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List similarity?

Hi,

I have an app where users can mark products. A user might generate a list of strings of products they marked like:

    truck
    wagon
    bottle
    dog

I want to find other users which have generated *similar* lists of marked product strings. I star "similar" because it's difficult for me to define.

This user would be the most similar:

    truck, wagon, bottle, dog

This user would be a little less similar:

    truck, wagon, bottle, dog, red

This user would also be a little less similar:

    truck, bottle, dog

This user has zero similarity:

    duck

This is probably some type of search, I just don't know the name. I want to find users that have similar tastes in the products they marked,

Thanks
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DJ_AM_Juicebox
Asked:
DJ_AM_Juicebox
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5 Solutions
 
themrrobertCommented:
You will have to create your own algorithm, Especially if you plan to have a dynamic list now or in the future, it could change the order, thus impeding efforts even further.

The perfect function for this would be worth a great deal of cash.

Here is a good starting point:

$arrayUser1 = split("\n",$UserData1);
$arrayUser2 = split("\n",$UserData2);

this creates two arrays.

Loop through both, in a for x, for y, next y, next x format and add 1 point for each entry that matches.

For each user, divide the number of matches by the number of items of user1. (dividing by user2 gives you the converse 'similarity').

This should give you a good start, let me know if you have any questions
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aburrCommented:
" I star "similar" because it's difficult for me to define. "
If you find "similar" difficult to define, how do you expect the computed to do better?
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DJ_AM_JuiceboxAuthor Commented:
>> The perfect function for this would be worth a great deal of cash.

Ha yeah I realize this has many useful applications.

I can't think of anything better than the solution you propose. Is there a common name for this problem though? I want to reference it a post I'm writing - like "the list similarity problem"? I doubt I'll find a solution for it, but if it's a known problem, I can site it as such and carry on!!

Thanks
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aburrCommented:
"Is there a common name for this problem though?"
Try The Similarity Problem"

Do not let my previous comment discourage you too much.
One current related problem is face recognition software
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DJ_AM_JuiceboxAuthor Commented:
@aburr
Yeah bad wording on my part. I meant I don't know what term to label this problem as.

If a user has 4 items in their list, and I want to find the most similar users, those users would each have a similarity of "1", generated by dividing the number of matched items by max(total items in their own lists, total number of items in the original user's list):

    userA { red, green, blue }

    user1 { red, blue }  2 / 3 = 0.66

    user2 { green, blue, red }    3 / 3  = 1

    user3 { red, green, blue, yellow } 3 / 4 = 0.75

so:

    similarity = matchedItems / max(setUserA.size(), setUserCurrent.size());  

Thanks
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aburrCommented:
"  similarity = matchedItems / max(setUserA.size(), setUserCurrent.size());"
It looks like you are well on your way especially if the users construct their lists by selecting from a menu so that the items have a well defined form.
ie red will not be confused with redish
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DJ_AM_JuiceboxAuthor Commented:
Yeah the solution by themrrobert gives the correct answer, the problem is hoping to do it at scale. For instance, if I have a database of 1 million users, there's no way I can cycle through all of them to do the computation. I would need to precompute some value each time they add a new item to their list and store it in their profile so I could at least get a rough set of users to iterate over in a single select statement.

Even better would be a single precomputed value to sort on which just orders the similarity for me in one select! The problem is that the criteria for similarity is variable and depends on the user the computation is being done for.

It's interesting and not easy to solve in 7 minutes!

Thanks
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ozoCommented:
> if I have a database of 1 million users
About how many products would be in the database, and about how many would you expect each user to mark?
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DJ_AM_JuiceboxAuthor Commented:
The number of products in the database is large, over 10 million.

The average number of items a user marks is 50.


 
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ozoCommented:
So this seems to be the model you are using
http://en.wikipedia.org/wiki/Standard_Boolean_model
But for a dataset that large, speed could be a problem
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PCableGuyCommented:

Is there any advantage of making everyone the same string length along with 'fake' characters for an item that they didn't choose? Does this make it easier or harder to compute, or roughly the same?
 
Database of Choices, Ascending Order
bottle, dog, duck, red, truck, wagon

New User, hasn't chosen anything yet.
######, ###, ####, ###, #####, #####

User 1
bottle, dog, ####, ###, truck, wagon

User A
bottle, dog, ####, red, truck, wagon

User B
bottle, dog, ####, ###, truck, #####

User C
######, ###, duck, ###, #####, #####
 
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DJ_AM_JuiceboxAuthor Commented:
@PCableGuy

Yeah was thinking of something like that, but the set of products is not fixed, new products are added every day. Also there are over 10 million products so the string would be huge (per user too). Also I would have to update every user's string when new products are added to keep it the 'missin' items sorted and up to date.

I can sort products as users mark a new product, I haven't thought of a way that that helps in a sql context though!

Thanks
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tliottaCommented:
Consider also that you might find value in tracking how many times each product is chosen for a listb. If 9M products have never been chosen, there's not much reason to pay attention to them in any algorithm.

Also, you might find value in weighting related to the number of times a product is chosen. If a customer lists four products that no one else has ever chosen, that list might have a total "weight" of zero, or something along those lines. Regardless, there will clearly be no other customers with a similar list -- it's not necessary even to check.

Tom
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