[Baypiggies] Fwd: manipulating lists question

Thu Dec 5 11:33:05 CET 2013

In the example i have given, the second and third elements of the larger
list (comp[7] and comp[8]) have a 1:1 mapping after the second element. So
i would like to keep the first element as it is and then collapse or merge
the second and third elements (comp[7] and comp[8]) into a single element:

>>> comp[6]
['6558', 'NM_001046.2', 'SLC12A2', '6037226', '2', 'chr5', '127502453',
'127502454', 'het-ref', 'snp', 'A', 'T', 'A', '185', '113', '184', '112',
'VQHIGH', 'VQHIGH', '', '', '', '', '259974', '9', '6', '6', '15',
'6558:NM_001046.2:SLC12A2:CDS:MISSENSE',
'6558:NM_001046.2:SLC12A2:CDS:NO-CHANGE', 'PFAM:PF01490:Aa_trans', '', '',
'', '0.99', '2', '0.99', '0.998', '1.01', '1.000', '0.5', '0.46', '0.5',
'1', '18', '18', '19', 'ref-identical;onlyA', 'snp', '0.072', '-1',
'SQHIGH']

>>> comp[7]
['1302', 'NM_080679.2', 'COL11A2', '6525172', '2', 'chr6', '33271374',
'33271376', 'het-ref', 'del', 'GT', '', 'GT', '542', '542', '458', '458',
'VQHIGH', 'VQHIGH', '', '', '', '', '71150', '34', '106', '106', '140',
'1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC',
'1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080681.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;6257:NM_021976.3:RXRB:CDS:NO-CHANGE',
'', '', '', '', '0.95', '2', '0.98', '0.998', '0.99', '1.000', '0.46',
'0.42', '0.5', '0', '102', '102', '102', 'ref-identical;onlyA', 'del',
'0.990', '6', 'SQHIGH']

>>> comp[8]
['1302', 'NM_080680.2', 'COL11A2', '6525172', '2', 'chr6', '33271374',
'33271376', 'het-ref', 'del', 'GT', '', 'GT', '542', '542', '458', '458',
'VQHIGH', 'VQHIGH', '', '', '', '', '71150', '34', '106', '106', '140',
'1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC',
'1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080681.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;6257:NM_021976.3:RXRB:CDS:NO-CHANGE',
'', '', '', '', '0.95', '2', '0.98', '0.998', '0.99', '1.000', '0.46',
'0.42', '0.5', '0', '102', '102', '102', 'ref-identical;onlyA', 'del',
'0.990', '6', 'SQHIGH']

After collapsing comp[7] and comp[8] i  get:

>>> collapsed = ['1302', 'NM_080679.2,NM_080680.2', 'COL11A2', '6525172',
'2', 'chr6', '33271374', '33271376', 'het-ref', 'del', 'GT', '', 'GT',
'542', '542', '458', '458', 'VQHIGH', 'VQHIGH', '', '', '', '', '71150',
'34', '106', '106', '140',
'1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC',
'1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080681.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;6257:NM_021976.3:RXRB:CDS:NO-CHANGE',
'', '', '', '', '0.95', '2', '0.98', '0.998', '0.99', '1.000', '0.46',
'0.42', '0.5', '0', '102', '102', '102', 'ref-identical;onlyA', 'del',
'0.990', '6', 'SQHIGH']

So in my larger list, after the modification, comp[6] is the first element
and collapsed the second element.
>>>

On Thu, Dec 5, 2013 at 5:22 AM, Martin Falatic <martin at falatic.com> wrote:

> Ah, genetics! Intriguing...
>
> Do you need anything beyond the third elements of each list? Does the
> third element always map 1:1 with the first, or could it vary? If so, what
> then?
>
> To refer to the simplified example, could you have this?
> x = [['cat', 'NM123', 12], ['cat', 'NM234', 43], ['dog', 'NM56', 65]]
>
> If so, what is the expected output?
>
>  - Marty
>
>
>
> On Thu, December 5, 2013 02:11, Vikram K wrote:
> > i am having some difficulty in applying this to my actual problem
> > although i love the dictionary method. Imagine the following three lists
> > are the first, second and third elements of a larger list:
> >
> >>>> comp[6]
> > ['6558', 'NM_001046.2', 'SLC12A2', '6037226', '2', 'chr5', '127502453',
> > '127502454', 'het-ref', 'snp', 'A', 'T', 'A', '185', '113', '184', '112',
> > 'VQHIGH', 'VQHIGH', '', '', '', '', '259974', '9', '6', '6', '15',
> > '6558:NM_001046.2:SLC12A2:CDS:MISSENSE',
> > '6558:NM_001046.2:SLC12A2:CDS:NO-CHANGE', 'PFAM:PF01490:Aa_trans', '',
> '',
> >  '', '0.99', '2', '0.99', '0.998', '1.01', '1.000', '0.5', '0.46', '0.5',
> >  '1', '18', '18', '19', 'ref-identical;onlyA', 'snp', '0.072', '-1',
> > 'SQHIGH']
> >
> >
> >>>> comp[7]
> > ['1302', 'NM_080679.2', 'COL11A2', '6525172', '2', 'chr6', '33271374',
> > '33271376', 'het-ref', 'del', 'GT', '', 'GT', '542', '542', '458', '458',
> > 'VQHIGH', 'VQHIGH', '', '', '', '', '71150', '34', '106', '106', '140',
> > '1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC',
> >
> '1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080680.2:COL11A
> > 2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080681.2:COL11A2:TSS-UPSTREAM:UNKNOWN-
> > INC;6257:NM_021976.3:RXRB:CDS:NO-CHANGE',
> > '', '', '', '', '0.95', '2', '0.98', '0.998', '0.99', '1.000', '0.46',
> > '0.42', '0.5', '0', '102', '102', '102', 'ref-identical;onlyA', 'del',
> > '0.990', '6', 'SQHIGH']
> >
> >
> >>>> comp[8]
> > ['1302', 'NM_080680.2', 'COL11A2', '6525172', '2', 'chr6', '33271374',
> > '33271376', 'het-ref', 'del', 'GT', '', 'GT', '542', '542', '458', '458',
> > 'VQHIGH', 'VQHIGH', '', '', '', '', '71150', '34', '106', '106', '140',
> > '1302:NM_080680.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC',
> >
> '1302:NM_080679.2:COL11A2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080680.2:COL11A
> > 2:TSS-UPSTREAM:UNKNOWN-INC;1302:NM_080681.2:COL11A2:TSS-UPSTREAM:UNKNOWN-
> > INC;6257:NM_021976.3:RXRB:CDS:NO-CHANGE',
> > '', '', '', '', '0.95', '2', '0.98', '0.998', '0.99', '1.000', '0.46',
> > '0.42', '0.5', '0', '102', '102', '102', 'ref-identical;onlyA', 'del',
> > '0.990', '6', 'SQHIGH']
> >
> >>>>
> >
> > ------
> > Can we apply the dictionary method to the problem where the key of the
> > dictionary is the first element of the three smaller lists
> ('6558','1302',
> >  '1302'). The second and third elements of the larger list (starting with
> >  '1302') need to be collapsed into a single element, based on their
> > second element ( 'NM_080679.2') and ('NM_080680.2') in a way similar to
> > how we had tackled the toy problem:
> >
> > x = [['cat', 'NM123', 12], ['cat', 'NM234', 12], ['dog', 'NM56', 65]]
> >
> >
> >
> >
> >
> >
> >
> > On Thu, Dec 5, 2013 at 4:18 AM, Michiel Overtoom <motoom at xs4all.nl>
> > wrote:
> >
> >
> >>
> >> On Dec 5, 2013, at 10:09, Vikram K wrote:
> >>
> >>
> >>> another option could have been to obtain a dictionary like so: {'dog':
> >>> ['NM56', 65], 'cat': ['NM123,NM234', 12]}
> >>>
> >>
> >> Oh, in that case the code can become somewhat simpler:
> >>
> >>
> >> x = [['cat', 'NM123', 12], ['cat', 'NM234', 12], ['dog', 'NM56', 65]]
> >>
> >> d = {} for key, label, quant in x: if key in d: d[key][0] += ", " +
> label
> >> else:
> >> d[key] = [label, quant]
> >>
> >> print d
> >>
> >>
> >> I agree with Michael that the problem is somewhat underspecified, but
> >> it's a starting point.
> >>
> >> Greetings,
> >>
> >>
> >> --
> >> "If you don't know, the thing to do is not to get scared, but to learn."
> >> -
> >> Ayn Rand
> >>
> >>
> >>
> >>
> >>
> > _______________________________________________
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> > https://mail.python.org/mailman/listinfo/baypiggies
>
>
>
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