Re: "Intelligent" array joining

by valentin (Abbot)
 on Feb 05, 2004 at 21:50 UTC ( #326900=note: print w/replies, xml ) Need Help??

in reply to "Intelligent" array joining

Let Tie::IxHash keep your order. I did not understand which order is prefered for your example, but put them in your order into the hash. That's what you get back.
```my @array1 = (1, 3, 4, 6);
my @array2 = (1, 2, 4, 6);
my @array3 = (1, 2, 3, 5);

tie my %seen, 'Tie::IxHash';

for (0..\$#array1) {
my @v =  sort ( \$array1[\$_], \$array2[\$_], \$array3[\$_] );
@seen{ @v } = 1;
}

print join ' ', keys %seen;

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Re: Re: "Intelligent" array joining
by ngomong (Sexton) on Feb 05, 2004 at 22:22 UTC
Hmmm... this appears to be a step in the right direction, though I don't need that "sort" in there. You see, numerical order is of no importance... only the order in which it appears in the array. It's like this:

```my @array1 = (1, 3, 4, 6);
my @array2 = (1, 2, 4, 6);
my @array3 = (1, 2, 3, 5);
So, the program looks as the first array, and notes its order. Then, it looks at the second array, and notices that "2" hasn't been seen before. It notes that it goes after "1" and before "4". At this point, the logical order could be either

(1, 2, 3, 4, 6) or
(1, 3, 2, 4, 6)

It doesn't matter which it chooses. Then, it examines the third array and realizes that "2" comes before "3". Plus, the "5" is new, and comes after "3". So, now the order could be:

(1, 2, 3, 4, 5, 6) or
(1, 2, 3, 4, 6, 5)

Does that make sense?

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