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Say you have a <- c(1,3,5,7,9) b <- c(3,6,8,9,10) c <- c(2,3,4,5,7,9) A straightforward way to do the job is: intersect(intersect(a,b),c) More cleverly, and more conveniently if you have a lot of arguments: Reduce(intersect, list(a,b,c)) The Redu…
What Is It? A hash table, or associative array, is a well known key-value data structure. In R there is no equivalent, but you do have some options. You can use a vector of any type, a list, or an environment. But as you’ll see with all of these opti…
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R语言实战实现基于用户的简单的推荐系统(数量较少) a<-c(1,1,1,1,2,2,2,2,3,3,3,4,4,4,5,5,5,5,6,6,7,7) b<-c(1,2,3,4,2,3,4,5,4,1,2,3,2,4,5,2,6,4,1,2,3,4) da<-data.frame(a,b) a<-c(1,1,2,2,3,3,3,3,3,4,4,5,5,5,6,6,7,7) b<-c(2,5,7,2,6,4,7,1,8,6,3,3,4,1,2,4,4,9) da2<-da…
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problem Iahub and Xors 题目大意 一个n*n的矩阵,要求支持两种操作. 操作1:将一个子矩阵的所有值异或某个数. 操作2:询问某个子矩阵的所以值的异或和. 解题分析 由于异或的特殊性,可以用二维树状数组来维护. 因为同一个值只有异或奇数次才有效,因此若单点修改某个点,那么其影响的点为与其行和列奇偶性相同的点,故可以开4个二维树状数组来维护. 如果是区间修改x1,y1,x2,y2,则只需单点修改(x1,y1).(x1,y2+1).(x2+1,y2).(x2+1,y2+1) 如…