Re: Problem with the sort() function

[Nick Coghlan]

clementine wrote:
Hi,
I have an array of arrays in the form of 
list = [[3,'fork',0.3,1],[2,'fork,0.1,2],[3,'exec',0.2,2]]

The in-built sort(),list.sort() sorts on the first element, if the first
elts are equal then it sorts on the second elt and so on...But i really
dont want to search on the second elt if the first elts are equal...the
1-D lists shud be left in the same position i.e. i want the sorted list to
be [[2,'fork',0.1,2],[3,'fork,0.3,1],[3,'exec',0.2,2]] and not
[[2,'fork',0.1,2],[3,'exec',0.2,2],[3,'fork,0.3,1]].
Try this:
Py> from operator import itemgetter
Py> list = [[3,'fork',0.3,1],[2,'fork',0.1,2],[3,'exec',0.2,2]]
Py> list.sort(key=itemgetter(0))
Py> list
[[2, 'fork', 0.10000000000000001, 2], [3, 'fork', 0.29999999999999999, 1], [3, '
exec', 0.20000000000000001, 2]]
If the 'key' argument isn't accepted (i.e. you aren't using Python 2.4), you'll 
need to do the decoration manually:

def mysort(iterable, cmp=None, key=None, reverse=False):
    "return a sorted copy of its input"
    if sys.version_info >= (2,4):
        return sorted(iterable, cmp, key, reverse)
    seq = list(iterable)
    if reverse:
        seq.reverse()        # preserve stability
    if key is not None:
        seq = [(key(elem), i, elem) for i, elem in enumerate(seq)]
    seq.sort(cmp)
    if key is not None:
        seq = [elem for (key, i, elem) in seq]
    if reverse:
        seq.reverse()
    return seq
list = mysort([[3,'fork',0.3,1],[2,'fork',0.1,2],[3,'exec',0.2,2]],
               key=lambda x: x[0])
(Taken from Raymond's code in:
http://mail.python.org/pipermail/python-list/2005-January/263275.html)
Cheers,
Nick.
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Nick Coghlan   |   [EMAIL PROTECTED]   |   Brisbane, Australia
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