Replacing zero values in table A with non-zero values in table B if available

user_8071608 0
Viewed 54

I have data as follows:

tableA <- structure(c(0L, 0L, 6L, 0L, 6L, 0L, 3L, 0L, 0L), dim = c(3L, 
3L), dimnames = structure(list(c("0.3", "0.4", 
"0.6"), c("A", "B", "C")), names = c("", "")), class = "table")

      A B C
  0.3 0 0 3
  0.4 0 6 0
  0.6 6 0 0

tableB <-structure(c(0L, 1L, 2L, 0L, 3L, 0L, 3L, 2L, 0L), dim = c(3L, 
3L), dimnames = structure(list(c("0.3", "0.4", 
"0.6"), c("A", "B", "C")), names = c("", "")), class = "table")

      A B C
  0.3 0 0 3
  0.4 1 3 2
  0.6 2 0 0

I would like to replace all zeroes in tableA with positive values in tableB if they are available and somehow mark them.

What would be the easiest way to achieve this?

Desired output:

      A B C
  0.3 0 0 3
  0.4 1 6 2
  0.6 6 0 0

The added issue is that I would somehow like to know which values were replaced, but that seems difficult to realise.

r
4 Answers

Create a logical index based on the 0 values in tableA with == and the positive values (> 0) from tableB and do the assignment

i1 <- tableA == 0 & tableB > 0
tableA[i1] <- tableB[i1]

-output

> tableA
     
      A B C
  0.3 0 0 3
  0.4 1 6 2
  0.6 6 0 0
user_37322710

We can use !tableA as a mask like below

> (!tableA) * tableB + tableA

      A B C
  0.3 0 0 3
  0.4 1 6 2
  0.6 6 0 0
user_121587570

Another possible solution:

ifelse(tableA == 0, pmax(tableB,0), tableA)

#>      
#>       A B C
#>   0.3 0 0 3
#>   0.4 1 6 2
#>   0.6 6 0 0
user_115644870

another approach

ifelse(tableA == 0 & tableB > 0 , tableB , tableA)
user_191983370
Related