How to change matrix entries using conditional if in R

Question

I have this example matrix and I want to change the entries of the matrix with "YES" or "NO" based on a conditional if statement.

a<-c(5,1,0,3,2,0.6,1.6,7,9,0)
b<-c(11,0,1,18,11,11,0,13,20,10)
c<-c(10,20,0.7,0.8,0.3,0.4,0,0.9,1,1)

MAT<-cbind(a,b,c)
MAT

for (i in 1:nrow(MAT)){
  for (j in 1:ncol(MAT)){
  if (MAT[i,j]>5){
    MAT[i,j]="YES"
    } else {
    MAT[i,j]="NO"
    }
  }
}
print(MAT)

The output I got is like this and its wrong. Please help tell me what's wrong and how to fix it?

      a     b    c   
[1,] "NO"  "NO" "NO"
[2,] "NO"  "NO" "NO"
[3,] "NO"  "NO" "NO"
[4,] "NO"  "NO" "NO"
[5,] "NO"  "NO" "NO"
[6,] "NO"  "NO" "NO"
[7,] "NO"  "NO" "NO"
[8,] "YES" "NO" "NO"
[9,] "YES" "NO" "NO"
[10,] "NO"  "NO" "NO"

Thoron

Answer 1

You do not need loops here. Just use the whole matrix in your call to x>5

ifelse(MAT>5, "YES", "NO")

This will do the logical operation over the entire matrix, and output a logical matrix.

You can reassign the VALUES from the output of ifelse() while keeping the STRUCTURE of MAT by using the empty brackets [], as in:

MAT[]<-ifelse(MAT>5, "YES", "NO")

Paralysis

It works well as I want and it is the simplest code among others so I accept this – Thoron

Empty brackets after MAT are not necessary here because ifelse returns a value with the same shape as its test argument. Since MAT>5 is a matrix, the value of ifelse will be a matrix as well. – Codi

Yes, you are right for this particular case, @Robert. But I think including the empty brackets is good practice here, as it is a more general approach that enforces the preservation of the class of the original data. Ifelse does return an object the same type of the 'test' argument. But the '<' is a function that coerces to a logical vector or matrix. If MAT was a data.frame, the empty brackets would be necessary, otherwise MAT would be coerced into a matrix. – Paralysis

@Paralysis This is a good point, but in your answer you state that the output from ifelse would be a vector (a matrix stripped from its dim attribute) which is not true. – Codi

Yes, the answer included an innacurate statement, thanks for spoting that out, @Robert. I corrected the answer, considering your observations. – Paralysis

Answer 2

Reason of Failure

The reason you failed in your attempt comes from this part:

  if (MAT[i,j]>5){
    MAT[i,j]="YES"
    } else {
    MAT[i,j]="NO"
    }
  }

You should have be aware of that MAT is numerical, but you are assigning characters to MAT with in if...else... statement, which will make MAT converted to a character matrix. In this case, when you run MAT[i,j] > 5, you are comparing a character with a numeric value, e.g., "18" > 5, which returns an undesired FALSE.

---

Workaround

A workaround is using another variable to store the values after if...else..., instead of replacing values in MAT:

a <- c(5, 1, 0, 3, 2, 0.6, 1.6, 7, 9, 0)
b <- c(11, 0, 1, 18, 11, 11, 0, 13, 20, 10)
c <- c(10, 20, 0.7, 0.8, 0.3, 0.4, 0, 0.9, 1, 1)

MAT <- cbind(a, b, c)
out <- MAT

for (i in 1:nrow(MAT)) {
  for (j in 1:ncol(MAT)) {
    if (MAT[i, j] > 5) {
      out[i, j] <- "YES"
    } else {
      out[i, j] <- "NO"
    }
  }
}

such that

> out
      a     b     c
 [1,] "NO"  "YES" "YES"
 [2,] "NO"  "NO"  "YES"
 [3,] "NO"  "NO"  "NO"
 [4,] "NO"  "YES" "NO"
 [5,] "NO"  "YES" "NO"
 [6,] "NO"  "YES" "NO"
 [7,] "NO"  "NO"  "NO"
 [8,] "YES" "YES" "NO"
 [9,] "YES" "YES" "NO"
[10,] "NO"  "YES" "NO"

---

Alternative

There are already many answers to this question, and below is another base R option

> `dim<-`(as.character(factor(MAT > 5, labels = c("NO", "YES"))), dim(MAT))
      [,1]  [,2]  [,3]
 [1,] "NO"  "YES" "YES"
 [2,] "NO"  "NO"  "YES"
 [3,] "NO"  "NO"  "NO"
 [4,] "NO"  "YES" "NO"
 [5,] "NO"  "YES" "NO"
 [6,] "NO"  "YES" "NO"
 [7,] "NO"  "NO"  "NO"
 [8,] "YES" "YES" "NO"
 [9,] "YES" "YES" "NO"
[10,] "NO"  "YES" "NO"

Gesner

For the alternative, it's also possible to write structure(as.character(...), dim = dim(MAT)) or array(as.character(...), dim = dim(MAT)). – Nadanadab

@Jean-ClaudeArbaut Yes, exactly. – Gesner

@Jean-ClaudeArbaut could you please check my post here #68793963 since it does not work when I use the codes inside the r function I made – Thoron

Answer 3

Using just logical matrix converted to numeric index

MAT[] <- c("NO", "YES")[1 + (MAT > 5)]

-ouptut

> MAT
      a     b     c    
 [1,] "NO"  "YES" "YES"
 [2,] "NO"  "NO"  "YES"
 [3,] "NO"  "NO"  "NO" 
 [4,] "NO"  "YES" "NO" 
 [5,] "NO"  "YES" "NO" 
 [6,] "NO"  "YES" "NO" 
 [7,] "NO"  "NO"  "NO" 
 [8,] "YES" "YES" "NO" 
 [9,] "YES" "YES" "NO" 
[10,] "NO"  "YES" "NO" 

Nightie

Answer 4

Try this instead:

apply(MAT, 2, function(x) ifelse(x > 5, "YES", "NO"))

      a     b     c    
 [1,] "NO"  "YES" "YES"
 [2,] "NO"  "NO"  "YES"
 [3,] "NO"  "NO"  "NO" 
 [4,] "NO"  "YES" "NO" 
 [5,] "NO"  "YES" "NO" 
 [6,] "NO"  "YES" "NO" 
 [7,] "NO"  "NO"  "NO" 
 [8,] "YES" "YES" "NO" 
 [9,] "YES" "YES" "NO" 
[10,] "NO"  "YES" "NO" 

Cacique

Answer 5

Update: After the helpful notes of Jean-Claude Arbaut, ThomasIsCoding and GuedesBF please note that the first answer is wrong, here is an alternative with dplyr:

We could use across after changing matrix to tibble class and rechange to matrix after our operation:

library(tibble)
library(dplyr)

MAT <- MAT %>% 
  as_tibble() %>% 
  mutate(across(everything(), ~ifelse(. > 5, "YES", "NO"))) %>% 
  as.matrix()

First answer: Warning!

Do not use this code

MAT[MAT>5] <- "yes"
MAT[MAT<=5] <- "no"

as Jean-Claude Arbaut, ThomasIsCoding and GuedesBF indicated, it will coerce to character after the first assignment, which could lead to unexpected outcomes in downstream operations.

Cinque

Highly buggy code. Note that after the first instruction, MAT is a character matrix. Then, for instance, "6e-12" <= 5 is false. Even if you don't change the type inadvertently, it's dangereous to write it this way. Replace "yes" with 1 and "no" with 0 and it fails again. – Nadanadab

This will coerce to character after the first assignment, which could lead to unexpected outcomes in dowstream operations as observed by @Jean-ClaudeArbaut and ThomasIsCoding – Paralysis

Thank you all for your input. I have updated my answer. Please tell me if I should delete the answer or maybe leave it to have an inverse learning effect. – Cinque