How to handle missing values in R by replacing it with mean? [closed]

Answer 1

If this is your data set

temp <- read.table(text = "row1 key1 10 
row2 key1 12
row3 key1 NA
row4 key2 2
row5 key2 3
row6 key2 NA", header = F)

library(data.table)
setDT(temp)[, V3 := as.numeric(V3)]
temp[, V3 := lapply(.SD, function(x) ifelse(is.na(x), mean(x, na.rm = T), V3)), .SDcols = "V3",  by = V2]

Answer 2

There will be tons of ways to do this. The library zoo has a built in function na.aggregate which does precisely what you want.

library(zoo)
d <- data.frame(key = rep(c("key1", "key2"), each = 3), 
    value = c(10, 12, NA, 2, 3, NA))  
with(d, na.aggregate(value, by = key))

Answer 3

You could try:

 dat <- read.table(text="row1 key1 10 
 row2 key1 12
 row3 key1 NA
 row4 key2 2
 row5 key2 3
 row6 key2 NA
 row7 key2 NA",sep="",header=F,stringsAsFactors=F)

 repl <- with(dat, table(is.na(V3),V2)[2,]) #to get the number of missing values per group
 dat1 <- dat
 indx <- is.na(dat$V3) #create a index of position of missing values
 dat$V3[indx] <- rep(with(dat, by(V3, V2, FUN=mean, na.rm=T)), repl) #replicate by `repl` in cases of multiple NAs per group

Or use ave

 dat1$V3[indx] <- with(dat1, ave(V3, V2, FUN=function(x) mean(x, na.rm=T)))[indx]
identical(dat1, dat)
#[1] TRUE

 dat1$V3
# [1] 10.0 12.0 11.0  2.0  3.0  2.5  2.5

Answer 4

This should do the trick, using your data:

 d <- data.frame(id = rep(c("key1", "key2"), each = 3), x = c(10, 12, NA, 2, 3, NA))  
 library(plyr)
 d2 <- ddply(d, .(id), transform, x=ifelse(is.na(x), mean(x, na.rm=T), x) )

Here, we split the dataframe by your key and transform x depending on whether or not it is NA. The result is

     id    x
 1 key1 10.0
 2 key1 12.0
 3 key1 11.0
 4 key2  2.0
 5 key2  3.0
 6 key2  2.5