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Replace the NA line with the previous line in R

I was wondering if anyone has a quick and dirty solution to the following problem: I have a matrix with NA rows and I would like to replace the NA rows with the previous row (unless it is also an NA row).

Suppose the first line is not an NAs string

Thank!

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6 answers


Adapted from answer to this question: An idiomatic way to copy the values ​​of cells "down" in a vector R



f <- function(x) {
  idx <- !apply(is.na(x), 1, all)
  x[idx,][cumsum(idx),]
}

x <- data.frame(a=c(1, 2, NA, 3, NA, NA), b=c(4, 5, NA, 6, NA, 7))

> x
   a  b
1  1  4
2  2  5
3 NA NA
4  3  6
5 NA NA
6 NA  7

> f(x)
     a b
1    1 4
2    2 5
2.1  2 5
4    3 6
4.1  3 6
6   NA 7
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Trying to think about times, you might have two NA strings per line.

#create a data set like you discuss (in the future please do this yourself)
set.seed(14)
x <- matrix(rnorm(10), nrow=2)
y <- rep(NA, 5)
v <- do.call(rbind.data.frame, sample(list(x, x, y), 10, TRUE))

One approach:



NArows <- which(apply(v, 1, function(x) all(is.na(x))))          #find all NAs
notNA <- which(!seq_len(nrow(v)) %in% NArows)                    #find non NA rows
rep.row <- sapply(NArows, function(x) tail(notNA[x > notNA], 1)) #replacement rows
v[NArows, ] <- v[rep.row, ]                                      #assign
v                                                                #view

It won't work if your first line is all NA.

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If m

is your matrix, this is your quick and dirty solution:

sapply(2:nrow(m),function(i){ if(is.na(m[i,1])) {m[i,] <<- m[(i-1),] } })

Note that it uses an ugly (and dangerous) operator <<-

.

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You can always use a loop, assuming 1 is not NA, as indicated:

fill = data.frame(x=c(1,NA,3,4,5))
for (i in 2:length(fill)){
  if(is.na(fill[i,1])){ fill[i,1] = fill[(i-1),1]}
 }
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Matthew's example:

x <- data.frame(a=c(1, 2, NA, 3, NA, NA), b=c(4, 5, NA, 6, NA, 7))
 na.rows <- which( apply( x , 1, function(z) (all(is.na(z)) ) ) )
 x[na.rows , ] <- x[na.rows-1, ]
 x
#---
   a b
1  1 4
2  2 5
3  2 5
4  3 6
5  3 6
6 NA 7

Obviously the first line with all NA will present problems.

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Here's a simple and conceptually, arguably the simplest one-liner:

x <- data.frame(a=c(1, 2, NA, 3, NA, NA), b=c(4, 5, NA, 6, NA, 7))

   a  b
1  1  4
2  2  5
3 NA NA
4  3  6
5 NA NA
6 NA  7

x1<-t(sapply(1:nrow(x),function(y) ifelse(is.na(x[y,]),x[y-1,],x[y,])))

     [,1] [,2]
[1,] 1    4   
[2,] 2    5   
[3,] 2    5   
[4,] 3    6   
[5,] 3    6   
[6,] NA   7

To get back the column names, just use the code names (x1) <-colnames (x)

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