Question

我在R中有一个三维数组。我想使用count函数将其总结为一个三维矩阵。我该怎么办？

# input
scm_test <- array(0,c(5,5,2))

sender_test        <- c(1,1,3,4,5)
reciever_test      <- c(2,2,1,5,2)
week_test          <- c(1,1,2,1,2)

df <- as.data.frame(cbind(sender_test,reciever_test, week_test))

sender_1    <- df$sender_test
reciever_1  <- df$reciever_test
week_1      <- df$week_test


sender  =df$sender
reciever=df$reciever
month   =df$week_1

# the output should look like scm_test 
scm_test[1,2,1] <- 2
scm_test[3,1,2] <- 1
scm_test[4,5,1] <- 1
scm_test[5,2,2] <- 1
scm_test

Answer 1

我们可以这样进行：首先通过对“ df”的所有列进行分组来获取计数，然后使用“ df”作为3D索引来分配新的计数“ val”

library(tidyverse)
val <- df %>% 
          group_by_all() %>% 
          mutate(n = n()) %>% 
          pull(n)

或者使用ave中的base R获取“ val”

val <- with(df, ave(seq_along(sender_test), sender_test, 
              reciever_test, week_test, FUN = length))
scm_test[as.matrix(df)] <- val
scm_test
#, , 1

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    2    0    0    0
#[2,]    0    0    0    0    0
#[3,]    0    0    0    0    0
#[4,]    0    0    0    0    1
#[5,]    0    0    0    0    0

#, , 2

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    0    0    0    0
#[2,]    0    0    0    0    0
#[3,]    1    0    0    0    0
#[4,]    0    0    0    0    0
#[5,]    0    1    0    0    0

或使用count

dfN <- df %>% 
           count(!!! rlang::syms(names(.)))
replace(scm_test, as.matrix(dfN[names(df)]), dfN$n)

数据

scm_test <- array(0,c(5,5,2))

Answer 2

我通常尝试避免for循环，但是在这种情况下，这感觉像是最简单的解决方案：

for(i in 1:nrow(df)) {
  scm_test[df[i,1],df[i,2],df[i,3]]<-scm_test[df[i,1],df[i,2],df[i,3]]+1
}

对3维矩阵中的3维数组求和

2 个答案:

数据