计算重复行的频率

Question

我有一个包含大量重复项的矩阵,并希望获得一个矩阵,其中包含唯一行和每个唯一行的频率计数.

下面显示的示例解决了这个问题,但速度很慢.

rowsInTbl <- function(tbl,row){
  sum(apply(tbl, 1, function(x) all(x == row) ))
}

colFrequency <- function(tblall){
  tbl <- unique(tblall)
  results <- matrix(nrow = nrow(tbl),ncol=ncol(tbl)+1)
  results[,1:ncol(tbl)] <- as.matrix(tbl)
  dimnames(results) <- list(c(rownames(tbl)),c(colnames(tbl),"Frequency"))

  freq <- apply(tbl,1,function(x)rowsInTbl(tblall,x))
  results[,"Frequency"] <- freq
  return(results)
}


m <- matrix(c(1,2,3,4,3,4,1,2,3,4),ncol=2,byrow=T)
dimnames(m) <- list(letters[1:nrow(m)],c("c1","c2"))
print("Matrix")
print(m)

[1] "Matrix"
  c1 c2
a  1  2
b  3  4
c  3  4
d  1  2
e  3  4

print("Duplicate frequency table")
print(colFrequency(m))


[1] "Duplicate frequency table"
  c1 c2 Frequency
a  1  2         2
b  3  4         3

以下是与我的例子相比,@ Heroka和@ m0h3n答案的速度测量值.将上面显示的基质重复1000次.Data.table显然是最快的解决方案.

[1] "Duplicate frequency table - my example"
   user  system elapsed 
   0.372   0.000   0.371 

[1] "Duplicate frequency table - data.table"
   user  system elapsed 
   0.008   0.000   0.008 

[1] "Duplicate frequency table - aggregate"
   user  system elapsed 
   0.092   0.000   0.089 

Answer 1

看起来像是一份工作data.table,因为你需要能够快速聚合的东西.

library(data.table)


m <- matrix(c(1,2,3,4,3,4,1,2,3,4),ncol=2,byrow=T)

mdt <- as.data.table(m)

res <- mdt[,.N, by=names(mdt)]
res
# > res
# V1 V2 N
# 1:  1  2 2
# 2:  3  4 3