Sorting a list created from a parsed xml message

[Thomas Jollans]

On 07/21/2010 03:38 PM, kaklis at gmail.com wrote:
> On Jul 21, 9:04 am, "kak... at gmail.com" <kak... at gmail.com> wrote:
>> On Jul 21, 8:58 am, Stefan Behnel <stefan... at behnel.de> wrote:
>>
>>
>>
>>> kak... at 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