Search code examples
rif-statementmatrixvector

How to return 1 columns value from a matrix on R?

I am trying to match values from one vector (1,7) to a matrix (2,21)

The goal here is to match the vector values to the matrix values as there are equal values within both, shown in example e.g.

if(Removed[,1] %like% T1PickProb[2]){
  print("yes")
} else{
  print("no match")
}

T1PickProb
[output]
0.01754386 0.06140351 0.13157895 0.23684211 0.28070175 0.17543860 0.09649123

Removed
[output]
           [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       [,7]       [,8]       [,9]      [,10]      [,11]     [,12]     [,13]     [,14]      [,15]     [,16]     [,17]      [,18]
[1,] 0.01754386 0.01754386 0.01754386 0.01754386 0.01754386 0.01754386 0.06140351 0.06140351 0.06140351 0.06140351 0.06140351 0.1315789 0.1315789 0.1315789 0.13157895 0.2368421 0.2368421 0.23684211
[2,] 0.06140351 0.13157895 0.23684211 0.28070175 0.17543860 0.09649123 0.13157895 0.23684211 0.28070175 0.17543860 0.09649123 0.2368421 0.2807018 0.1754386 0.09649123 0.2807018 0.1754386 0.09649123
         [,19]      [,20]      [,21]
[1,] 0.2807018 0.28070175 0.17543860
[2,] 0.1754386 0.09649123 0.09649123

This should print out "yes" as T1PickProb[2] is 0.06140351 which is in Removed[,1] which has two values 0.01754386 & 0.06140351. However it prints out "no match"

So my question is, is there a way I can match values to specific columns within the matrix? I have tried the following

Removed[,1]

but this returns the following error when running the about if statement:

In if (Removed[, 1] %like% T1PickProb[2]) { :
  the condition has length > 1 and only the first element will be used

I have also tried

Removed[1,2]

However this return the first value in the second column. What else can I do?

Solution

  • Just to illustrate: 

    if(any(removed[,1] %in% T1PickProb[2])){
  print("yes")
} else{
  print("no match")
}

> "yes"

    That is because 

    > removed[,1] %in% T1PickProb[2]
[1] FALSE  TRUE

    yields a vector with two elements. To see if any of these two is true, you can use any:

    > any(removed[,1] %in% T1PickProb[2])
[1] TRUE

    You can also vectorise this further, e.g.

    Using the fact that:

    removed %in% T1PickProb
 [1]  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE
[21]  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE
[41]  TRUE  TRUE

    you could do 

    ifelse(removed %in% T1PickProb, "yes", "no match")
 [1] "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"     
[12] "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"      "yes"     
[23] "no match" "no match" "no match" "no match" "no match" "yes"      "yes"      "yes"      "no match" "no match" "no match"
[34] "yes"      "yes"      "yes"      "no match" "yes"      "yes"      "yes"      "yes"      "yes"