# [Baypiggies] Fwd: manipulating lists question

Martin Falatic martin at falatic.com
Thu Dec 5 11:32:57 CET 2013

```Actually, I'll just throw this out there since it's late and I have to get
up early...

I didn't post my solution originally because the other solutions were so
much more elegant and a skosh more Pythonic too. The list split/join is a
little overcomplicated but you can easily meld this with the other
solutions to get something a bit more concise. However, a minor tweak to
isn't a concern (if it is, it's pretty easy to fix it)

y = dict()
for part in x:
if part[0] in y:
mylist = y[part[0]][0].split(",")
if not part[1] in mylist:
mylist.append(part[1])
y[part[0]][0] = ",".join(mylist)
else:
y[part[0]] = part[1:3] # truncate the rest of the data subset
print y

I got this, which seems like what you wanted:
{'6558': ['NM_001046.2', 'SLC12A2'], '1302': ['NM_080679.2,NM_080680.2',
'COL11A2']}

when I tested this against this:
x=[\
['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']\
,\
['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']\
,\
['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']\
]

On Thu, December 5, 2013 02:22, Martin Falatic 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:COL1
>> 1A
>> 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:COL1
>> 1A
>> 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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>
>

```