Use ifelse to extract information from a matrix

KGeor Source

I am trying to extract some information from a table and I am trying to avoid any for loops or apply type functions.

Assume a vector m

 m=c(1:20)  
 m
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20

and a matrix g

x1=c(0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5)   
x2=c(1,0,2,1,0,3,2,1,0,4,3,2,1,0,5,4,3,2,1,0)  
u=.4*x1^.5+.6*x2^.5  
g=cbind(x1,x2,u)  
 g
      x1 x2         u  
 [1,]  0  1 0.6000000  
 [2,]  1  0 0.4000000  
 [3,]  0  2 0.8485281  
 [4,]  1  1 1.0000000    
 [5,]  2  0 0.5656854  
 [6,]  0  3 1.0392305  
 [7,]  1  2 1.2485281  
 [8,]  2  1 1.1656854  
 [9,]  3  0 0.6928203  
[10,]  0  4 1.2000000  
[11,]  1  3 1.4392305  
[12,]  2  2 1.4142136  
[13,]  3  1 1.2928203  
[14,]  4  0 0.8000000  
[15,]  0  5 1.3416408  
[16,]  1  4 1.6000000  
[17,]  2  3 1.6049159  
[18,]  3  2 1.5413485  
[19,]  4  1 1.4000000  
[20,]  5  0 0.8944272  

I want for every element of m, to check if the sum g[,1]+g[,2] is equal to that element. For all the cases where the condition is TRUE, I want my code to return the position of the one which has the highest value of g[,3]. For example when m=5, the condition x[,1]+x[,2]==5 is TRUE at 15,16,17,18,19 and 20. Out of this 6, entry 17 has the highest value, so I want my code to return the value 17.

So at the end, I would expect a vector of length=length(m), that will indicate for each element of m, where is the maximum value of g[,3] that satisfies the above condition. The vector should be something like this:

1,4,7,11,17,0,0,0,0,0,0...,0

where when m=1, the row where g[,1]+g[,2]==1 is row 1 when m=2, the row where g[,1]+g[,2]==2 is row 4 and so on.

Currently I am using apply over each row of g, but this process is repeated thousands of times, and my code is very slow. To speed up things I resorted to ifelse in an effort to do things in a vectorised way.

What I am trying to do is the following:

ifelse(m>0,which(g[,3]==max(g[which(g[,1]+g[,2]==m),3])),0)

When I am running this I am getting

 [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

while if I substitute m with 5, it returns a vector of 17s. It seems that it only uses the first element of m instead of the whole vector. Any suggestions of how I can make this work, or an alternative that could do the same job is more than welcome.

rif-statementmatrixvectorization

Answers

answered 1 year ago akrun #1

We could use outer with max.col

m1 <- t(outer( g[,1] + g[,2], m,  `==`)* g[,3])
max.col(m1) * (rowSums(m1!= 0) !=0)
#[1]  1  4  7 11 17  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

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