Python sorting a dictionary whose value is a multidimensional list

Question

I have the following dictionary:

DICT = {'ABC': [[u'2012-07-19', 1.22, 1, 1.22, 1, 0], [u'2012-04-26', 1.33, 210, 1.33, 210, 0], [u'2011-11-14', 1.25, 171, 1.32, 241, 70]], 
'XYZ': [[u'2012-02-17', 1.82, 59, 1.25, 182, 123], [u'2011-04-06', 1.50, 30, 1.18, 54, 24], [u'2011-01-06', 1.21, 38, 1.17, 41, 3], [u'2010-06-21', 1.28, 26, 1.28, 26, 0]]}

Values are two dimensional list.

How can I sort keys based on the third column of values.

Let me explain: The third column minimum value of key 'ABC' is 1.22 (among 1.22,1.33 and 1.32). The third column minimum value of key 'XYZ' is 1.17 (among 1.25,1.18,1.17 and 1.28). Since 1.17 > 1.22 'XYZ' will be first and then 'ABC'.

Are you trying to sort the dictionary? If so, that is impossible on the account of being represented most likely as a binary tree, thereby not being sortable. — Nino
– Nino, Commented Aug 1, 2012 at 1:29
Your code example appears as though it will be very difficult for others to read or maintain. I highly recommend creating a class, in this case. — Cryo
– Cryo, Commented Aug 1, 2012 at 1:35
@kojiro Thanks for your answer. I was trying with itemgetter in the itertools module — Vinod
– Vinod, Commented Aug 1, 2012 at 9:58

kojiro · Accepted Answer · 2012-08-01 01:37:00Z

2

You can use the python builtin sorted with a key function to pick out the specific characteristic of the iterable you want to use as a sort key:

>>> [l[1] for l in DICT['XYZ']]
[1.82, 1.5, 1.21, 1.28]
>>> min(l[1] for l in DICT['ABC'])
1.22
>>> sorted(DICT, key=lambda k: min(l[1] for l in DICT[k]))
['XYZ', 'ABC']

edited Aug 1, 2012 at 1:37

answered Aug 1, 2012 at 1:30

kojiro

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