Re: DeleteCases

• To: mathgroup at smc.vnet.net
• Subject: [mg98207] Re: DeleteCases
• From: "andre.robin3" <andre.robin3 at wanadoo.fr>
• Date: Thu, 2 Apr 2009 04:49:17 -0500 (EST)
• References: <gqsn69\$3qp\$1@smc.vnet.net>

1) A solution for the first problem :

data00 = {{0, 0, 1, 1}, {0, 0, 2, 2}, {0, 0, 3, 3}, {1, 1, 0, 0}, {1, 1, 2,
2},
{1, 1, 3, 3}, {2, 2, 0, 0}, {2, 2, 1, 1}, {2, 2, 3, 3}, {3, 3, 0, 0},
{3, 3, 1, 1}, {3, 3, 2, 2}};

data00 //. {x1___, a : {a1_, a1_, b1_, b1_}, x2___, {b1_, b1_, a1_, a1_},
x3___} -> {x1, a, x2, x3}

It returns :{{0, 0, 1, 1}, {0, 0, 2, 2}, {0, 0, 3, 3}, {1, 1, 2, 2}, {1, 1,
3, 3}, {2, 2, 3, 3}} as you want.

2)  For the more general second point :

data01 = {{a, b, c, d}, {b, a, c, d}, {a, b, d, c}};

testQ[a_, b_] := (Sort[a] === Sort[b])  (* put here any relation you want,
so far i
understand it *)

data01 //. ({x1___, a : {_, _, _, _}, x2___, b : {_, _, _, _}, x3___} /;
testQ[a, b])  ->  {x1, a, x2, x3}

It returns : {{a, b, c, d}} as you want, I think.
It works with data00 too.

The key command is  "//." (alias "PeplaceAll[  ]") or anything of the family

The solutions I suggest here are slow (not to use on big data sets).

"Filippo Miatto" <sottosupremo at gmail.com> a écrit dans le message de news:
gqsn69\$3qp\$1 at smc.vnet.net...
> Dear all,
> Given this example list:
>
> {{0, 0, 1, 1}, {0, 0, 2, 2}, {0, 0, 3, 3}, {1, 1, 0, 0}, {1, 1, 2, 2},
> {1, 1, 3, 3}, {2, 2, 0, 0}, {2, 2, 1, 1}, {2, 2, 3, 3}, {3, 3, 0, 0},
> {3, 3, 1, 1}, {3, 3, 2, 2}}
>
> I need the elements {a,a,b,b} and {b,b,a,a} to be the same, i.e. i
> want to delete one of the two occurrences from the list. (or all but
> one if the condition is more complicated, say {a,b,c,d} and {b,a,c,d}
> or {a,b,c,d} and {a,b,d,c})
> How can i do that? Can i use DeleteCases? if so, what conditional
> should I use to compare different elements in the list?
> I tried without success for all the night.. (it's 6.15AM now!)
> Thank you,
>
> Filippo
>

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