ranking

How do I find the above without removing the largest element and searching again? Is there a more efficient way to do this? It does not matter if the these elements are duplicates. for (e: all elements) { if (e > largest) { second = largest; largest = e; } else if (e > second) { second = e; } } You could either initialize largest and second to an appropriate lower bound, or to the first two items in the list (check which one is bigger, and don't forget to check if the list has at least two items) using partial_sort ? std::partial_sort(aTest.begin(), aTest.begin() + 2, aTest.end(), Functor); An

When developing a database of articles in a Knowledge Base (for example) - what are the best ways to sort and display the most relevant answers to a users' question? Would you use additional data such as keyword weighting based on whether previous users found the article of help, or do you find a simple keyword matching algorithm to be sufficient? Perhaps the easiest and most naive approach that will give immediately useful results would be to implement *tf-idf : Variations of the tf–idf weighting scheme are often used by search engines as a central tool in scoring and ranking a document's

For a school project, we'll have to implement a ranking system. However, we figured that a dumb rank average would suck: something that one user ranked 5 stars would have a better average that something 188 users ranked 4 stars, and that's just stupid. So I'm wondering if any of you have an example algorithm of "smart" ranking. It only needs to take in account the rankings given and the number of rankings. Thanks! You can use a method inspired by Bayesian probability . The gist of the approach is to have an initial belief about the true rating of an item, and use users' ratings to update your

1.字符串（string） 1.1设置值 set key value [ex seconds] [px milliseconds] [nx|xx] 例如: 127.0.0.1:6379> set hello world OK 1.2获取值 127.0.0.1:6379> get hello "world" 字符串类型的内部编码有3种： int：8个字节的长整型。 embstr：小于等于39个字节的字符串。 raw：大于39个字节的字符串。 Redis会根据当前值的类型和长度决定使用哪种内部编码实现。 2.哈希(hash) hash由多个field构成，适合存储拥有多个属性的对象。 2.1 设置值 hset key field value 例如添加一个名字为tom的用户，user为key，name为field 127.0.0.1:6379> hset user name tom (integer) 1 再为这个hash添加一个名为age的field 127.0.0.1:6379> hset user age 13 (integer) 1 2.2 获取值 hget key field 例如获取用户的名字 127.0.0.1:6379> hget user name "tom" 获取用户的年龄 127.0.0.1:6379> hget user age 13 3.list

CREATE TABLE `players` ( `pid` int(2) NOT NULL AUTO_INCREMENT, `name` varchar(50) NOT NULL, `team` varchar(20) NOT NULL, `age` int(2) NOT NULL, PRIMARY KEY (`pid`), UNIQUE KEY `name` (`name`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `players` (`pid`, `name`, `age`, `team`) VALUES (1, 'Samual', 25, 'aa'), (2, 'Vino', 20, 'bb'), (3, 'John', 20, 'dd'), (4, 'Andy', 22, 'cc'), (5, 'Brian', 21, 'dd'), (6, 'Dew', 24, 'xx'), (7, 'Kris', 25, 'qq'), (8, 'William', 26, 'cc'), (9, 'George', 23, 'nn'), (10, 'Peter', 19, 'aa'), (11, 'Tom', 20, 'aa'), (12, 'Andre', 20, 'aa'); In the above table, I

I'm trying to compute a medal table for a sports event. My data looks like this: test <- data.frame("ID" = c("1_1", "1_2", "1_3", "1_4","1_5","1_6"), "gold"=c(10, 4, 1, 7, 7, 1), "silver"=c(1, 3, 2, 19, 19, 2), "bronze"=c(1, 8, 2, 0, 0, 2)) First, I want to order the data based on number of "gold", "silver", and "bronze", like this: (test_ordered <- with(test, test[order(-gold, -silver, -bronze), ])) Then compute the final medal rank. This is how the final rank column should like: (test_ordered$rank<-c(1, 2, 2, 4, 5, 5)) # ID gold silver bronze rank # 1 1_1 10 1 1 1 # 4 1_4 7 19 0 2 # 5 1_5 7