sorting tuples in list, using lambda function

Question

What I am trying to do is sort a list of tuples, that the first cell in the tuple can be many nested tuples, but the second cell is always a int value,

My code:

t = [('d',2), ('b', 1), (('a', 'c'), 2)]
sorted(t, key=lambda x: x[1], reverse=True)

But this is not working, the last nested tuple stays in the end. any ideas?

I cannot reproduce your issue; I get [('d', 2), (('a', 'c'), 2), ('b', 1)]. Or did you expect (('a', 'c'), 2) to be sorted first? — Martijn Pieters
– Martijn Pieters, Commented Jan 2, 2014 at 18:30
And note that sorted() returns a new, sorted list. If you wanted t to be sorted in-place, call t.sort() instead. — Martijn Pieters
– Martijn Pieters, Commented Jan 2, 2014 at 18:33
that was the issue, sort and not sorted(), i didn't understand why the list wont change! — user2918984
– user2918984, Commented Jan 2, 2014 at 18:36

Buddhima Gamlath · Accepted Answer · 2014-01-02 18:31:55Z

4

The sorted method return a new sorted list. It does not modify the existing list. On the other hand, list.sort() method sort the list in-place.

t = [('d',2), ('b', 1), (('a', 'c'), 2)]

To sort this use either

t = sorted(t, key=lambda x: x[1], reverse=True)

or

t.sort(t, key=lambda x: x[1], reverse=True)

answered Jan 2, 2014 at 18:31

Buddhima Gamlath

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