matrix dimension and for loop

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

x <- matrix(rnorm(50),10,5) # collapsing your first two commands
y <- 1:5
Z <- matrix(0,nrow(x),length(y)) # case sensitive, and y is a vector 
Z[,length(y)] <- x %*% y

It isn't clear that this is what you want, though.  It fills up
one column of Z.

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command
without using if (NCOL(x)==1), or something like that?
(in case I want to include them in my function.

I think it will still work.  So long as you pay attention to what
is a matrix - to which dim() applies - and what is a vector, to
which length() applies.

Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron
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sonchawan tamkaew <psu17772 at yahoo.com> writes:

Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

Yes.  If you multiply (i.e. the '*' operator) a matrix and a vector
you will do operations columnwise on the matrix.  Here you want to
multiply the rows so you could transpose x, multiply by y and
transpose the result.

x <- array(rnorm(50), dim = c(10,5))
dim(x)

[1] 10  5

y <- 1:5
Z <- t(y * t(x))
dim(Z)

[1] 10  5

x[1,]

[1]  0.23144399 -0.25393369  0.06469486 -0.13116405  0.48534804

Z[1,]

[1]  0.2314440 -0.5078674  0.1940846 -0.5246562  2.4267402

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command
without using if (NCOL(x)==1), or something like that?
(in case I want to include them in my function.

The above will work in that case too.

x <- rnorm(10)
y <- 2
t(y * t(x))

[,1]
 [1,]  1.5755705
 [2,]  0.3811325
 [3,] -0.4629773
 [4,]  1.7612785
 [5,]  0.7881174
 [6,] -0.2815202
 [7,]  0.4315893
 [8,]  1.1609736
 [9,]  1.8737894
[10,]  5.3069982
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Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command
without using if (NCOL(x)==1), or something like that?
(in case I want to include them in my function.

Example of x is a vector.

x<-rnorm(10)
y=2
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

Error in x[, j] : incorrect number of dimensions

Could anyone please help me to solve these problem?
Thank you very much for your kindness.

x %*% diag(y, ncol(x))

Uwe Ligges
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Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

Z <- x * y

seems to do the job....

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command
without using if (NCOL(x)==1), or something like that?
(in case I want to include them in my function.

still seems to do the job...

Example of x is a vector.

x<-rnorm(10)
y=2
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

Error in x[, j] : incorrect number of dimensions

note: I discovered above that assignment like 'y=2' where now valid....
(learn something new evryday on r-help.... ;-) )




 where now valid....
(learn something new evryday on r-help.... ;-) )




L

Could anyone please help me to solve these problem? 
Thank you very much for your kindness.


Sonchawan Tamkaew 

__________________________________________________



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Laurent Gautier			CBS, Building 208, DTU
PhD. Student			D-2800 Lyngby,Denmark	
tel: +45 45 25 24 85		http://www.cbs.dtu.dk/laurent
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Laurent Gautier <laurent at genome.cbs.dtu.dk> writes:

On Tue, Apr 09, 2002 at 05:46:40AM -0700, sonchawan tamkaew wrote:

Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

Z <- x * y

seems to do the job....

Nope. Try it with x <- matrix(1,10,5) and see.

O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907
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Laurent Gautier <laurent at genome.cbs.dtu.dk> writes:

On Tue, Apr 09, 2002 at 05:46:40AM -0700, sonchawan tamkaew wrote:

Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

Z <- x * y

seems to do the job....

Nope. Try it with x <- matrix(1,10,5) and see.

...right... it was 'y[j]' and not 'y[,j]' (blurry eyes... looking too long at the screen)


this is probably what everybody already gave:
(I receive r-help few hours before the first ones to receive it)


 x * matrix(y, ncol=ncol(x), nrow=nrow(x), byrow=T)


(I hope it work for good now... :)  )




L.

-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907

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Dear all,

My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

Z <- t( t(x) * y )

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command
without using if (NCOL(x)==1), or something like that?
(in case I want to include them in my function.

Example of x is a vector.

x<-rnorm(10)
y=2
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

Error in x[, j] : incorrect number of dimensions

The answer already given to theprevious
question automatically handles the case 
where x is a vector so its not necessary to 
provide special code for it.

However, even though its not needed here,
just for interest:

x <- as.matrix( x ) 

will turn vector x into a 1 column matrix and leave it
as is if its already a matrix.


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My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command

I thought I would collect together some of the solutions that have
been posted as well as mention a few others.  I have also added
some discussion particularly with reference to the second question.
There appears to be a myriad of ways to address this problem in R.
Since I am learning R I found it useful for myself to explore various
solutions and maybe others will find this helpful too.

The problem is to multiply the jth column of a matrix x by y[j] for each j.
Its also possible for x to be a vector in which case y may be a scalar.

The first 4 solutions regard x as a one column matrix if x is a vector and
so always return a matrix.  The 5th solution returns a vector if x is a 
vector.


1. Matrix multiplication. Note that if x is a vector then it is treated as a
   one column matrix.  Also note that NROW(x) is the same as nrow(x), if x is a
   matrix; however, if x is a vector, then NROW(x) is its length while nrow(x) 
   is NULL.  Thus it is important here to use NROW and not nrow. This is 
   discussed in the help on nrow.

	x %*% diag(y,NROW(x))  

2. Scalar multiplication and transpose.  Note that if x is a vector then t(x) 
   is a row matrix made out of it, so no special processing is necessary.  

	t(t(x)*y)  

3. Sweep.  Note that as.matrix(x) is the same as x, if x is a matrix, but is a 
   column matrix made out of x, if x is a vector.

	sweep(as.matrix(x),2,y,FUN="*")  

4. Scalar multiplication and col.  col(m) is a matrix the same size as
   matrix m.  Every entry in column j equals j (for all columns).

	as.matrix(x) * y[col(as.matrix(x))]   

5. Variant of #4. The following is the same as #4 except that it returns a 
   vector if x is a vector.

	x * y[col(as.matrix(x))]   


as.matrix(x) can be abbreviated to just x, in #3, #4 and #5 
if x is known always to be a matrix.  For example, the last two solutions 
would be  just x*y[col(x)].


I had not expected the last 2 solutions to work in the case that x is a vector
and y is a scalar, since they involve subscripting of a scalar, but apparently
that works in R.

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My questions are that if I have

x<-rnorm(50)
dim(x)<-c(10,5)
y=1:5
Z<-matrix(0,NROW(x),NROW(y))
for (j in 1:NROW(y)) Z[,j]<-x[,j]*y[j]

1. Is there any other way to write this without 'for'
loop?

2. and if I don't know the dimension of x, which could
be only 1 column, how could I write the command

I thought I would collect together some of the solutions that have
been posted as well as mention a few others.  I have also added
some discussion particularly with reference to the second question.
There appears to be a myriad of ways to address this problem in R.
Since I am learning R I found it useful for myself to explore various
solutions and maybe others will find this helpful too.

The problem is to multiply the jth column of a matrix x by y[j] for each j.
Its also possible for x to be a vector in which case y may be a scalar.

The first 4 solutions regard x as a one column matrix if x is a vector and
so always return a matrix.  The 5th solution returns a vector if x is a
vector.

1. Matrix multiplication. Note that if x is a vector then it is treated as a
   one column matrix.  Also note that NROW(x) is the same as nrow(x), if x is a
   matrix; however, if x is a vector, then NROW(x) is its length while nrow(x)
   is NULL.  Thus it is important here to use NROW and not nrow. This is
   discussed in the help on nrow.

        x %*% diag(y,NROW(x))

Not quite. The number of *columns* of x, ncol(x), is interesting here,
not NROW(x),
or just use length(y) instead of ncol(x) ....


Uwe Ligges
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ggrothendieck at yifan.net wrote:

1. Matrix multiplication. Note that if x is a vector then it is treated as a
   one column matrix.  Also note that NROW(x) is the same as nrow(x), if x is a
   matrix; however, if x is a vector, then NROW(x) is its length while nrow(x)
   is NULL.  Thus it is important here to use NROW and not nrow. This is
   discussed in the help on nrow.

        x %*% diag(y,NROW(x))

Not quite. The number of *columns* of x, ncol(x), is interesting here,
not NROW(x),
or just use length(y) instead of ncol(x) ....

Thanks for pointing out this error.  Note, however, that NROW(x) should
be NCOL(x), and not ncol(x), since x can be a vector.   length(y) is
probably best.

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