Phil Dawes' Stuff - Latest Comments in More factor: tabular to triples

Re: More factor: tabular to triples

Greg M — Mon, 15 Oct 2007 23:01:39 -0000

I prefer the Haskell list comprehension version:

[(i,col,cell)|(i,row)<-zip [startid..] rows, (col,cell) <- zip cols row]

Now I wonder how this will get formatted? Need a preview button.

Re: More factor: tabular to triples

Slava Pestov — Mon, 15 Oct 2007 14:56:26 -0000

Hi,

I had a go at implementing this. The key, as you've noticed, is to decompose the problem into steps where you're working with no more than a handful of values at once.

Don'write a big loop, instead break it up into multiple iterations over the data and use library words where possible.

USE: math.ranges

: (column-names) ( cols row -- pairs )
2array flip ;

: column-names ( cols rows -- seqs-of-pairs )
[ (column-names) ] curry* map ;

: (number-rows) ( pairs n -- triples )
[ add* ] curry map ;

: number-rows ( seqs-of-pairs -- triples )
tuck length [a,b] [ (number-rows) ] 2map concat ;

: tabular>triples ( start-rowid cols rows -- triples )
column-names number-rows ;

Re: More factor: tabular to triples

Phil Dawes — Sat, 13 Oct 2007 15:32:33 -0000

Thanks Christopher - it was a missing close brace in the factor link preventing it from being displayed.
I'll definitely check out both Cat and ripple - thanks for the links

Re: More factor: tabular to triples

Christopher Diggins — Fri, 12 Oct 2007 16:21:52 -0000

Hi Phil,

Small typo in the first sentence: "I’ve been playing with for a couple of weeks.", I believe you meant "factor" instead of "for".

If you are interested in concatenative languages and the semantic web have you seen Ripple? http://ripple.fortytwo.net/

There is also my own relatively immature language Cat (http://www.cat-language.com), which will be stable "real soon now".

Cheers,
Christopher Diggins

Re: More factor: tabular to triples

paddy3118 — Fri, 12 Oct 2007 00:55:43 -0000

I thought from the input format and the output that a Python list comprehension would be the way I'd solve it and came up with this:

>>> from pprint import pprint as pp
>>> tabular = [["col1","col2","col3"],[["a","b","c"],["e","f","g"]]]
>>> triples = [(row[0], col[1], row[1][col[0]])
for row in enumerate(tabular[1])
for col in enumerate(tabular[0])]
>>> pp(triples)
[(0, 'col1', 'a'),
(0, 'col2', 'b'),
(0, 'col3', 'c'),
(1, 'col1', 'e'),
(1, 'col2', 'f'),
(1, 'col3', 'g')]
>>>

(I added a couple of newlines to the list comprehension to get it into the comment box).

If I were using it then I might wrap it in a function and add a docstring to test/explain it.

- Paddy.

Re: More factor: tabular to triples

Neil Bartlett — Thu, 11 Oct 2007 17:58:48 -0000

Just for a laugh, here it is in Haskell. You can type this at the interpreter.

let insertList x = map (insert x) where insert x (y,z) = (x,y,z)

let cols = ["col1","col2","col3"]

let rows = [["a","b","c"],["e","f","g"]]

concat $ zipWith insertList [0..] $ map (zip cols) rows

This prints [(0,"col1","a"),(0,"col2","b"),(0,"col3","c"),(1,"col1","e"),(1,"col2","f"),(1,"col3","g")]

No preview available so I don't know if your blog software is going to mangle that!