Reorder a Python list of lists according to the second element of embedded lists

Question

I need to reorder a Python list according to the second element of embedded lists:

So this:

[[1, 'A'], [2, 'B'], [3, 'B'], [4, 'A'], [5, 'C']]

Should turn into that:

[[1, 'A'], [4, 'A'], [2, 'B'], [3, 'B'], [5, 'C']]

I am struggling with this piece of code right now:

values = set(map(lambda x:x[1], l))
l = [[y[0] for y in l if y[1]==x] for x in values]

What the most straightforward way to accomplish this?

What does *"struggling with"* mean, exactly? A [mcve] with example inputs and expected and actual outputs is useful. — jonrsharpe, Feb 05 '16 at 00:02
You could have found several answers to your question by Googling its exact title. — TigerhawkT3, Feb 05 '16 at 02:03

score 4 · Accepted Answer · answered Feb 04 '16 at 23:55

4

Using the built-in function sorted:

>>> l = [[1, 'A'], [2, 'B'], [3, 'B'], [4, 'A'], [5, 'C']]
>>> from operator import itemgetter
>>> sorted(l, key=itemgetter(1))
[[1, 'A'], [4, 'A'], [2, 'B'], [3, 'B'], [5, 'C']]

answered Feb 04 '16 at 23:55

eskaev

1,108
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score 2 · Answer 2 · answered Feb 04 '16 at 23:55

2

Like this:

>>> def getKey(item):
...     return item[1]
>>> l = [[1, 'A'], [2, 'B'], [3, 'B'], [4, 'A'], [5, 'C']]
>>> sorted(l, key=getKey)
[[1, 'A'], [4, 'A'], [2, 'B'], [3, 'B'], [5, 'C']]

answered Feb 04 '16 at 23:55

FBruynbroeck

499
3
15

score 1 · Answer 3 · answered Feb 05 '16 at 00:00

1

First sort in reverse by the first, and the sort by the second...

L = [[1, 'A'], [2, 'B'], [3, 'B'], [4, 'A'], [5, 'C']]
L.sort(key=lambda sublist: sublist[0], reverse=True)
L.sort(key=lambda sublist: sublist[1])

answered Feb 05 '16 at 00:00

TheLazyScripter

2,541
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Reorder a Python list of lists according to the second element of embedded lists

3 Answers3