On Jul 22, 12:56 pm, Thomas Jollans <tho...@jollans.com> wrote:
> On 07/21/2010 03:38 PM, kak...@gmail.com wrote:
>
>
>
> > On Jul 21, 9:04 am, "kak...@gmail.com" <kak...@gmail.com> wrote:
> >> On Jul 21, 8:58 am, Stefan Behnel <stefan...@behnel.de> wrote:
>
> >>> kak...@gmail.com, 21.07.2010 14:36:
>
> >>>> From the subject of my message it's clear that i get an xml message
> >>>> from a socket,
>
> >>> Not at all, but now that you say it...
>
> >>>> i parse it and the result is a list like the one that
> >>>> follows:
> >>>> ID_Col
> >>>> 4 Server ak ip OFFLINE
>
> >>>> 29 Server and2 ip OFFLINE
>
> >>>> 5 Proxy l34e ip OFFLINE
>
> >>>> 6 Proxy barc ip ONLINE
>
> >>>> 41 Proxy proxy-2 ip ONLINE
>
> >>>> 53 Server server-4 ip ONLINE
>
> >>>> 52 Server server-3 ip ONLINE
>
> >>> Doesn't look like a Python list to me...
>
> >>>> What i want is to print this list sorted by ID_Col?
> >>>> Any Suggestions?
>
> >>> Assuming that the above is supposed to represent a list of tuples, you can
> >>> use the .sort() method on the list and pass operator.itemgetter(0) as
> >>> 'key'
> >>> argument (see the sort() method and the operator module).
>
> >>> Stefan
>
> >> No it is not a Python list at all. This the way i print the parsed
> >> items 'like a list'.
> >> But i want them to be sorted.
>
> > Well i did this:
>
> > SortedServers = []
>
> > for session in sessions:
> > for IP in session.getElementsByTagName("ipAddress"):
> > for iphn in session.getElementsByTagName("hostName"):
> > tempTuple = session.getAttribute("id"),
> > session.getAttribute("type"), iphn.childNodes[0].data,
> > IP.childNodes[0].data, session.getAttribute("status")
>
> Please try to persuade your mail client to not mess up python code, if
> you could. It would make this *so* much easier to read
>
>
>
> > SortedServers.append(tempTuple)
>
> > Sorted = sorted(SortedServers, key=lambda id: SortedServers[0])
>
> Anyway, let's look at that key function of yours:
>
> key=lambda id: SortedServers[0]
>
> translated to traditional function syntax:
>
> def key(id):
> return SortedServers[0]
>
> No matter which item sorted() examines, the key it sorts by is always
> the same (the first item of the WHOLE LIST).
> You want something more like this:
>
> def key(row):
> return row[0]
>
> ergo, what you want, all in all, is either of these:
>
> Sorted = sorted(SortedServers, key=(lambda row: row[0])) # option 1
> SortedServers.sort(key=(lambda row: row[0])) # option 2
>
> option 2, the in-place sort, might be faster.
>
> (and, as Stefan noted, as you probably want a numeric sort, you'll want
> your key to be an int)
>
> > for item in Sorted:
> > print item
>
> > but the list is still unsorted and with u' in front of each item
>
> > (u'4', u'Server', u'aika74', u'ip', u'OFFLINE')
> > (u'29', u'Server', u'ando', u'ip2', u'OFFLINE')
>
> > How do i remove the u'
>
> > Antonis
>
>
Thank you so much for your detailed response!
Antonis K.
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