if there is at least one Na in the line, make the line Na-CodePudding

my data:

    structure(list(col1 = 1:6, col2 = c(NA, NA, 3L, 4L, 5L, 6L), 
    col3 = c(1L, 1L, NA, 1L, NA, 1L), col4 = c(2L, 2L, 2L, 2L, 
    NA, 2L), col5 = c(3L, 3L, 3L, NA, 3L, 3L), GROUP = c(1L, 
    1L, 1L, 2L, 2L, 2L)), class = "data.frame", row.names = c(NA, -6L))

I want to get a new date frame in which, if NA occurs, then replace the entire line with NA from the combination list(c(1,2),c(3,4,5)).

What I want to get:

data= structure(list(col1 = c(NA, NA, 3L, 4L, 5L, 6L), col2 = c(NA, 
NA, 3L, 4L, 5L, 6L), col3 = c(1L, 1L, NA, NA, NA, 1L), col4 = c(2L, 
2L, NA, NA, NA, 2L), col5 = c(3L, 3L, NA, NA, NA, 3L), GROUP = c(1L, 
1L, 1L, 2L, 2L, 2L)), class = "data.frame", row.names = c(NA, 
-6L))

UPDATE: I tried to do so:

list_v <- list(c(1,2),c(3,4,5))
data1 <- do.call(qpcR:::cbind.na, lapply(list_v, function(i)data[complete.cases(data[c(i)]),i]))

I need not to delete the line with Where there is at least one NA, but to replace the remaining elements with Na where there is at least one Na

CodePudding user response：

You can do:

l <- lapply(list(c(1,2), c(3,4,5)), \(i) df[, i])

cbind.data.frame(lapply(l, \(y) t(apply(y, 1, \(x) {
  if(any(is.na(x))) x[1:length(x)] <- NA
  x
}))))

#  col1 col2 col3 col4 col5
#1   NA   NA    1    2    3
#2   NA   NA    1    2    3
#3    3    3   NA   NA   NA
#4    4    4   NA   NA   NA
#5    5    5   NA   NA   NA
#6    6    6    1    2    3

Using apply and an if else statement, you can do:

t(apply(df, 1, function(x){
  if(any(is.na(x[1:2])))
    x[1:2] <- NA
  else if(any(is.na(x[3:5])))
    x[3:5] <- NA
  x
}))

#     col1 col2 col3 col4 col5 GROUP
#[1,]   NA   NA    1    2    3     1
#[2,]   NA   NA    1    2    3     1
#[3,]    3    3   NA   NA   NA     1
#[4,]    4    4   NA   NA   NA     2
#[5,]    5    5   NA   NA   NA     2
#[6,]    6    6    1    2    3     2

Here, apply repeats a function (defined in the third parameter) on each rows (MARGIN = 1) of df.