Re: Creating a new data structure while filtering its data origin.
On Mar 28, 4:44 pm, <[EMAIL PROTECTED]> wrote:
> Hi everyone.
>
> I'm trying to work with very simple data structures but I'm stuck in the very
> first steps. If someone has the luxury of a few minutes and can give an
> advice how to resolve this, I'll really appreciate it.
>
> 1- I have a list of tuples like this:
> lista= [(162, 141, 3), (162, 141, 3), (162, 141, 3), (168, 141, 2), (168,
> 141, 2), (168, 141, 2), (201, 141, 1), (213, 141, 1), (203, 141, 1), (562,
> 142, 4), (562, 142, 4), (562, 142, 4), (568, 142, 2), (568, 142, 2), (568,
> 142, 2), (501, 142, 1), (513, 142, 1), (503, 142, 1)]
> and I want to end with a dict like this:
> {141: {1: [203, 213, 201], 2: [168, ], 3: [162, ]}, 142: {1: [503, 513, 501],
> 2: [568, ], 4: [562, ]}}
> the logic of the final output:
> a) the outer dict's key is a set() of the 2rd value of the input.
> b) the inner dict's key is a set() of the 3th value for tuples which 3rd
> value equals a).
> c) the inner list will be fill up with the 1st value of every tuple which
> 3rd value equals b) and its 2rd value equals a).
>
> So far, the only thing it seems I can achieve is the first part:
> outer_dict = dict([(x,dict()) for x in set(row[1] for row in lista)])
>
> >From then on, I'm starting to get tired after several successful failures (I
> >tried with itertools, with straight loops ...) and I don't know which can be
> >the easier way to get that final output.
>
> Thanks in advance.
dict([[b,dict([[k, dict([[x, None] for (x,y,z) in lista if y==b and
z==k]).keys()] for (i,j,k) in lista if j==b])] for (a,b,c) in lista])
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Re: I can't get multi-dimensional array to work...
On Mar 30, 4:56 pm, "erikcw" <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm trying to create a multidimensional data structure. However, id
> doesn't seem to work past the 2nd dimension.
>
> Here is my method:
>
> def endElement(self, name):
> if name == 'row' :
> if not self.data.has_key(self.parent):
> self.data[self.parent] = {}
> elif not self.data[self.parent].has_key(self.child):
> self.data[self.parent][self.child] = []
> self.data[self.parent]
> [self.child].append((self.creativeid, self.clicks, self.imps))
>
> I've tried all kinds of different variations, and I keep getting the
> same result:
>
> Traceback (most recent call last):
> File "sax.py", line 123, in ?
> parser.parse(open('report.xml'))
> File "/usr/lib/python2.4/site-packages/_xmlplus/sax/expatreader.py",
> line 109, in parse
> xmlreader.IncrementalParser.parse(self, source)
> File "/usr/lib/python2.4/site-packages/_xmlplus/sax/xmlreader.py",
> line 123, in parse
> self.feed(buffer)
> File "/usr/lib/python2.4/site-packages/_xmlplus/sax/expatreader.py",
> line 216, in feed
> self._parser.Parse(data, isFinal)
> File "/usr/lib/python2.4/site-packages/_xmlplus/sax/expatreader.py",
> line 315, in end_element
> self._cont_handler.endElement(name)
> File "sax.py", line 51, in endElement
> self.data[self.parent][self.child].append((self.creativeid,
> self.clicks, self.imps))
> KeyError: u'Pickup Trucks'
>
> I have a lot of php experience - am I accidentally doing a "php thing"
> in my code?
>
> Thanks so much for your help!
>
> Erik
I haven't tested it, but superficially I'd suggest giving this a try:
def endElement(self, name):
if name == 'row' :
if not self.data.has_key(self.parent):
self.data[self.parent] = {}
if not self.data[self.parent].has_key(self.child):
self.data[self.parent][self.child] = []
self.data[self.parent]
[self.child].append((self.creativeid, self.clicks, self.imps))
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