我有以下数据集:
day <- c(rep(17,4), rep(18,2))
beep <- c(74.50, 77.50, 89.50, 75.25, 58.25, 81.25)
m <- cbind(day, beep)
m
day beep
[1,] 17 74.50
[2,] 17 77.50
[3,] 17 89.50
[4,] 17 75.25
[5,] 18 58.25
[6,] 18 81.25
我想要的是将此数据集转换为具有天数(在本例中为2)作为列数的矩阵。这就是它想要的:
[,1] [,2]
[1,] 74.50 58.25
[2,] 77.50 81.25
[3,] 89.50 NA
[4,] 75.25 NA
由于这个人在第1天有4次哔哔声,第2天有2次哔哔声,因此必须有2个NAs必须在矩阵内。我很想知道如何在这里打开上面的数据集,而不像我现在做的那样手动调整它。
答案 0 :(得分:1)
我同意@ flodel的评论,但这是一种方式:
m2 <- unstack(m, beep~day)
nrow <- max(sapply(m2, length))
m2 <- sapply(m2, function(x) {
length(x) <- nrow
x
})
# 17 18
#[1,] 74.50 58.25
#[2,] 77.50 81.25
#[3,] 89.50 NA
#[4,] 75.25 NA
答案 1 :(得分:1)
您还可以使用reshape
包中的stats
功能,但您需要将matrix
转换为data.frame
并将data.frame
转换为matrix
}。但我认为它是更灵活的方式,因为你为变量创建了你想要负责列的id(在你的情况下是一天)。
m.df<-as.data.frame(m) ## convert to data.frame
m.df$id<-ave(m.df$day,m.df$day,FUN=seq_along) ### create index with ave function (more solutions: http://stackoverflow.com/questions/8997638/numbering-by-groups )
m2<-reshape(m.df,idvar='id',timevar='day',direction='wide') ## reshape data timevar is responsible for columns and with direction you tell how data set should be expaned.
as.matrix(m2) ### convert back to matrix
答案 2 :(得分:0)
假设矩阵的暗度已知,这可能比其他方法简单。
n <- m[, "beep"]
length(n) <- 8 # Since n is a vector with 6 elements, this will make it 8 element-
# vector with two NAs attached.
dim(n) <- c(4, 2) # change the dimension of the vector to a 4 by 2 matrix
print(n)
[,1] [,2]
[1,] 74.50 58.25
[2,] 77.50 81.25
[3,] 89.50 NA
[4,] 75.25 NA