[erlang-questions] qlc expression help.
Adam Bozanich
adam.boz@REDACTED
Mon Feb 11 04:09:47 CET 2008
Hi all,
I'm having trouble creating a qlc expression which does what I want. I'm
basically trying to create a filter for QH1 with a predicate that depends on
QH2.
In the example below, test2/0 gives me the result I'm looking for, but it
requires the second QH to be eval'd. I'd like the expression to look like
what's in the test1/0 function. Any suggestions?
Thanks,
-Adam
session:
8> test_qlc:test2().
[{1,6,7},{2,8,9}]
9> test_qlc:test1().
[{1,4,5},
{1,2,3},
{1,6,7},
{1,6,7},
{2,4,5},
{2,2,3},
{2,8,9},
{2,8,9},
{3,4,5},
{3,2,3}]
code:
-module(test_qlc).
-include_lib("stdlib/include/qlc.hrl").
-export([test1/0,test2/0,handroll/0]).
-define(DB,[
{1,4,5},{1,2,3},{1,6,7},
{2,4,5},{2,2,3},{2,8,9},
{3,4,5},{3,2,3}
]).
-define(OBJ,[{2,3},{4,5}]).
test1() ->
qlc:eval(qlc:q([ {O1,C1,V1} ||
{O1,C1,V1} <- ?DB,
{C2,V2} <- ?OBJ,
C1 /= C2,
V1 /= V2])).
test2() ->
qlc:eval(qlc:q([ {O1,C1,V1} ||
{O1,C1,V1} <- ?DB,
lists:member({C1,V1},?OBJ) == false ])).
handroll() ->
S = mkset(?DB),
O = mkobj(?OBJ),
to_table(difference(O,related(O,S))).
related(S1,D) ->
dict:filter(fun(_,S2) ->
sets:is_subset(S1,S2)
end,D).
difference(S1,D) ->
dict:filter(fun(_,S2) ->
sets:size(S2) > 0
end,
dict:map(fun(_,S2) ->
sets:subtract(S2,S1)
end,D)).
to_table(D) ->
lists:flatmap(fun({K,V}) ->
lists:map(fun({X,Y}) ->
{K,X,Y}
end,
sets:to_list(V))
end,
dict:to_list(D)).
mkset(DB) ->
dict:map(fun(_,V) ->
sets:from_list(V)
end,
lists:foldl(fun({O,C,V},Acc) ->
dict:append(O,{C,V},Acc)
end,
dict:new(),DB)).
mkobj(OBJ) ->
sets:from_list(OBJ).
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