ifelse与for循环

Question

I would like to traverse through rows of a matrix and perform some operations on data entries based on a condition.

Below is my code

m = matrix(c(1,2,NA,NA,5,NA,NA,1,NA,NA,NA,NA,4,5,NA,NA,NA,NA,NA,NA), nrow = 5, ncol = 4)
if (m[,colSums(!is.na(m)) > 1, drop = FALSE]){
        for(i in 1:4){
              a = which(m[i,] != "NA") - mean(which(!is.na(m[i,])))
                for(j in 2:5){
                       b = which(m[j,] != "NA") - mean(which(!is.na(m[j,])))
                       prod(a,b)
                }
        }
}

I get a warning message as below in my "if" condition

Warning message:
In if (m[, colSums(!is.na(m)) > 1, drop = FALSE]) { :
  the condition has length > 1 and only the first element will be used

I know it returns a vector and I should be using ifelse block. How to incorporate for loops inside ifelse block? It seems to be a basic question, I am new to R.

Answer 1

Based on your description, you want to check the number of non NA in matrix by column and then do something dependent on this results (that why you need "if"/"ifelse" statement). So, you can implemented as below, and write inner loops in a specific function.

yourFunc <- function(x, data) {
 # do what your want / your loops on "data"
 # sample, you can check the result in here
 if(x > 1)  1
 else       0
}

m = matrix(c(1,2,NA,NA,5,NA,NA,1,NA,NA,NA,NA,4,5,NA,NA,NA,NA,NA,NA), nrow = 5, ncol = 4)
# use "apply" series function in here
sapply(colSums(!is.na(m)), yourFunc, data=m)
#[1] 1 0 1 0

Actually, I think you need to re-organize your problem and optimize the code, the "ifelse with for loop" may be totally unnecessary.

Answer 2

As you are new to R, I assume that some of the terminology is maybe a bit confusing. So here is a little explanation regarding the if statement .

Lets look at the if condition :

m[,colSums(!is.na(m)) > 1, drop = FALSE]

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

This is nothing that if can work with as an if condition has to be boolean (evaluate to TRUE/FALSE). So why the result? Well the result of

colSums(!is.na(m))
[1] 3 1 2 0

is a vector of counts of entries that are not NA ! (= number of TRUE's in each column). Be carful as this is not the same as

colSums(m, na.rm = TRUE)
[1] 8 1 9 0

which returns a vector of sums over all five rows for each column, excluding NA 's. My guess is that the latter is what you are looking for. In any case: be aware of the difference!
By asking which of those sums is greater than 1 you do get a boolean vector

colSums(!is.na(m)) > 1
[1]  TRUE FALSE  TRUE FALSE

However, using that boolean vector as a criteria for selecting columns, you correctly get a matrix which is obviously not boolean:

m[,colSums(!is.na(m)) > 1]

Note: drop = FALSE is unnecessary here as there are no dimensions to be dropped potentially. See ? [ or ?drop . You can verify this using identical :

identical(m[,colSums(!is.na(m)) > 1, drop = FALSE],
          m[,colSums(!is.na(m)) > 1])

Now to the loop. You find tons of discussions on avoiding for loops and using the apply family of functions . I suspect you have to take some time togo through all that. Note however, that using apply - contrary to common belief - is not necessarily superior to a for loop in terms of speed, as it is actually just a fancy wrapper around a for loop (check the source code!). It is, however, clearly superior in terms of code clarity as it is compact and clear about what it is doing. So do try to use apply functions if possible!

In order to rewrite your loop it would be helpful if you could verbally describe what you actually want to do, since I assume that what the loop is doing right now is probably not what you want. As which() returns the index/posistion of an element in a vector or matrix what you are basically doing is:

indices of the i'th row that are not NA (for a given column) - mean over these indices

While this is theoretically possible, this usually doesnt make much sense. So with all my notes at hand: clearly state your problem so we can think of a fix.