I wrote:
> Is there a better way to find indices of a subset within a greater list?
This is my hacky solution
Sorry, please disregard my question. I figured out a cleaner implementation
using the table adverb
(; > L:0 I. each <"1 (find =/ list)) { list
NB. 13 : '(; > L:0 I. each <"1 (x =/ y)) { y'
in=:] {~ [: ; [: >L:0 [: I.&.> [: <"1 =/
] find in list
┌─┬─┬─┬─┐
│a│a│b│q│
└─┴─┴─┴─┘
] (<'a') in ('a';'b')
┌─┐
│a│
└─┘
] ('a';'b') in ('a';'b';'c';'a')
┌─┬─┬─┐
│a│a│b│
└─┴─┴─┘
idx=:13 : '(; > L:0 I. each <"1 (x =/ y))'
Which then lets us do things like the following to merge/join tables
(find idx list) { list
My previous hack of namesTenRows=.(; L:0 > (3 : '(<y) ([: I. =) names' )
each namesTen) { data
Becomes: namesTenRows=. (namesTen idx names) { data
Also quite a bit faster than the hacky solution
]1e5 * 1e5 (6!:2) '; > L:0 (3 : ''(<y) ([: I. =) list'' ) each find'
0.589178
]1e5 * 1e5 (6!:2) '(find idx list)'
0.224204
Fun fun
http://www.jsoftware.com/help/dictionary/d420.htm
http://www.jsoftware.com/help/learning/18.htm
I'd be happy to contribute to a wiki page on slicing/dicing tables. I don't
have access to modify the wiki at present.
On Tue, Oct 8, 2013 at 9:20 AM, Joe Bogner <[email protected]> wrote:
> One more -- matching to a list (joining tables)
>
> NB. Get the names of everyone from the frequency table with exactly 10
> matches
>
> ]namesTen=. > 0}"1 (10 ([: I. =) (> 1{"1 nameFrequency)) { nameFrequency
>
> ┌─────┬─────┬─────┬─────┬─────┬──────┬─────┬──────┬────────┬─────┐
>
> │Ruben│Paddy│Cecil│Chico│Edwin│Martin│Kelly│Junior│Jonathan│Duane│
>
> └─────┴─────┴─────┴─────┴─────┴──────┴─────┴──────┴────────┴─────┘
>
>
>
> namesTenRows=.(; L:0 > (3 : '(<y) ([: I. =) names' ) each namesTen) { data
>
>
> NB. Get the unique names in our subset to validate it matches the above
>
> ] ~. (colIdx <'nameFirst') {"1 namesTenRows
>
> ┌─────┬─────┬─────┬─────┬─────┬──────┬─────┬──────┬────────┬─────┐
>
> │Ruben│Paddy│Cecil│Chico│Edwin│Martin│Kelly│Junior│Jonathan│Duane│
>
> └─────┴─────┴─────┴─────┴─────┴──────┴─────┴──────┴────────┴─────┘
>
>
>
> Is there a better way to find indices of a subset within a greater list?
> This is my hacky solution
>
>
> find=.('a';'b';'q')
>
> list=.('a';'b';'z';'c';'a';'q')
>
>
> ] (; > L:0 (3 : '(<y) ([: I. =) list' ) each find) { list
>
> ┌─┬─┬─┬─┐
>
> │a│a│b│q│
>
> └─┴─┴─┴─┘
>
>
> find xxx list
>
>
>
> On Tue, Oct 8, 2013 at 7:14 AM, Ric Sherlock <[email protected]> wrote:
>
>> A relatively simple adverb like the one below can be useful if you a
>> repeatedly trying to access different fields in the table:
>>
>> NB.*getFields a Returns fields (columns) from y specified in x
>>
>> NB. EG: 'fieldname' tablehdr getFields table
>>
>> NB. EG: ('Lastname';'Firstname') hdr getFields dat
>>
>> NB. y is: table of records
>>
>> NB. m is: list of boxed field names in y
>>
>> NB. x is: list of boxed field names to return
>>
>> getFields=: 1 : 0
>>
>> :
>>
>> flds=. boxopen x
>>
>> idx=. m i. flds
>>
>> idx {"1 y
>>
>> )
>>
>>
>> require 'tables/csv'
>>
>> 'hdr dat'=: split readcsv 'c:/tmp/master.csv'
>>
>>
>> 4{. dat hdr getFields~ 'nameFirst';'birthYear';'weight'
>>
>>
>> ┌──────┬────┬───┐
>>
>> │Hank │1934│180│
>>
>> ├──────┼────┼───┤
>>
>> │Tommie│1939│190│
>>
>> ├──────┼────┼───┤
>>
>> │Don │1954│190│
>>
>> ├──────┼────┼───┤
>>
>> │Andy │1972│184│
>>
>> └──────┴────┴───┘
>>
>>
>>
>>
>> On Tue, Oct 8, 2013 at 11:39 PM, Joe Bogner <[email protected]> wrote:
>>
>> > Here are a couple other examples:
>> >
>> > NB. Convert years to a flat list of numbers
>> >
>> > birthYears=. , L:0 (> ". each ((colIdx <'birthYear') {"1 data))
>> >
>> >
>> > NB. Find players born between 1930 and 1950 named Hank
>> >
>> > (I. (birthYears > 1930) * (birthYears < 1950) * (names = <'Hank')) {
>> data
>> >
>> >
>> > NB. Find players born between 1930 and 1950 named Hank or Tommie
>> >
>> > (I. (birthYears > 1930) * (birthYears < 1950) * ( (names = <'Hank') +
>> > (names = <'Tommie'))) { data
>> >
>> >
>> >
>> > NB. http://www.jsoftware.com/jwiki/Essays/DataStructures
>> >
>> > diff=: -. [
>> >
>> > intersect=: e. # [
>> >
>> > union=: ~.@,
>> >
>> >
>> > in=: ] ([: I. =)~ [: < [ NB. 4 : '(<x) ([: I. =)
>> y'
>> >
>> > inT=: 4 : 'x in (colIdx <y) {"1 data'
>> >
>> >
>> > NB. Another style
>> >
>> > NB. Hank or Tommie - the union of indices
>> >
>> > (('Hank' inT 'nameFirst') union ('Tommie' inT 'nameFirst')) { data
>> >
>> >
>> > NB. Hank or Tommie AND born in Cuba
>> >
>> > ((('Hank' inT 'nameFirst') union ('Tommie' inT 'nameFirst')) intersect
>> > ('Cuba' inT 'birthCountry')) { data
>> >
>> >
>> >
>> > On Tue, Oct 8, 2013 at 3:50 AM, R.E. Boss <[email protected]> wrote:
>> >
>> > > So 'data' was plural?
>> > >
>> > >
>> > > R.E. Boss
>> > >
>> > > (Add your info to
>> http://www.jsoftware.com/jwiki/Community/Demographics)
>> > >
>> > >
>> > > > -----Original Message-----
>> > > > From: [email protected] [mailto:programming-
>> > > > [email protected]] On Behalf Of Henry Rich
>> > > > Sent: dinsdag 8 oktober 2013 3:52
>> > > > To: [email protected]
>> > > > Subject: Re: [Jprogramming] Slicing and dicing tables
>> > > >
>> > > > That is what he meant, but it's a minority opinion.
>> > > >
>> > > > Henry Rich
>> > > >
>> > > > On 10/7/2013 8:56 PM, Dan Bron wrote:
>> > > > > You mean " 'data' is plural " .
>> > > > >
>> > > > > -Dan
>> > > > >
>> > > > > PS: http://en.m.wikipedia.org/wiki/Muphry's_law
>> > > > >
>> > > > > (Yes, that link is spelled correctly :)
>> > > > >
>> > > > > Please excuse typos; composed on a handheld device.
>> > > > >
>> > > > > On Oct 7, 2013, at 8:00 PM, Eldon Eller <[email protected]>
>> wrote:
>> > > > >
>> > > > >> Data are plural.
>> > > > >>
>> > > > >> On 10/07/2013 02:04 PM, Dan Bron wrote:
>> > > > >>> Ganesh Rapolu
>> > > > >>>> Because the data is boxed, all comparisons must be boxed.
>> > > > >>> This was very well put (one short sentence which both clarifies
>> the
>> > > > >>> problem and justifies the solution).
>> > > > >>>
>> > > > >>> -Dan
>> > > > >>>
>> > > > >>>
>> > > > >>>
>> > > > >>>
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>
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