我正在努力在数据集中输入缺失值。我有一个预测模型,可以生成值来估算缺失。当我使用dfm [is.na(dfm)]< -impute进行插补时,它会以列方式输入值。但是我需要对行进行估算,因此我转换了数据矩阵。我的问题是,有没有一种优雅的方法可以做到这一点而无需转置矩阵?这是一个带有可重复示例的rcode。
set.seed(1)
r=5
c=4
df<-matrix(runif(r*c), ncol=c)
df
[,1] [,2] [,3] [,4]
[1,] 0.2655087 0.89838968 0.2059746 0.4976992
[2,] 0.3721239 0.94467527 0.1765568 0.7176185
[3,] 0.5728534 0.66079779 0.6870228 0.9919061
[4,] 0.9082078 0.62911404 0.3841037 0.3800352
[5,] 0.2016819 0.06178627 0.7698414 0.7774452
d=dim(df)
p=0.30
#### generate missing data matrix by replacing some values by NAs
dfm<-df
dfm[matrix(rbinom(prod(d), size=1,prob=p)==1,nrow=d[1])]<-NA
dfm
[,1] [,2] [,3] [,4]
[1,] NA 0.89838968 0.2059746 0.4976992
[2,] 0.3721239 0.94467527 0.1765568 NA
[3,] 0.5728534 0.66079779 0.6870228 0.9919061
[4,] 0.9082078 NA 0.3841037 NA
[5,] 0.2016819 0.06178627 NA 0.7774452
# generate values to impute the missing
impute<-rgamma(sum(is.na(dfm)),shape=1,scale=0.5)
impute
[1] 0.6804725 0.6029941 0.2770577 0.6035013 0.7812393
#imputes columnwise
dfm[is.na(dfm)]<-impute
dfm
[,1] [,2] [,3] [,4]
[1,] 0.6804725 0.89838968 0.2059746 0.4976992
[2,] 0.3721239 0.94467527 0.1765568 0.6035013
[3,] 0.5728534 0.66079779 0.6870228 0.9919061
[4,] 0.9082078 0.60299409 0.3841037 0.7812393
[5,] 0.2016819 0.06178627 0.2770577 0.7774452
#impute rowwise
tdfm<-t(dfm)
tdfm[is.na(tdfm)]<-impute
tdfm
[,1] [,2] [,3] [,4] [,5]
[1,] 0.6804725 0.3721239 0.5728534 0.9082078 0.20168193
[2,] 0.8983897 0.9446753 0.6607978 0.2770577 0.06178627
[3,] 0.2059746 0.1765568 0.6870228 0.3841037 0.78123933
[4,] 0.4976992 0.6029941 0.9919061 0.6035013 0.77744522
dfm.fill<-t(tdfm)
dfm.fill
[,1] [,2] [,3] [,4]
[1,] 0.6804725 0.89838968 0.2059746 0.4976992
[2,] 0.3721239 0.94467527 0.1765568 0.6029941
[3,] 0.5728534 0.66079779 0.6870228 0.9919061
[4,] 0.9082078 0.27705769 0.3841037 0.6035013
[5,] 0.2016819 0.06178627 0.7812393 0.7774452
答案 0 :(得分:4)
使用which
代替arr.ind
,以便您可以先按行排序。
示例:
test1 <- matrix(1:12, 3, 4, byrow = TRUE)
test1[c(1, 3, 8, 6, 10)] <- NA
test2 <- test1
impute <- c(-1, -4, -7, -9, -10)
## What you are currently doing--column-wise
test1[is.na(test1)] <- impute
test1
# [,1] [,2] [,3] [,4]
# [1,] -1 2 3 -10
# [2,] 5 6 -9 8
# [3,] -4 -7 11 12
## What it sounds like you want--row-wise
nas <- which(is.na(test2), arr.ind = TRUE)
test2[nas[order(nas[, "row"]), ]] <- impute
test2
# [,1] [,2] [,3] [,4]
# [1,] -1 2 3 -4
# [2,] 5 6 -7 8
# [3,] -9 -10 11 12