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Equivalent to matlab ".*" operator in R

13 messages · Ruima E., Chel Hee Lee, Berend Hasselman +7 more

Original
Ruima E.
# 
Hi,

I have this:

y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
z = matrix(c(12, -6),ncol=2)

In matlab I would do this
I would get this in matlab
0    -0
6    -3
12   -6

What is the equivalent in R?

Thanks
Chel Hee Lee
# 
[,1] [,2]
[1,]    0    0
[2,]    6   -3
[3,]   12   -6

I hope this helps.

Chel Hee Lee
On 14-11-19 08:22 AM, Ruima E. wrote:
Ruima E.
# 
Thank you Chel Hee.

Isn't there a simpler way to do so?
On Wed, Nov 19, 2014 at 3:35 PM, Chel Hee Lee <chl948 at mail.usask.ca> wrote:
Berend Hasselman
#
On 19-11-2014, at 15:22, Ruima E. <ruimaximo at gmail.com> wrote:

            

      
      
      
    

        
One way of doing this could be:

y * rep(z,1,each=nrow(y))

Berend

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Chel Hee Lee
    

    

      
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Another (simpler) way that I can think is that

            

      
      
      
    

        
[,1] [,2]
[1,]    0    0
[2,]    6   -3
[3,]   12   -6

I hope this helps.

Chel Hee Lee

        
On 14-11-19 08:43 AM, Ruima E. wrote:

            

      
      
      
    

        

      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Ista Zahn
    

    

      
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On Wed, Nov 19, 2014 at 9:50 AM, Berend Hasselman <bhh at xs4all.nl> wrote:

            

      
      
      
    

        
another is

t(t(y) * as.vector(z))

--Ista

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Dénes Tóth
    

    

      
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Hi,

It is better to use sweep() for these kinds of problems, see ?sweep

y <- matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
z <- matrix(c(12, -6),ncol=2)
sweep(y, 2, z, "*")


Best,
   Denes

        
On 11/19/2014 03:50 PM, Berend Hasselman wrote:

            

      
      
      
    

        

      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Jeff Newmiller
    

    

      
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When your matrices are the same size, the "*" operator does what you want. The problem is that you have to make a conforming version of z before you can use that operator.

y*matrix(rep(z,3),ncol=2,byrow=TRUE)

or

y*matrix(rep(z,each=3),ncol=2)

To interpret this, just keep in mind that matrices are folded vectors in R... every matrix can be thought of as a linear vector of columnwise data with dimension attributes.

---------------------------------------------------------------------------
Jeff Newmiller                        The     .....       .....  Go Live...
DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live Go...
                                      Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
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Sent from my phone. Please excuse my brevity.

        
On November 19, 2014 6:43:55 AM PST, "Ruima E." <ruimaximo at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Dénes Tóth
    

    

      
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Hi,

just for the records, your original code seems incorrect, see inline.

        
On 11/19/2014 03:22 PM, Ruima E. wrote:

            

      
      
      
    

        
Here you wrote 'x' which I guess refers to 'z', and should be repmat(z, 
size(y, 1), 1) in matlab (assuming 'z' is a row vector)

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Benno Pütz
    

    

      
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It can be simplified a bit, though, as the second operand in the multiplication does not need to be a matrix:

		y * rep(z,each=3)

        
On 19 Nov 2014, at 16:24 , Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:

        

            

      
      
      
    

        

      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Boris Steipe
    

    

      
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Or ... if you mean "simpler" as in "less to type", you can define your own binary operator by enclosing it in "%" signs, and the assign any of the previously proposed solutions, e.g.

y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
z = matrix(c(12, -6),ncol=2)
'%.*%' <- function(a,b) {a * rep(b, each=3)}


y %.*% z


     [,1] [,2]
[1,]    0    0
[2,]    6   -3
[3,]   12   -6

  
  


  


      

    

    

      
      

        


  

        

      John C Frain
    

    

      
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What you have written does not work in Matlab -

            

      
      
      
    

        
y =

         0         0
    0.5000    0.5000
    1.0000    1.0000

            

      
      
      
    

        
z =

    12    -6

.

            

      
      
      
    

        
Error using  .*
Matrix dimensions must agree.

When dimensions agree it you get the same result in R as in Matlab by using
the * operator

            

      
      
      
    

        
[,1] [,2]
[1,]  0.0  0.0
[2,]  0.5  0.5
[3,]  1.0  1.0

            

      
      
      
    

        
[,1] [,2]
[1,]    1    4
[2,]    2    5
[3,]    3    6

            

      
      
      
    

        
[,1] [,2] [,3]
[1,]    0    3  2.5
[2,]    1    0  6.0

            

      
      
      
    

        
[,1] [,2]
[1,]    0  0.0
[2,]    1  2.5
[3,]    3  6.0

            

      
      
      
    

        
John C Frain
3 Aranleigh Park
Rathfarnham
Dublin 14
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:frainj at tcd.ie
mailto:frainj at gmail.com

        
On 19 November 2014 16:48, Boris Steipe <boris.steipe at utoronto.ca> wrote:

        

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Peter Dalgaard
    

    

      
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Not sure you really want the overhead of sweep() for this, but logically, you want z as a vector or maybe 1d array since sweep() is designed as complimentary to apply(). I.e., you can sweep out marginal means using, say,

m <- matrix(c(5, 7, 9, 13), 2)
colmean <- apply(m, 2, mean)
sweep(m, 2, colmean, "-")

in which colmean is a vector. In general, apply() yields an array with fewer extents, and sweep expects one.

-pd

        
On 19 Nov 2014, at 16:00 , D?nes T?th <toth.denes at ttk.mta.hu> wrote: