How to have two different sort orders within the same list?

Question

I have a sorted list such as

[[1, 'start'], [13, 'start'], [13, 'end'], [15, 'end']]

which is the output of my_list.sort(key=lambda x: x[0]). However, if there is a tie with the 0th element of each sub-list, as is the case with the middle two elements, then I need the sort to have the lists with end come first. So the desired output would be

[[1, 'start'], [13, 'end'], [13, 'start'], [15, 'end']]

`sorted(my_list)` is sufficient here (try it yourself). It works because `end` is before `start` alphabetically. — jpp, Jan 29 '19 at 16:07

score 1 · Accepted Answer · edited Oct 29 '19 at 14:25

1

You can do it sorting by another element, a boolean, indicating whether or not the second element is end:

my_list.sort(key=lambda x: (x[0], x[1] != 'end'))
[[1, 'start'], [13, 'end'], [13, 'start'], [15, 'end']]

This will be using the following list of tuples in order to sort the list:

[(x[0], x[1] != 'end') for x in my_list]
[(1, True), (13, True), (13, False), (15, False)]

Where the second element will be False if it is not end in my_list. Now sorting this array will yield the expected result, given that False i.e. (0) will come before True (1) when the first elements match:

sorted([(x[0], x[1] != 'end') for x in my_list])
[(1, True), (13, False), (13, True), (15, False)]

As @jpp points out you could simply sort by the second element lexicographically in this case:

my_list.sort()
[[1, 'start'], [13, 'end'], [13, 'start'], [15, 'end']]

edited Oct 29 '19 at 14:25

wjandrea

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answered Jan 29 '19 at 16:02

yatu

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Can you explain how the tuple in the lambda works? – TheRealFakeNews Jan 29 '19 at 16:03
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In the case where you only have `start` and `end`, `(x[0], x[1])` is sufficient. – jpp Jan 29 '19 at 16:05
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Yes good point @jpp lexicographic order is good enough here – yatu Jan 29 '19 at 16:09
@yatu I think you have a mistake in your example. Shouldn't the output of `[(x[0], x[1] != 'end') for x in my_list]` be `[(1, True), (13, False), (13, False), (15, False)]`? – TheRealFakeNews Jan 29 '19 at 16:25
No, you can check it yourself, it gives the mentioned output which is then used to sort – yatu Jan 29 '19 at 16:27
Oh oops ><. Thanks! – TheRealFakeNews Jan 29 '19 at 16:31

score 0 · Answer 2 · answered Jan 29 '19 at 16:14

0

try this

my_list.sort(key = lambda x:(x[0],x[1]))

answered Jan 29 '19 at 16:14

jamil

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`key=lambda x: (x[0], x[1])` is redundant. – wjandrea Oct 29 '19 at 14:26

How to have two different sort orders within the same list?

2 Answers2