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Find the max entry in column 2 - that satisfies a condition given a fixed entry in column 1

5 messages · Vangelis Litos, Bert Gunter, Rui Barradas +1 more

Original
Vangelis Litos
# 
I am using R and I created based on a vector x, a grid (named: plain) - where zero is included. This gives me a 9x2 ``matrix''.
My aim is to focus on column 1 and take i.e the first entry (then the second, the third entry etc through a loop) of the unique (c (0, x)) vector (==0) [rows 1, 4 and 7] and find the maximum value (entry) in the second column of the matrix that satisfies a condition.

Let say that the condition is satisfied when at column 2 the value is 2. That is I want row 7 to be selected

Then I want to create a column vector b (9x1) that has zero everywhere apart from row 7.

Can anybody help me with this?

Thank you in advance.
Bert Gunter
# 
"Can anybody help me with this?"

Probably not. Your explanation is as clear as mud -- to me anyway. Showing
us the actual data or code of a simple example, what you want to do and
what you have tried, and what you would like to get would likely help
(assuming it's not just me being dense, always a distinct possibility!).

And please *post in **plain text** not html* to avoid opaque text
formatting in this plain text list.

Cheers,
Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Jun 20, 2019 at 4:50 PM Vangelis Litos <Redondo5_447 at hotmail.com>
wrote:

  
  


  


      

    

    

      
      

        


  

        

      Rui Barradas
    

    

      
      #
    

  


  

      
Hello,

Inline.

?s 10:15 de 20/06/19, Vangelis Litos escreveu:

            

      
      
      
    

        
This code works but there are several issues with it.

1) c(1) is exactly the same as just 1. Try running

identical(c(1), 1)

2) You end the line with a semi-colon. This is not needed.
In fact,

the semi-colon is the INSTRUCTIONS SEPARATOR

so what you end up with is 2 instructions, one on the left of the ; and 
the other, the NULL instruction, on the right. What the parser sees and 
processes is

instruction;NULL

3) Then in the next line you transpose the matrix you have created. Much 
  simpler:

y <- rbind(1, 2)
identical(x, y)    # TRUE

            

      
      
      
    

        
4) Now you combine the matrix you have created with a new element, 0.

c(0, x)

The output of this is no longer a matrix, it's a vector without the dim 
attribute.

identical(c(0, x), c(0, 1, 2))    # TRUE

            

      
      
      
    

        
The following function does this.


fun <- function(X, val){
   u <- sort(unique(X))
   plain <- expand.grid(u, u)
   w <- which(plain[, 1] == val)
   w[which.max(plain[w, 2])]
}

fun(c(0, x), 0)
#[1] 7
fun(c(x, 0, x), 0)
#[1] 7

            

      
      
      
    

        
And what would be the non-zero value? The max in column 2?

b <- numeric(nrow(plain))    # creates a vector of 9 zeros
i <- fun(c(0, x), 0)
b[i] <- new_value

# or maybe

b[i] <- plain[i, 2]


The first and the last instructions have an obvious problem, 'plain' 
only exists inside the function fun(), it will have to be created elsewhere.


Hope this helps,

Rui Barradas

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Rui Barradas
    

    

      
      #
    

  


  

      
Hello,

Thinking again, if all you want is vector b the following function fun2 
will do it without having to create plain. Less memory and (much) faster.

fun2 <- function(X, val){
   u <- sort(unique(X))
   n <- length(u)
   b <- numeric(n^2)
   b[n*(n - 1) + match(val, u)] <- u[n]
   b
}

fun2(c(0, x), 0)
fun2(c(0, x), 1)
fun2(c(0, x), 2)


Hope this helps,

Rui Barradas

?s 06:47 de 21/06/19, Rui Barradas escreveu:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    
1 day later

    

      
      

        


  

        

      Richard O'Keefe
    

    

      
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I have read your message four times in the last couple of days,
but I still have very little idea what you want.
Let's try some things I've gleaned.
You have a matrix with 9 rows and 2 columns, and there is a 2
somewhere in column 1.

            

      
      
      
    

        
[,1] [,2]
 [1,]    1   10
 [2,]    2   11
 [3,]    3   12
 [4,]    2   13
 [5,]    5   14
 [6,]    6   15
 [7,]    2   16
 [8,]    2   17
 [9,]    9   18
You want to select rows where column 1 is 2.

            

      
      
      
    

        
[1] FALSE  TRUE FALSE  TRUE FALSE FALSE  TRUE  TRUE FALSE
You want to select the corresponding elements from column 2.

            

      
      
      
    

        
[1] 11 13 16 17
You want the maximum of these elements.

            

      
      
      
    

        
[1] 17

You can do pretty amazing things with indexing in R without any extra
packages.

Of course, I have almost certainly misunderstood you.
Perhaps you can ask again, maybe with an example?

On Fri, 21 Jun 2019 at 11:50, Vangelis Litos <Redondo5_447 at hotmail.com>
wrote: