1, 1], [1, 5, 3], [2, 5, 4], [7, 5, 8]]

You can also sort in-place with pop.sort(key=sum). Unless you definitely want to preserve the original list, you should prefer in-pace sorting.

edited Aug 28 '13 at 00:57

answered Aug 27 '13 at 23:06

arshajii

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in-place sort is recommended unless you absolutely need a new list (+1 anyway) – inspectorG4dget Aug 27 '13 at 23:09
@inspectorG4dget Sure, I added an approach for that as well. – arshajii Aug 27 '13 at 23:13

Óscar López · Answer 2 · 2013-08-27T23:14:10.977

Try this:

sorted(pop, key=sum)

Explanation:

The sorted() procedure sorts an iterable (a list in this case) in ascending order
Optionally, a key parameter can be passed to determine what property of the elements in the list is going to be used for sorting
In this case, the property is the sum of each of the elements (which are sublists)

So essentially this is what's happening:

[[1,5,3], [1,1,1], [7,5,8], [2,5,4]]              # original list
[sum([1,5,3]), sum([1,1,1]), sum([7,5,8]), sum([2,5,4])] # key=sum
[9, 3, 20, 11]                                    # apply key
sorted([9, 3, 20, 11])                            # sort
[3, 9, 11, 20]                                    # sorted
[[1,1,1], [1,5,3], [2,5,4], [7,5,8]]              # elements coresponding to keys

in-place sort is recommended unless you absolutely need a new list (+1 anyway) — inspectorG4dget, Aug 27 '13 at 23:10

score 1 · Accepted Answer · edited May 23 '17 at 11:51

@arshajii beat me to the punch, and his answer is good. However, if you would prefer an in-place sort:

>>> pop = [[1,5,3],[1,1,1],[7,5,8],[2,5,4]]
>>> pop.sort(key=sum)
>>> pop
[[1, 1, 1], [1, 5, 3], [2, 5, 4], [7, 5, 8]]

I have to look up Python's sorting algorithm -- I think it's called Timsort, bit I'm pretty sure an in-place sort would be less memory intensive and about the same speed.

Edit: As per this answer, I would definitely recommend x.sort()

If you wanted to sort the lists in a less traditional way, you could write your own function (that takes one parameter.) At risk of starting a flame war, I would heavily advise against lambda.

For example, if you wanted the first number to be weighted more heavily than the second number more heavily than the third number, etc:

>>> def weightedSum(listToSum):
...     ws = 0
...     weight = len(listToSum)
...     for i in listToSum:
...         ws += i * weight
...         weight -= 1
...     return ws
... 
>>> weightedSum([1, 2, 3])
10
>>> 1 * 3 + 2 * 2 + 3 * 1
10
>>> pop
[[1, 5, 3], [1, 1, 1], [7, 5, 8], [2, 5, 4]]
>>> pop.sort(key=weightedSum)
>>> pop
[[1, 1, 1], [1, 5, 3], [2, 5, 4], [7, 5, 8]]
>>> pop += [[1, 3, 8]]
>>> pop.sort(key=weightedSum)
>>> pop
[[1, 1, 1], [1, 5, 3], [1, 3, 8], [2, 5, 4], [7, 5, 8]]

sort 2-D list python

3 Answers3