I would like to use a multinomial logistic regression to get win probabilities for each of the 5 horses that participate in any given race using each horses previous average speed.
RACE_ID H1_SPEED H2_SPEED H3_SPEED H4_SPEED H5_SPEED WINNING_HORSE
1 40.482081 44.199627 42.034929 39.004813 43.830139 5
2 39.482081 42.199627 41.034929 41.004813 40.830139 4
I am stuck on how to handle the independent variables for each horse given that any of the 5 horses average speed can be placed in any of H1_SPEED through H5_SPEED.
Given the fact that for each race I can put any of the 5 horses under H1_SPEED meaning there is no real relationship between H1_SPEED from RACE_ID 1 and H1_SPEED from RACE_ID 2 other than the arbitrary position I selected.
Would there be any difference if the dataset looked like this -
For RACE_ID 1 I swapped H3_SPEED and H5_SPEED and changed WINNING_HORSE from 5 to 3
For RACE_ID 2 I swapped H4_SPEED and H1_SPEED and changed WINNING_HORSE from 4 to 1
RACE_ID H1_SPEED H2_SPEED H3_SPEED H4_SPEED H5_SPEED WINNING_HORSE
1 40.482081 44.199627 43.830139 39.004813 42.034929 3
2 41.004813 42.199627 41.034929 39.482081 40.830139 1
Is this an issue, if so how should this be handled? What if I wanted to add more independent features per horse?
You cannot change in that way your dataset, because each feature (column) has a meaning and probably it depends on the values of the other features. You can imagine it as a six dimensional hyperplane, if you change the value of a feature the position of the point in the hyperplane changes, it does not remain stationary.
If you deem that a feature is useless to solve your problem (i.e. it is independent from the target), you can drop it or avoid to use it during the training phase of your model.
Edit
To solve your specific problem you may add a parameter for each speed column that takes care of the specific horse which is running with that speed. It is a sort of data augmentation, in order to add more problem related features to your model.
RACE_ID H1_SPEED H1_HORSE H2_SPEED H2_HORSE ... WINNING_HORSE
1 40.482081 1 44.199627 2 ... 5
2 39.482081 3 42.199627 5 ... 4
I've invented the number associated to each horse, but it seems that this information is present in your dataset.
Supposing I have 1000 numbers from 1 -> 1000, and a user can have any of the 1000 combination (eg: 4, 25, 353..).
How can I efficiently store that combination in a MySQL DB.
What I thought. I can use the power of 2, and store each number in a really large int, like:
1 -> 01
2 -> 10
4 -> 100
etc.
So if I happen to get the number 6 (110) I know the user has the combination of numbers 2, 4 (2 | 4 = 6) .
So we can have 2^1000 combinations, 125byte. But that is not efficient at all since bigint has 8bytes and I cant store
that in MySQL without using vachars etc. Nodejs cant handle that big number either (and I dont as well) with 2^53-1 being the max.
Why I am asking this question; can I do the above with base 10 instead of 2 and minimize the max bytes that the int can be. That was silly and I think making it to base10 or another base out of 2 changes nothing.
Edit: Additional thoughts;
So one possible solution is to make them in sets of 16digit numbers then convert them to strings concat them with a delimiter, and store that instead of numbers. (Potentially replace multiple 1's or 0's with a certain character to make it even smaller. Though I have a feeling that falls into the compression fields, but nothing better has come to my mind.)
Based on your question I am assuming you are optimizing for space
If most users have many numbers from the set then 125 bytes the way you described is the best you can do. You can store that in a BINARY(125) column though. In Node.js you could just a Buffer (you could use a plain string but should use a Buffer) to operate on the 125 byte bit-field.
If most users have only a few elements in the set then it will take less space to have a separate table with two columns such as:
user_id | has_element (SMALLINT)
---------------------
1 | 4
1 | 25
1 | 353
2 | 7
2 | 25
2 | 512
2 | 756
2 | 877
This will also make queries cleaner and more efficient for doing simple queries like SELECT user_id FROM user_elements WHERE has_element = 25;. You should probably add an index on has_element if you do queries like that to make them many times more efficient than storing a bitfield in a column.
ok, i'm not sure if i can explain this right.
Lets say i have a table with three columns (id, price, maxcombo)
maybe there's like 5 rows in this table with random numbers for price. 2. id is just incremental unique key)
maxcombo specified if that price can be in a combination of up to whatever number it is.
If x was 3, i would need to find the combination that has the maximum value of the sum 1-3 columns.
So say the table had:
1 - 100 - 1
2 - 50 - 3
3 - 10 - 3
4 - 15 - 3
5 - 20 - 2
the correct answer with be just row id 1.
since 100 alone (and can only be alone based on the maxcombo number)
is greater than say 50 + 20 + 15 or 20 + 15 or 10 + 20 etc.
Does that make sense?
I mean i could just calculate all the diff combinations and see which has the largest value, but i would imagine that would take a very long time if the table was larger than 5 rows.
Was wondering any math genius or super dev out there had some advice or creative way to figure this out in a more efficient manner.
Thanks ahead of time!
I built this solution to achieve the desired query. However, it hasn't been tested in terms of efficiency.
Following the example of colums 1-3:
SELECT max(a+b+c) FROM sample_table WHERE a < 3;
EDIT:
Looking at:
The correct answer will be just row id 1
...I considered maybe I misunderstood your question, and you want the query just obtain the rowid. So, I made this other one:
SELECT a FROM sum_combo WHERE a+b+c=(
SELECT max(a+b+c) FROM sum_combo WHERE a > 3
);
Which would for sure take too long in larger tables than just 5 rows.
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I'm looking for a way to weight my results to get the "best" highest rated result.
I have a table consisting of rating (0-5), mentions and name.
I.E.
RATING MENTIONS NAME
2.5 15 Bob
4.4 14 Susan
1 60 John
5 2 Steve
Both mentions and rating are important so sorting by just rating won't get the desired results.
For this example; while Steve has the highest rating he has very little mentions so i'm not very confident that he is the "best" highest rated person. Susan has several mentions and a high rating so she should surpass Steve. John has a very low rating but lots of mentions, he should only surpass any of the other people if he has a ridiculous amount of mentions.
The ideal result would be something similar to
RATING MENTIONS NAME
4.4 14 Susan
5 2 Steve
2.5 15 Bob
1 60 John
Appreciate the help!
Simplest way to do this is
RATING * RATING * Mentions
Which would provide the following:
RATING MENTIONS NAME SCORE
2.5 15 Bob 93.75
4.4 14 Susan 271.04
1 60 John 60
5 2 Steve 50
It is a pretty simple way to 'weight' the value of the rating.
Obviously you can go more complex but I would think the above is sufficient and the Query is easy so I will let you try and work that out yourself if you like the method!
Obviously you can just add another RATING if you want a LOT of weight on the rating OR multiply it by a a fixed amount - but the squaring / POWER is key (you could try RATING ^ 2.5) (^ is POWER)
When I encounter this problem, I often take the approach of reducing the rating by one standard error. The formula for the standard error is:
standard deviation for the group / sqrt(group size)
If you had the standard deviation for each group, I would order them using:
order by (case when mentions > 1 then stdev / sqrt(mentions) end)
This is not as "punishing" as Evan Miller's suggestion (pointed to by Juergen). This is essentially taking a confidence interval more like 60% than 95%. Admittedly, my preference is a bit empirical (based on experience). However, there is an issue with multiple comparisons and you don't need to estimate the exact confidence interval -- you just need to know the relative ordering of them.
You can calculate the standard deviation using the function stdev().
Well, I'm not very good in statistic, but from your expected result, I believe you need to find the importance of each property.. Which one is more important than the other one, I think you can use equation below:
values = weight * RATING + (1-weight) * MENTIONS
You can play around with the weight value, till you got what you want.. For me 0.8 kind a make sense..
RATING MENTIONS NAME SCORE
4.4 14 Susan 6.32
2.5 15 Bob 5
5 2 Steve 4.4
1 60 John 2
I'm currently ranking videos on a website using a bayesian ranking algorithm, each video has:
likes
dislikes
views
upload_date
Anyone can like or dislike a video, a video is always views + 1 when viewed and all videos have a unique upload_date.
Data Structure
The data is in the following format:
| id | title | likes | dislikes | views | upload_date |
|------|-----------|---------|------------|---------|---------------|
| 1 | Funny Cat | 9 | 2 | 18 | 2014-04-01 |
| 2 | Silly Dog | 9 | 2 | 500 | 2014-04-06 |
| 3 | Epic Fail | 100 | 0 | 200 | 2014-04-07 |
| 4 | Duck Song | 0 | 10000 | 10000 | 2014-04-08 |
| 5 | Trololool | 25 | 30 | 5000 | 2014-04-09 |
Current Weighted Ranking
The following weighted ratio algorithm is used to rank and sort the videos so that the best rated are shown first.
This algorithm takes into account the bayesian average to give a better overall ranking.
Weighted Rating (WR) = ((AV * AR) + (V * R))) / (AV + V)
AV = Average number of total votes
AR = Average rating
V = This items number of combined (likes + dislikes)
R = This items current rating (likes - dislikes)
Example current MySQL Query
SELECT id, title, (((avg_vote * avg_rating) + ((likes + dislikes) * (likes / dislikes)) ) / (avg_vote + (likes + dislikes))) AS score
FROM video
INNER JOIN (SELECT ((SUM(likes) + SUM(dislikes)) / COUNT(id)) AS avg_vote FROM video) AS t1
INNER JOIN (SELECT ((SUM(likes) - SUM(dislikes)) / COUNT(id)) AS avg_rating FROM video) AS t2
ORDER BY score DESC
LIMIT 10
Note: views and upload_date are not factored in.
The Issue
The ranking currently works well but it seems we are not making full use of all the data at our disposal.
Having likes, dislikes, views and an upload_date but only using two seems a waste because the views and upload_date are not factored in to account how much weight each like / dislike should have.
For example in the Data Structure table above, items 1 and 2 both have the same amount of likes / dislikes however item 2 was uploaded more recently so it's average daily views are higher.
Since item 2 has more likes and dislikes in a shorter time than those likes / dislikes should surely be weighted stronger?
New Algorithm Result
Ideally the new algorithm with views and upload_date factored in would sort the data into the following result:
Note: avg_views would equal (views / days_since_upload)
| id | title | likes | dislikes | views | upload_date | avg_views |
|------|-----------|---------|------------|---------|---------------|-------------|
| 3 | Epic Fail | 100 | 0 | 200 | 2014-04-07 | 67 |
| 2 | Silly Dog | 9 | 2 | 500 | 2014-04-06 | 125 |
| 1 | Funny Cat | 9 | 2 | 18 | 2014-04-01 | 2 |
| 5 | Trololool | 25 | 30 | 5000 | 2014-04-09 | 5000 |
| 4 | Duck Song | 0 | 10000 | 10000 | 2014-04-08 | 5000 |
The above is a simple representation, with more data it gets a lot more complex.
The question
So to summarise, my question is how can I factor views and upload_date into my current ranking algorithm in a style to improve the way that videos are ranked?
I think the above example by calculating the avg_views is a good way to go but where should I then add that into the ranking algorithm that I have?
It's possible that better ranking algorithms may exist, if this is the case then please provide an example of a different algorithm that I could use and state the benefits of using it.
Taking a straight percentage of views doesn't give an accurate representation of the item's popularity, either. Although 9 likes out of 18 is "stronger" than 9 likes out of 500, the fact that one video got 500 views and the other got only 18 is a much stronger indication of the video's popularity.
A video that gets a lot of views usually means that it's very popular across a wide range of viewers. That it only gets a small percentage of likes or dislikes is usually a secondary consideration. A video that gets a small number of views and a large number of likes is usually an indication of a video that's very narrowly targeted.
If you want to incorporate views in the equation, I would suggest multiplying the Bayesian average you get from the likes and dislikes by the logarithm of the number of views. That should sort things out pretty well.
Unless you want to go with multi-factor ranking, where likes, dislikes, and views are each counted separately and given individual weights. The math is more involved and it takes some tweaking, but it tends to give better results. Consider, for example, that people will often "like" a video that they find mildly amusing, but they'll only "dislike" if they find it objectionable. A dislike is a much stronger indication than a like.
I can point you to a non-parametric way to get the best ordering with respect to a weighted linear scoring system without knowing exactly what weights you want to use (just constraints on the weights). First though, note that average daily views might be misleading because movies are probably downloaded less in later years. So the first thing I would do is fit a polynomial model (degree 10 should be good enough) that predicts total number of views as a function of how many days the movie has been available. Then, once you have your fit, then for each date you get predicted total number of views, which is what you divide by to get "relative average number of views" which is a multiplier indicator which tells you how many times more likely (or less likely) the movie is to be watched compared to what you expect on average given the data. So 2 would mean the movie is watched twice as much, and 1/2 would mean the movie is watched half as much. If you want 2 and 1/2 to be "negatives" of each other which sort of makes sense from a scoring perspective, then take the log of the multiplier to get the score.
Now, there are several quantities you can compute to include in an overall score, like the (log) "relative average number of views" I mentioned above, and (likes/total views) and (dislikes / total views). US News and World Report ranks universities each year, and they just use a weighted sum of 7 different category scores to get an overall score for each university that they rank by. So using a weighted linear combination of category scores is definitely not a bad way to go. (Noting that you may want to do something like a log transform on some categories before taking the linear combination of scores). The problem is you might not know exactly what weights to use to give the "most desirable" ranking. The first thing to note is that if you want the weights on the same scale, then you should normalize each category score so that it has standard deviation equal to 1 across all movies. Then, e.g., if you use equal weights, then each category is truly weighted equally. So then the question is what kinds of weights you want to use. Clearly the weights for relative number of views and proportion of likes should be positive, and the weight for proportion of dislikes should be negative, so multiply the dislike score by -1 and then you can assume all weights are positive. If you believe each category should contribute at least 20%, then you get that each weight is at least 0.2 times the sum of weights. If you believe that dislikes are more important that likes, then you can say (dislike weight) >= c*(like weight) for some c > 1, or (dislike_weight) >= c*(sum of weights) + (like weight) for some c > 0. Similarly you can define other linear constraints on the weights that reflect your beliefs about what the weights should be, without picking exact values for the weights.
Now here comes the fun part, which is the main thrust of my post. If you have linear inequality constraints on the weights, all of the form that a linear combination of the weights is greater than or equal to 0, but you don't know what weights to use, then you can simply compute all possible top-10 or top-20 rankings of movies that you can get for any choice of weights that satisfy your constraints, and then choose the top-k ordering which is supported by the largest VOLUME of weights, where the volume of weights is the solid angle of the polyhedral cone of weights which results in the particular top-k ordering. Then, once you've chosen the "most supported" top-k ranking, you can restrict the scoring parameters to be in the cone that gives you that ranking, and remove the top k movies, and compute all possibilities for the next top-10 or top-20 ranking of the remaining movies when the weights are restricted to respect the original top-k movies' ranking. Computing all obtainale top-k rankings of movies for restricted weights can be done much, much faster than enumerating all n(n-1)...(n-k+1) top-k possible rankings and trying them all out. If you have two or three categories then using polytope construction methods the obtainable top-k rankings can be computed in linear time in terms of the output size, i.e. the number of obtainable top-k rankings. The polyhedral computation approach also gives the inequalities that define the cone of scoring weights that give each top-k ranking, also in linear time if you have two or three categories. Then to get the volume of weights that give each ranking, you triangulate the cone and intersect with the unit sphere and compute the areas of the spherical triangles that you get. (Again linear complexity if the number of categories is 2 or 3). Furthermore, if you scale your categories to be in a range like [0,50] and round to the nearest integer, then you can prove that the number of obtainable top-k rankings is actually quite small if the number of categories is like 5 or less. (Even if you have a lot of movies and k is high). And when you fix the ordering for the current top group of movies and restrict the parameters to be in the cone that yields the fixed top ordering, this will further restrict the output size for the obtainable next best top-k movies. The output size does depend (polynomially) on k which is why I recommended setting k=10 or 20 and computing top-k movies and choosing the best (largest volume) ordering and fixing it, and then computing the next best top-k movies that respect the ordering of the original top-k etc.
Anyway if this approach sounds appealing to you (iteratively finding successive choices of top-k rankings that are supported by the largest volume of weights that satisfy your weight constraints), let me know and I can produce and post a write-up on the polyhedral computations needed as well as a link to software that will allow you to do it with minimal extra coding on your part. In the meantime here is a paper http://arxiv.org/abs/0805.1026 I wrote on a similar study of 7-category university ranking data where the weights were simply restricted to all be non-negative (generalizing to arbitrary linear constraints on weights is straightforward).
A simple approach would be to come up with a suitable scale factor for each average - and then sum the "weights". The difficult part would be tweaking the scale factors to produce the desired ordering.
From your example data, a starting point might be something like:
Weighted Rating = (AV * (1 / 50)) + (AL * 3) - (AD * 6)
Key & Explanation
AV = Average views per day:
5000 is high so divide by 50 to bring the weight down to 100 in this case.
AL = Average likes per day:
100 in 3 days = 33.33 is high so multiply by 3 to bring the weight up to 100 in this case.
AD = Average dislikes per day:
10,000 seems an extreme value here - would agree with Jim Mischel's point that dislikes may be more significant than likes so am initially going with a negative scale factor of twice the size of the "likes" scale factor.
This gives the following results (see SQL Fiddle Demo):
ID TITLE SCORE
-----------------------------
3 Epic Fail 60.8
2 Silly Dog 4.166866
1 Funny Cat 1.396528
5 Trololool -1.666766
4 Duck Song -14950
[Am deliberately keeping this simple to present the idea of a starting point - but with real data you might find linear scaling isn't sufficient - in which case you could consider bandings or logarithmic scaling.]
Every video have:
likes
dislikes
views
upload_date
So we can deduct the following parameters from them:
like_rate = likes/views
dislike_rate = likes/views
view_rate = views/number_of_website_users
video_age = count_days(upload_date, today)
avg_views = views/upload_age
avg_likes = likes/upload_age
avg_dislikes = dislikes/upload_age
Before we can set the formula to be used, we need to specify how different videos popularity should work like, one way is to explain in points the property of a popular video:
A popular video is a recent one in most cases
The older a video gets, the higher avg_views it requires to become popular
A video with a like_rate over like_rate_threshold or a dislike_rate over dislike_rate_threshold, can compete by the difference from its threshold with how old it gets
A high view_rate of a video is a good indicator to suggest that video to a user who have not watched it before
If avg_likes or avg_dislikes make most of avg_views, the video is considered active in the meantime, in case of active videos we don't really need to check how old it's
Conclusion: I don't have a formula, but one can be constructed by converting one unit into another's axis, like cutting a video age by days based on a calculation made using avg_likes, avg_dislikes, and avg_views
Since no one has pointed it out yet (and I'm a bit surprised), I'll do it. The problem with any ranking algorithm we might come up with is that it's based on our point of view. What you're certainly looking for is an algorithm that accomodates the median user point of view.
This is no new idea. Netflix had it some time ago, only they personalized it, basing theirs on individual selections. We are looking - as I said - for the median user best ranking.
So how to achieve it? As others have suggested, you are looking for a function R(L,D,V,U) that returns a real number for the sort key. R() is likely to be quite non-linear.
This is a classical machine learning problem. The "training data" consists of user selections. When a user selects a movie, it's a statement about the goodness of the ranking: selecting a high-ranked one is a vote of confidence. A low-ranked selection is a rebuke. Function R() should revise itself accordingly. Initially, the current ranking system can be used to train the system to mirror its selections. From there it will adapt to user feedback.
There are several schemes and a huge research literature on machine learning for problems like this: regression modeling, neural networks, representation learning, etc. See for example the Wikipedia page for some pointers.
I could suggest some schemes, but won't unless there is interest in this approach. Say "yes" in comments if this is true.
Implementation will be non-trivial - certainly more than just tweaking your SELECT statement. But on the plus side you'll be able to claim your customers are getting what they're asking for in very good conscience!