Efficient way to match unordered arrays?

[The_Kez]

I have a function which I am entering an array, and the job of the function is to match the entered array to an array within the list. It looks a little something like this...

PHP Code:

function Test( arry ) {
  temp.list = {
  {"Test One", "Test Two"},"String One",
  {"Test Three", "Test One"},"String Two",
  {"Test Three", "Test Two"},"String Three"
  };
  //matching script should go here
  return appropriate string;
} 


Thing is I can't compare them directly. Because the fed array may be in a different order. For example, if 'arry' is {"Test Two", "Test One"} I still need it two return "String One".
If there are no matches, OR if 'arry' contains a match but has extra items which are not in the list (i.e {"Test One", "Test Two", "Test Three"}) it should return false.

I could probably do some magic and come up with a way to match these, but I'd like to see some of your ideas to match them as there is probably a much better way then what I would do.

Thanks

[DustyPorViva]

PHP Code:

function comparearrays(arr1,arr2) {
  temp.checklist = arr2.size();
  for (temp.i : arr1) {
    if (i in arr2) {
      checklist--;
      arr2.remove(i);
    } else break;
  }
  return (checklist == 0);
} 


? I dunno, I didn't test it... seems to make sense in my head, however.

[[email protected]]

function onCreateds() {
  temp.arr = {"test", "foo", "test"};
  
  temp.list = {
    {"test", "test", "foo"}, "string a",
    {"test", "test", "foos"}, "string b",
    {"test", "testa", "foo"}, "string c",
    {"foo", "test", "test"}, "string d",
    {"test", "teste"}, "string e"
  };
  
    //Let's first match the sizes
  temp.size = temp.arr.size();
  
  for (temp.i: temp.list) {
      //If it's a totally different size, ignore it
    if (temp.size != temp.i.size()) continue;

      //Now let's clear it the counter
    temp.count = 0;

    for (temp.e: temp.i) {
        //If it's in the list, let's increase the counter
      if (temp.e in temp.arr) temp.count++;
    }

      //If it's the same amount let's add to the list
    if (temp.count == temp.size) {
      //Now, let's find the index and increase it
      temp.found.add(temp.list[temp.list.index(@ temp.i) + 1]);
    }
  }
  
  return temp.found;
}

This returns "string a","string d"

[The_Kez]

Stupid of me to post this and then come up with a solution for myself 10 minutes later.
Modifying your script to how for what I wanted worked perfectly. So thanks for the help. I also came up with this script awhile after I made the thread...

PHP Code:

function Compare( combo ) {
  temp.list = {
  {"Test One","Test Two","Test Three"},"Test String Zero",
  {"Test One","Test Two"},"Test String One",
  {"Test Two","Test Three"},"Test String Two"
  };
  for ( temp.i : list ) {
    if ( i.type() == 3 ) {
      temp.match = 0;
      for ( temp.j : combo ) {
        if ( j in i ) match++;
      }
      if ( match == i.size() || combo == i.size() ) return list[list.index(i)+1];
    }
  }
} 


This also works, I've tested it a few times. :]

[The_Kez]

I think either of your two ways would be better to use since they're skipping cycles which are unnecessary instead of running through them and comparing the sizes afterwards.

[fowlplay4]

An array is pretty much just a string with comma's in it, so as long as you aren't doing crazy stuff this should get you by.

PHP Code:

function onCreated() {
  echo(onTestFunc({"asdf", "jkl;"})); // output = rawr
  echo(onTestFunc({"rawr", "asdf"})); // output = !
}
function onTestFunc(arr) {
  temp.stuff = {
    {"asdf", "jklm"}, 123,
    {"asdf", "jkl."}, 124,
    {"asdf", "jkl;"}, "rawr"
  };
  arr = arr.substring(0);
  for (temp.i = 0; temp.i < stuff.size(); temp.i += 2) {
    if (stuff[i].substring(0) == arr) return stuff[i+1];
  }
  return "!";
} 


If the stuff array was properly sorted you could probably write a binary search instead of that simple for loop.