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f00biebletch
Posted: Wed Nov 14, 2007 12:53 am Reply with quote
Joined: 13 Nov 2007 Posts: 3
I am yet another java/c#/ruby engineer teaching myself erlang, and, am trying to write a fun to do polynomial multiplication. I can take 2 tuple lists representing a polynomial (tuples being terms like 3x^2) and am down to a list like this:

Code:

[{term,1,x,2},{term,2,x,1},{term,2,x,1},{term,4,x,0}]

where element 2 is the coefficient, 3 is the var name, and 4 is the exponent. I just need to group like terms now, "like" being the same exponent.

I can (and have) folded the list with a dict and aggregated by finding and storing the term with the exponent for the key, but that seems less than optimal.

Are there any elegant list comprehension solutions out there?

Thanks, this is really a vibrant site that helps a ton with Erlang.
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francesco
Posted: Wed Nov 14, 2007 9:41 am Reply with quote
User Joined: 07 Jul 2006 Posts: 249 Location: London
I don't get your question, but will try to reply anyhow. Are you trying to order something like

Code:
[{term,7,x,0},{term,2,x,2},{term,2,x,1},{term,4,x,3}]


to something like

Code:
[{term,4,x,3},{term,2,x,2},{term,2,x,1},{term,7,x,0}]


If so, use lists:reverse(lists:keysort(4,List)).

If the above is not what you are looking for, could you pls give a before and after example.

Francesco
--
http://www.erlang-consulting.com
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f00biebletch
Posted: Wed Nov 14, 2007 1:34 pm Reply with quote
Joined: 13 Nov 2007 Posts: 3
Sorry for the confusion, I'm trying to get from

Code:

[{term,1,x,2},{term,2,x,1},{term,2,x,1},{term,4,x,0}]


to

Code:

[{term,1,x,2},{term,4,x,1},{term,4,x,0}]


The last element of the tuple is the exponent, so I am just adding the coefficients of terms with like exponents. If I had something like:

Code:

[{term,5,x,2},{term,1,x,2},{term,2,x,0},{term,4,x,0}]


I would want to get to
Code:

[{term,6,x,2},{term,6,x,0}]


Hope that helps clarify, thanks.
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francesco
Posted: Thu Nov 15, 2007 8:43 am Reply with quote
User Joined: 07 Jul 2006 Posts: 249 Location: London
Try
Code:

F = fun({term,Val,x,Exp}, [{term,Val2,x,Exp}|Tail]) ->
       [{term, Val + Val2,x,Exp}|Tail];
       ({term,Val,x,Exp}, Acc) ->
       [{term,Val,x,Exp}|Acc]
    end.


L = lists:keysort(4,[{term,1,x,2},{term,2,x,1},{term,2,x,1},{term,4,x,0}]).


lists:foldl(F, hd(L), tl(L)).


Regards,
Francesco
--
http://www.erlang-consulting.com
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f00biebletch
Posted: Thu Nov 15, 2007 4:47 pm Reply with quote
Joined: 13 Nov 2007 Posts: 3
Brilliant, I should have suspected pattern matching would be the key, I like the "peek" functionality by splitting the "rest" of the list as well. Many thanks for your time and help.
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