在没有For循环的情况下替换或估算R中的NA值

Question

Is there a better way to go through observations in a data frame and impute NA values? I've put together a 'for loop' that seems to do the job, swapping NAs with the row's mean value, but I'm wondering if there is a better approach that does not use a for loop to solve this problem -- perhaps a built in R function?

# 1. Create data frame with some NA values. 

rdata <- rbinom(30,5,prob=0.5)
rdata[rdata == 0] <- NA
mtx <- matrix(rdata, 3, 10)
df <- as.data.frame(mtx)  
df2 <- df

# 2. Run for loop to replace NAs with that row's mean.

for(i in 1:3){            # for every row
x <- as.numeric(df[i,])   # subset/extract that row into a numeric vector
y <- is.na(x)             # create logical vector of NAs
z <- !is.na(x)            # create logical vector of non-NAs
result <- mean(x[z])      # get the mean value of the row 
df2[i,y] <- result        # replace NAs in that row
}

# 3. Show output with imputed row mean values.

print(df)  # before
print(df2) # after 

Answer 1

Here's a possible vectorized approach (without any loop)

indx <- which(is.na(df), arr.ind = TRUE)
df[indx] <- rowMeans(df, na.rm = TRUE)[indx[,"row"]]

---

Some explanation

We can identify the locations of the NA s using the arr.ind parameter in which . Then we can simply index df (by the row and column indexes) and the row means (only by the row indexes) and replace values accordingly

Answer 2

One possibility, using impute from Hmisc , which allows for choosing any function to do imputation,

library(Hmisc)
t(sapply(split(df2, row(df2)), impute, fun=mean))

Also, you can hide the loop in an apply

t(apply(df2, 1, function(x) {
    mu <- mean(x, na.rm=T)
    x[is.na(x)] <- mu
    x
}))

Answer 3

Data:

set.seed(102)
rdata <- matrix(rbinom(30,5,prob=0.5),nrow=3)
rdata[cbind(1:3,2:4)] <- NA
df <- as.data.frame(rdata)

This is a little trickier than I'd like -- it relies on the column-major ordering of matrices in R as well as the recycling of the row-means vector to the full length of the matrix. I tried to come up with a sweep() solution but didn't manage so far.

rmeans <- rowMeans(df,na.rm=TRUE)
df[] <- ifelse(is.na(df),rmeans,as.matrix(df))