how to address zip list in sort lambda

Question

I have a class with 2 attributes which are lists themselves:

class data...:
    list1 = [["g1", 2.0], ["x1", 3.0]...] # n elements
    list2 = [[2, 4, 5],[3, 2, 1]...] # n elements

I need to zip sort both lists, based on value of the second element of list2.

zipped = zip(dataobj.list1, dataobj.list2)
zipped.sort(cmp = lambda k: dataobj.list2[2])

This seems to not work.

How do I reference the second element of dataobj.list2[2] as this is not working and gave me the following error:

TypeError: <lambda>() takes exactly 1 argument (2 given)

zipped.sort(key = lambda k: k[1][1])?

Julien
– Julien

2016-08-10 06:34:21 +00:00
Commented Aug 10, 2016 at 6:34 — Julien
– Julien, Commented Aug 10, 2016 at 6:34
zipped.sort(key = lambda k: k[1])

Asif
– Asif

2016-08-10 06:41:11 +00:00
Commented Aug 10, 2016 at 6:41 — Asif
– Asif, Commented Aug 10, 2016 at 6:41

Mureinik · Accepted Answer · 2016-08-10 06:38:09Z

2

cmp should be a reference to a function that compares two values. Instead, you need something much simpler - a key field.

The easiest way would be to reference the value directly from zipped instead of going back to the original value in list2. Note, BTW, that lists in python are zero-based, so the second element would be [1], not [2]. To make a long story short:

zipped.sort(key = lambda k : k[1][1])

answered Aug 10, 2016 at 6:38

Mureinik

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