I'm researching how to implement leaderboards for my game with the parse.com SDK, my plan is to submit a score for the user every time they finish a level, attached to a "parent" leaderboard. I need to submit all scores because I need to retrieve leaderboards within time ranges (eg "all time", "last week", "last month", etc). The problem is, there'll be multiple scores for each user on the same leaderboard, and I only need to highest one. Is there a way to drop duplicate keys from a query? Is this the correct strategy? Everything else (sorting, paging, etc) seems to be in place.
Thanks.
You just need a table with 'User_id', 'Score', 'Level', and 'Date' (or whatever you need). Each time that a player finishes a level, you put the score into the table.
Then you have to calculate each (all time, last week, etc) in the query.
E.g.
10 Highs of the day:
SELECT TOP 10 User_id, Score, Date FROM Scores
WHERE Date = getdate()
ORDER BY Score DESC
I don't know if I understood the question. Let me know if I didn't.
Hope it helps.
As far as I understand from your question you want to retrieve data from Parse class. At the same time you want to eliminate the duplicate entry because user has multiple scores in different days. So to get the highest one, you have to query the class via query (based on SDK Android,iOS) and order by descending(based on your criteria), then obtain the first item in the result.
Or you can get the user all scores and create a structure where you can store the user scores as array list day by day. Based on day you can get the latest max or min scores. I hope I understand your question.
Hope this helps.Regards
Related
I've looked at many threads regarding COUNT and COUNTA, but I can't seem to figure out how to use it correctly.
I am new to DAX and am learning my way around. I have attempted to look this up and have gotten a little ways to where I need to be but not exactly. I think I am confused about how to apply a filter.
Here's the situation:
Four separate queries used to generate the data in the report; but only need to use two for the DAX function (Products and Display).
I have three columns I need to filter by, as follows:
Customer (Display or Products query; can do either)
Brand (Products query)
Location (Display query)
I want to count the columns based on if the data is unique.
Here's an example:
Customer: Big Box Buy;
Item: Lego Big Blocks;
Brand: Lego;
Location: Toys;
BREAK
Customer: Big Box Buy;
Item: Lego Star Wars;
Brand: Lego;
Location: Toys;
BREAK
Customer: Big Box Buy;
Item: Surface Pro;
Brand: Microsoft;
Location: Electronics;
BREAK
Customer: Little Shop on the Corner;
Item: Red Bicycle;
Brand: Trek;
Location: Racks;
In this example, no matter the fact that the items are different, we want to look at just the customer, the brand, and the location. We see in the first two records, the customer is "Big Box Buy" and the brand is "Lego" and the location is "Toys". This appears twice, but I want to count it distinct as "1". The next "Big Box Buy" store has the brand "Microsoft" and the location is "Electronics". It appears once and only once, and thus the distinct count is "1" anyway. This means that there are two separate entries for "Big Box Buy", both with a count of 1. And lastly there is "Little Shop on the Corner" which appears just once and is counted just once.
The "skeleton" of the code I have is basically just to see if I can get a count to work at all, which I can. It's the FILTER that I think is the problem (not used in the below example) judging by other threads I've read.
TotalDisplays = CALCULATE(COUNTA(products[Brand]))
Obviously I can't just count the amount of times a brand appears as that would give me duplicates. I need it unique based on if the following conditions are met:
Customer must be the same
Brand must be the same
Location must be the same
If so, we distinctly count it as one.
I know I ranted a bit and may seem to have gone in circles, but I was trying to figure out how to explain it. Please let me know if I need to edit this post or post clarification.
Many thanks in advance as I go through my journey with DAX!
I believe I have the answer. I used a NATURALINNERJOIN in DAX to create a new, merged table since I needed to reference all values in the same query (couldn't figure out how to do it otherwise). I also created an "unique identity" calculated column that combined data from multiple rows, but was hidden behind the scenes (not actually displayed on the report) so I could then take a measure of the unique values that way.
TotalDisplays = COUNTROWS(DISTINCT('GD-DP-Merge'[DisplayCountCalcCol]))
My calculated column is as follows:
DisplayCountCalcCol = 'GD-DP-Merge'[CustID] & 'GD-DP-Merge'[Brand] & 'GD-DP-Merge'[Location] & 'GD-DP-Merge'[Order#]
So the measure TotalDisplays now reports back the distinct count of rows based on the unique value of the customer ID, the brand, and the location of the item. I also threw in an order number just in case.
Thanks!
I am semi new to DAX and was struggling with Count and CountA formula, you post has helped me with answers. I would like to add the solution which i got for my query: Wanted count for Right Time start Achieved hence if anyone is looking for this kind of answer use below, filter will be selecting the table and adding string which you want to
RTSA:=calculate(COUNTA([RTS]),VEO_Daily_Services[RTS]="RTSA")
I'm new to Elasticsearch and I need to execute a complex query, but I need some help.
Here is my use case:
I would like to recommend a new place to each of my users everyday.
However:
The place must be opened this week day
The chosen place must be near of the user (closer places have higher score)
The place should not be one of the last 10 places a user have already been/suggested (if a place has already been visited by a user in his last 10 visits, this place should have a lower score)
My first guess is to have 2 documents types as follow:
user_history
user_id
place_id
date
place
place_id
opening_days (array with week days the place opens)
location geo position of the place
Given a user with position [lat, lon] and id user_1, what could be the search query to execute to retrieve X places sorted by score? (better score is near of user and not in the 10 last places a user have already been).
This query seems to be a basic but I can't figure out how to "mix" data from user_history and from place to get places I want.
But that's not all!
With this query, if I want to attribute to each user a place I need 3 steps:
retrieve all users (with their position)
for each user, search for the best place
once I have this place, add it to the user_history
This seems very time consuming task. Is it possible to simplify it with less Elasticsearch queries?
For instance, having something like this:
retrieving for each user his best place (with 1 query, search for all users and find them the best place)
add the place to the history
Or event better:
retrieving for each user his best place and add it to the history (with 1 query, perform all the 3 tasks above)
I don't know if it's possible to create queries that complex. That's why I need your help to tell me if it's possible and how it could be accomplished.
My app has a class that saves picture that users upload. Each object in the class has a city property that holds the name of the city that the picture was taken at, and a like property that tracks the number of likes.
I want to be able to send a query that returns one picture per city and each picture should have the highest ranking of likes in the city it belongs to. How can I do that?
One way which I first thought about is doing multiple queries by fetching the most liked picture of a city and save it in an array, and then do the same to other cities.
However, each country has more than one city, thus it's not that efficient.
Parse doesn't support the ordinary operations used in databases. Besides, I tried to use a compound query. Unfortunately, I can't set limit or ordering on the subqueries. Any good solution for this?
It would be easy using group by. Unfortunately, Parse does not support "select distinct" or "group by" features.
As you've suggested you need to fetch for each country all the cities, and for each one get the top most rated photo.
BUT, since Parse has strict restrictions on the duration time execution of a request ( 3 sec for an event listener, 7 sec for a custom function ), I suggest you to do this in a background job, saving in a new table the top rated photo for each city. In this way you can easily query the db from client. The Background jobs can be executed up to 15 minuted before parse drop them, so you could make that kind of queries without timeouts.
Hope it helps
Suppose I have a large (300-500k) collection of text documents stored in the relational database. Each document can belong to one or more (up to six) categories. I need users to be able to randomly select documents in a specific category so that a single entity is never repeated, much like how StumbleUpon works.
I don't really see a way I could implement this using slow NOT IN queries with large amount of users and documents, so I figured I might need to implement some custom data structure for this purpose. Perhaps there is already a paper describing some algorithm that might be adapted to my needs?
Currently I'm considering the following approach:
Read all the entries from the database
Create a linked list based index for each category from the IDs of documents belonging to the this category. Shuffle it
Create a Bloom Filter containing all of the entries viewed by a particular user
Traverse the index using the iterator, randomly select items using Bloom Filter to pick not viewed items.
If you track via a table what entries that the user has seen... try this. And I'm going to use mysql because that's the quickest example I can think of but the gist should be clear.
On a link being 'used'...
insert into viewed (userid, url_id) values ("jj", 123)
On looking for a link...
select p.url_id
from pages p left join viewed v on v.url_id = p.url_id
where v.url_id is null
order by rand()
limit 1
This causes the database to go ahead and do a 1 for 1 join, and your limiting your query to return only one entry that the user has not seen yet.
Just a suggestion.
Edit: It is possible to make this one operation but there's no guarantee that the url will be passed successfully to the user.
It depend on how users get it's random entries.
Option 1:
A user is paging some entities and stop after couple of them. for example the user see the current random entity and then moving to the next one, read it and continue it couple of times and that's it.
in the next time this user (or another) get an entity from this category the entities that already viewed is clear and you can return an already viewed entity.
in that option I would recommend save a (hash) set of already viewed entities id and every time user ask for a random entity- randomally choose it from the DB and check if not already in the set.
because the set is so small and your data is so big, the chance that you get an already viewed id is so small, that it will take O(1) most of the time.
Option 2:
A user is paging in the entities and the viewed entities are saving between all users and every time user visit your page.
in that case you probably use all the entities in each category and saving all the viewed entites + check whether a entity is viewed will take some time.
In that option I would get all the ids for this topic- shuffle them and store it in a linked list. when you want to get a random not viewed entity- just get the head of the list and delete it (O(1)).
I assume that for any given <user, category> pair, the number of documents viewed is pretty small relative to the total number of documents available in that category.
So can you just store indexed triples <user, category, document> indicating which documents have been viewed, and then just take an optimistic approach with respect to randomly selected documents? In the vast majority of cases, the randomly selected document will be unread by the user. And you can check quickly because the triples are indexed.
I would opt for a pseudorandom approach:
1.) Determine number of elements in category to be viewed (SELECT COUNT(*) WHERE ...)
2.) Pick a random number in range 1 ... count.
3.) Select a single document (SELECT * FROM ... WHERE [same as when counting] ORDER BY [generate stable order]. Depending on the SQL dialect in use, there are different clauses that can be used to retrieve only the part of the result set you want (MySQL LIMIT clause, SQLServer TOP clause etc.)
If the number of documents is large the chance serving the same user the same document twice is neglibly small. Using the scheme described above you don't have to store any state information at all.
You may want to consider a nosql solution like Apache Cassandra. These seem to be ideally suited to your needs. There are many ways to design the algorithm you need in an environment where you can easily add new columns to a table (column family) on the fly, with excellent support for a very sparsely populated table.
edit: one of many possible solutions below:
create a CF(column family ie table) for each category (creating these on-the-fly is quite easy).
Add a row to each category CF for each document belonging to the category.
Whenever a user hits a document, you add a column with named and set it to true to the row. Obviously this table will be huge with millions of columns and probably quite sparsely populated, but no problem, reading this is still constant time.
Now finding a new document for a user in a category is simply a matter of selecting any result from select * where == null.
You should get constant time writes and reads, amazing scalability, etc if you can accept Cassandra's "eventually consistent" model (ie, it is not mission critical that a user never get a duplicate document)
I've solved similar in the past by indexing the relational database into a document oriented form using Apache Lucene. This was before the recent rise of NoSQL servers and is basically the same thing, but it's still a valid alternative approach.
You would create a Lucene Document for each of your texts with a textId (relational database id) field and multi valued categoryId and userId fields. Populate the categoryId field appropriately. When a user reads a text, add their id to the userId field. A simple query will return the set of documents with a given categoryId and without a given userId - pick one randomly and display it.
Store a users past X selections in a cookie or something.
Return the last selections to the server with the users new criteria
Randomly choose one of the texts satisfying the criteria until it is not a member of the last X selections of the user.
Return this choice of text and update the list of last X selections.
I would experiment to find the best value of X but I have in mind something like an X of say 16?
Consider I have two collections in MongoDB. One for products with documents like:
{'_id': ObjectId('lalala'), 'title': 'Yellow banana'}
And another stores price changes with documents like:
{'product': DBRef('products', ObjectId('lalala')),
'since': datetime(2011, 4, 5),
'new_price': 150 }
One product may have many price changes. The price lasts until a new change with later time stamp. I guess you've caught idea.
Say, I have 100 products. I want to query my DB to get know what's the price of each product at the moment of June 9, 2011. What is the most efficient (quick) way to perform this query in MongoDB? Suppose I have no cache solution or cache is empty.
I thought about group statement on prices collection, where reduce function would select last since before a date provided, grouping by product.$id. But in this case I would not benefit from an index on since field and all documents would be scanned.
Any ideas?
I had a similar problem, but for GPS locations. I found the fastest way was to set up a query for each item, which is rather counter-intuitive if your used to SQL databases.
Query for the item where it's timestamp is less or equal than the date your looking for, and limit the result to 1. Repeat for each item. To really speed things up, run multiple querys in parallel to utilise all the cores on the MongoDB server.