I have been trying to get my hands dirty with Information Retrieval.My professor gave us this problem to solve, but I can't get my way around it. The matrix given, if it is a distance matrix, the diagonal elements should all be 0. But in the table, they're given as 1. The other entries are also less than 1. How is this possible? Can someone please explain?
Please see question 5.c. I could not enter the table manually and apologize for that.
In every similarity measurement, 1 means totally similar and 0 means there is no similarity between documents.
Related
I'm really hoping someone here can help.
I have performed a chi-square test of independence, looking at men/women and early/late drop out from therapy. I have a p-value of 0.047. Do I need to do any post hoc testing on this? Men drop out almost 50:50 early:late whereas women drop out almost 25:75 early:late. Do I need post hoc testing for this and a Bonferonni correction, or is the answer simply:
The frequency of retention rates was compared across gender, finding a significant interaction (X2 (1) = 3.94, p = 0.047), indicating that females were more likely to be retained past the third CBT session than men.
Any help would be greatly appreciated, stats hurt my head and I can't continue past this problem.
Since there's only one test performed, with a single degree of freedom, there's no way (or need) to do any multiple comparison correction.
I have some question on BLUE Score calculation for machine translation. I realized they may have a different metrics for BLEU. I found the code reports five value for BLEU, namely BLEU-1, BLEU-2, BLEU-3, BLEU-4 and finally BLEU, which seems to be an exponential average of the previous four BLEUs. Still it is not clear to me what the difference between those is. Do you have any ideas? Thanks
P.s. At first I thought that this question is more of a theoretical content and posted it on meta stackexange. A moderator has closed and commented it as a stackoverflow type question . So please don't punish me again. =)
source: http://www.statmt.org/book/slides/08-evaluation.pdf
I haven't heard of BLEU-1 and BLEU-2 but I guess it means 1-gram, 2-gram, 3-gram and 4-gram in the formula of BLEU score, I mean in the formula precision[i] = BLEU-i in your question:
Actually, BLEU-n doesn't use the n-gram scores only. It computes the 1-gram through n-gram scores and gives them equal weight to compute a final score. See the "Cumulative N-Gram Scores" section at this link for more info.
I have a very small question which has been baffling me for a while. I have a dataset with interesting features, but some of them are dimensionless quantities (I've tried using z-scores) on them but they've made things worse. These are:
Timestamps (Like YYYYMMDDHHMMSSMis) I am getting the last 9 chars from this.
User IDs (Like in a Hash form) How do I extract meaning from them?
IP Addresses (You know what those are). I only extract the first 3 chars.
City (Has an ID like 1,15,72) How do I extract meaning from this?
Region (Same as city) Should I extract meaning from this or just leave it?
The rest of the things are prices, widths and heights which understand. Any help or insight would be much appreciated. Thank you.
Timestamps can be transformed into Unix Timestamps, which are reasonable natural numbers
User IF/Cities/Regions are nominal values, which has to be encoded somehow. The most common approach is to create as much "dummy" dimensions as the number of possible values. So if you have 100 ciries, than you create 100 dimensions and give "1" only on the one representing a particular city (and 0 on the others)
IPs should rather be removed, or transformed into some small group of them (based on the DNS-network identification and nominal to dummy transformation as above)
My platform here is Ruby - a webapp using Rails 3.2 in particular.
I'm trying to match objects (people) based on their ratings for certain items. People may rate all, some, or none of the same items as other people. Ratings are integers between 0 and 5. The number of items available to rate, and the number of users, can both be considered to be non-trivial.
A quick illustration -
The brute-force approach is to iterate through all people, calculating differences for each item. In Ruby-flavoured pseudo-code -
MATCHES = {}
for each (PERSON in (people except USER)) do
for each (RATING that PERSON has made) do
if (USER has rated the item that RATING refers to) do
MATCHES[PERSON's id] += difference between PERSON's rating and USER's rating
end
end
end
lowest values in MATCHES are the best matches for USER
The problem here being that as the number of items, ratings, and people increase, this code will take a very significant time to run, and ignoring caching for now, this is code that has to run a lot, since this matching is the primary function of my app.
I'm open to cleverer algorithms and cleverer databases to achieve this, but doing it algorithmically and as such allowing me to keep everything in MySQL or PostgreSQL would make my life a lot easier. The only thing I'd say is that the data does need to persist.
If any more detail would help, please feel free to ask. Any assistance greatly appreciated!
Check out the KD-Tree. It's specifically designed to speed up neighbour-finding in N-Dimensional spaces, like your rating system (Person 1 is 3 units along the X axis, 4 units along the Y axis, and so on).
You'll likely have to do this in an actual programming language. There are spatial indexes for some DBs, but they're usually designed for geographic work, like PostGIS (which uses GiST indexing), and only support two or three dimensions.
That said, I did find this tantalizing blog post on PostGIS. I was then unable to find any other references to this, but maybe your luck will be better than mine...
Hope that helps!
Technically your task is matching long strings made out of characters of a 5 letter alphabet. This kind of stuff is researched extensively in the area of computational biology. (Typically with 4 letter alphabets). If you do not know the book http://www.amazon.com/Algorithms-Strings-Trees-Sequences-Computational/dp/0521585198 then you might want to get hold of a copy. IMHO this is THE standard book on fuzzy matching / scoring of sequences.
Is your data sparse? With rating, most of the time not every user rates every object.
Naively comparing each object to every other is O(n*n*d), where d is the number of operations. However, a key trick of all the Hadoop solutions is to transpose the matrix, and work only on the non-zero values in the columns. Assuming that your sparsity is s=0.01, this reduces the runtime to O(d*n*s*n*s), i.e. by a factor of s*s. So if your sparsity is 1 out of 100, your computation will be theoretically 10000 times faster.
Note that the resulting data will still be a O(n*n) distance matrix, so strictl speaking the problem is still quadratic.
The way to beat the quadratic factor is to use index structures. The k-d-tree has already been mentioned, but I'm not aware of a version for categorical / discrete data and missing values. Indexing such data is not very well researched AFAICT.
Firstly , For those of your who dont know - Anytime Algorithm is an algorithm that get as input the amount of time it can run and it should give the best solution it can on that time.
Weighted A* is the same as A* with one diffrence in the f function :
(where g is the path cost upto node , and h is the heuristic to the end of path until reaching a goal)
Original = f(node) = g(node) + h(node)
Weighted = f(node) = (1-w)g(node) +h(node)
My anytime algorithm runs Weighted A* with decaring weight from 1 to 0.5 until it reaches the time limit.
My problem is that most of the time , it takes alot time until this it reaches a solution , and if given somthing like 10 seconds it usaully doesnt find solution while other algorithms like anytime beam finds one in 0.0001 seconds.
Any ideas what to do?
If I were you I'd throw the unbounded heuristic away. Admissible heuristics are much better in that given a weight value for a solution you've found, you can say that it is at most 1/weight times the length of an optimal solution.
A big problem when implementing A* derivatives is the data structures. When I implemented a bidirectional search, just changing from array lists to a combination of hash augmented priority queues and array lists on demand, cut the runtime cost by three orders of magnitude - literally.
The main problem is that most of the papers only give pseudo-code for the algorithm using set logic - it's up to you to actually figure out how to represent the sets in your code. Don't be afraid of using multiple ADTs for a single list, i.e. your open list. I'm not 100% sure on Anytime Weighted A*, I've done other derivatives such as Anytime Dynamic A* and Anytime Repairing A*, not AWA* though.
Another issue is when you set the g-value too low, sometimes it can take far longer to find any solution that it would if it were a higher g-value. A common pitfall is forgetting to check your closed list for duplicate states, thus ending up in a (infinite if your g-value gets reduced to 0) loop. I'd try starting with something reasonably higher than 0 if you're getting quick results with a beam search.
Some pseudo-code would likely help here! Anyhow these are just my thoughts on the matter, you may have solved it already - if so good on you :)
Beam search is not complete since it prunes unfavorable states whereas A* search is complete. Depending on what problem you are solving, if incompleteness does not prevent you from finding a solution (usually many correct paths exist from origin to destination), then go for Beam search, otherwise, stay with AWA*. However, you can always run both in parallel if there are sufficient hardware resources.