how to subset part of data using for-loop

Question

I am a newbie to R programming recently.

I want to transform the existing matrix form to new one.

The matrix that I have as follows,

data()

    1       2       3       4       5       6
1   0.60!   0.29!   1.42!   0.64    1.06    1.02 
2   1.33!   0.68!   1.44!   1.44    1.05    1.03 
3   0.60    0.67@   0.64    1.37    2.40    1.00 
4   1.37    0.29    1.43    0.63    1.05    2.32 
5   1.39    0.68    1.40    1.46    1.06    1.02 
6   0.63    0.68    0.61    1.46    1.74    1.05 
7   1.39    0.68    1.01    0.65    2.43    2.27 
.... 

I've listed the first line with '!' And '@' for easy understanding.

What I want to is,

vari()

    1       2       3       4       5       6       7
1   0.60!   0.29!   1.42!   1.33!   0.68!   1.44!   0.67@ 
2   0.29    1.42    0.64    0.68    1.44    1.44    0.64 
3   1.42    0.64    1.06    1.44    1.44    1.05    1.37 
4   0.64    1.06    1.02    1.44    1.05    1.03    2.40 
5   1.33    0.68    1.44    0.60    0.67    0.64    0.29                        
6                           
….      

I would like to list the data in the 1:2 rows and 1:3 columns in one line and locate the (3,2)value to (1,7) in vari matrix.

I think it would be work with the for-loop. I made the code, however, it is my first time to make it. So, it doesn't work well.

This is my trial....

x=matrix(nrow = nrow(data)-2, ncol = 8);

for ( i in 1:(nrow(data)))
{ for ( j in 1:(ncol(data)))
{
  x[i,j]<-((c(as.vector(t(as.matrix(data[i:i+1,j:j+2]))),data[i+2,j+1])))
}
}
x

however, it represent the following sentence

"Error in data[i:i + 1, j:j + 2] : subscript out of bounds"

Please help me!!!

Answer 1

You needed to improve your indexing into the second data frame, I have added the n counter for this.

x=matrix(nrow = nrow(data)-2, ncol = 7);

n=1
for ( i in 1:(nrow(data)-3)){ 
  for ( j in 1:(ncol(data)-2)){
    x[n,]<-c(t(data[i:(i+1),j:(j+2)]),data[i+2,j+1])
    n=n+1
  }
}

 x
     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 0.60 0.29 1.42 1.33 0.68 1.44 0.67
[2,] 0.29 1.42 0.64 0.68 1.44 1.44 0.64
[3,] 1.42 0.64 1.06 1.44 1.44 1.05 1.37
[4,] 0.64 1.06 1.02 1.44 1.05 1.03 2.40
[5,] 1.33 0.68 1.44 0.60 0.67 0.64 0.29

Also be careful i:i+1is different to i:(i+1) which is what you were after.