,",
header = TRUE,
stringsAsFactors = F,
row.names = 1
))
-Original Message-
From: Val
Sent: Friday, August 4, 2023 2:03 PM
To: avi.e.gr...@gmail.com
Cc: r-help@r-project.org
Subject: Re: [R] Multiply
Thank you, Avi and Ivan. Worked for this particular Example.
Yes, I am lo
Às 19:03 de 04/08/2023, Val escreveu:
Thank you, Avi and Ivan. Worked for this particular Example.
Yes, I am looking for something with a more general purpose.
I think Ivan's suggestion works for this.
multiplication=as.matrix(dat1[,-1]) %*% as.matrix(dat2[match(dat1[,1],
dat2[,1]),-1])
Res=d
Thank you, Avi and Ivan. Worked for this particular Example.
Yes, I am looking for something with a more general purpose.
I think Ivan's suggestion works for this.
multiplication=as.matrix(dat1[,-1]) %*% as.matrix(dat2[match(dat1[,1],
dat2[,1]),-1])
Res=data.frame(ID = dat1[,1], Index = multipl
Val,
A data.frame is not quite the same thing as a matrix.
But as long as everything is numeric, you can convert both data.frames to
matrices, perform the computations needed and, if you want, convert it back
into a data.frame.
BUT it must be all numeric and you violate that requirement by havin
В Fri, 4 Aug 2023 09:54:07 -0500
Val пишет:
> I want to multiply two data frames as shown below,
>
> dat1 <-read.table(text="ID, x, y, z
> A, 10, 34, 12
> B, 25, 42, 18
> C, 14, 20, 8 ",sep=",",header=TRUE,stringsAsFactors=F)
>
> dat2 <-read.table(text="ID, weight, weiht2
> A, 0.25, 0
Why this works does not become clear until you actually pay attention to how
matrices are laid out in memory as a vector, and how vector replication works.
Those ideas are not that difficult to learn, but they feel different than in
other languages (e.g. matlab) and they make a huge difference
> Mat * c(3, 1, 0.5)
[,1] [,2] [,3]
[1,] 3.06 9.0
[2,] 4.05 6.0
[3,] 3.54 4.5
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Feb 21, 2017 at 8:23 AM, wrote:
> If we have the following matrix:
>
> Mat<-matrix(1:9, byrow=TRUE, nrow=3)
> Mat
>
> [,1] [,2] [,3]
>
Niloofar.Javanrouh yahoo.com> writes:
>
>
> hello,
> i want to differentiate of L with respect to b
> when:
>
> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k))
>#(negative binomial ln likelihood)
> and
> ln(mu/(mu+k)) = a+bx #link function
>
> how can i do it in R?
> thank you.
>
On Mon, May 5, 2014 at 10:16 PM, David Winsemius wrote:
>
> On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
>
>> On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
>> wrote:
>>>
>>>
>>> hello,
>>> i want to differentiate of L with respect to b
>>> when:
>>>
>>> L= k*ln (k/(k+mu)) + sum(y) *
On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
> On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
> wrote:
>>
>>
>> hello,
>> i want to differentiate of L with respect to b
>> when:
>>
>> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln
>> likelihood)
>> and
>>
On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
wrote:
>
>
> hello,
> i want to differentiate of L with respect to b
> when:
>
> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln
> likelihood)
> and
> ln(mu/(mu+k)) = a+bx #link function
>
> how can i do it in R?
Try th
g] on behalf of
arun [smartpink...@yahoo.com]
Sent: Tuesday, November 13, 2012 15:25
To: Haris Rhrlp
Cc: R help
Subject: Re: [R] multiply each row in a matrix with the help of the for loop
HI,
May be this helps:
list1<-lapply(lapply(1:3,function(i) {aa[1:i,,i]<-a[1:i,]*-1
return(aa[,,i])})
HI,
May be this helps:
list1<-lapply(lapply(1:3,function(i) {aa[1:i,,i]<-a[1:i,]*-1
return(aa[,,i])}),function(x) apply(x,2,function(i) ifelse(i==0,1,x)))
res<-array(unlist(list1),dim=c(nrow(list1[[1]]),ncol(list1[[1]]),length(list1)))
res
#, , 1
#
# [,1] [,2] [,3]
#[1,] -1 -1 -1
#[2
On Fri, Jun 1, 2012 at 5:23 PM, David Winsemius wrote:
>
> On Jun 1, 2012, at 2:37 PM, Sarah Goslee wrote:
>
>> There are several ways. The easiest to understand is probably using
>> if() statements: see ?if for help and examples.
>>
>> Sarah
>>
> I would have thought ifelse() to be the necessary
On Jun 1, 2012, at 2:37 PM, Sarah Goslee wrote:
There are several ways. The easiest to understand is probably using
if() statements: see ?if for help and examples.
Sarah
I would have thought ifelse() to be the necessary function, but for
such simple cases I find boolean math to be clearer.
There are several ways. The easiest to understand is probably using
if() statements: see ?if for help and examples.
Sarah
On Fri, Jun 1, 2012 at 11:34 AM, jfca283 wrote:
> Hi
> I need to do something very simple. I have 2 variables, Y and M. I need to
> multiply Y by 1 if M=1, by 2 if M=3 and by
I still do not get the point for what task this expansion of data may be
useful, by I guess you want (in this case probably very inefficient, but
other can work out how to improve if interested) to insert after
k <- lapply(h, function (x) x*0)
the lines:
for(i in seq_along(k)){
temp <- re
Sorry, forgot to quote:
Hi,
I am trying to use the objects from the list below to create more objects.
For each year in h I am trying to create as many objects as there are B's
keeping only the values of B. Example for 1999:
$`1999`$`8025`
B
B 8025 8026 8027 8028 8029
802511
Since this example is not reproducible (and you have not quuoted any
former code) I can only give advice "in principle":
1. Never use 1:length(x) since this will seriously fail if x is a length
0 object. Instead, use seq_along(x)
2. If k is a list, then you probably want to use doubled brackets
I am still thinking about this problem. The solution could look something
like this (it's net yet working):
k<-lapply(h, function (x) x*0) # I keep the same format as h, but set all
values to 0
years<-c(1997:1999) # I define the years
for (t in 1:length(years))
{
year = as.characte
It's interesting that sweep is the slowest one comparing to replicate and rep
:)
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/Multiply-each-depth-level-of-an-array-with-another-vector-element-tp2315537p2316586.html
Sent from the R help mailing list archive a
?sweep
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Maurits Aben
> Sent: Thursday, August 05, 2010 2:00 PM
I have only achieved a half improvement.
x <- array(1:2400*1, dim = c(200,300,400))
y <- 1:400*1
ptm <- proc.time()
z <- x*as.vector(t(replicate(dim(x)[1]*dim(x)[2], y[1:dim(x)[3]])))
"replicate:"
proc.time() - ptm
x <- array(1:2400*1, dim = c(200,300,400))
y <- 1:400*1
ptm <- proc.time
Dear Bert,
The easiest thing would be to merge both datasets and then multiply the
corresponding columns.
both <- merge(df1, df2)
both[, 3:22] * both[, 23:42]
HTH,
Thierry
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org namens Bert Jacobs
Verzonden: ma 7-6-2010 20:29
Aan: r
R -helpers
i have been trying to do this problem without must success,i managed to do a
graph for x, but it is not what i want to define,(i want to specify number of
observations as well). I have also been able to do simple rendom sample.
data.frame(ID=c(1,2,3),mu=c(1,34000,5),sigma=c
Try
lapply(abc, function(x) x*3)
Peter Ehlers
Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would be a
straightforward way to do this, but the best solution I found so far has a
for loop.
On Aug 24, 2009, at 10:58 AM, Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would
be a
straightforward way to do this, but the best solution I found so far
has a
for loop.
Everything e
?sweep
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> [EMAIL PROTECTED]
> Sent: Thursday, November 15, 2007 10:50 AM
> To: r-
On Thu, 2007-11-15 at 17:50 +, [EMAIL PROTECTED] wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
> A_ij
If i understand your question, you can do:
x <- matrix(1:10, 2)
y <- sample(10,5)
apply(x, 1, function(.x)mapply(y, .x, FUN="*"))
On 15/11/2007, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would
?sweep
b
On Nov 15, 2007, at 12:50 PM, [EMAIL PROTECTED] wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with
> jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
On Nov 15, 2007 12:50 PM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
> A_ij * v_j
>
>
Guolian Kang wrote on 10/24/2007 01:37 PM:
> Dear All,
>
> Is there a conmand to calculate the multiplication of the elements in a
> vector? For example:
>
> a=c(1,2,3,4)
> I want to get 1*2*3*4=24.
> Because the dimension of the vector is high, I want to know there is a
> command for this task
?prod
On 24 Oct 2007, at 05:37, Guolian Kang wrote:
> Dear All,
>
> Is there a conmand to calculate the multiplication of the elements
> in a vector? For example:
>
> a=c(1,2,3,4)
> I want to get 1*2*3*4=24.
> Because the dimension of the vector is high, I want to know there
> is a command fo
look at ?prod(), e.g.,
prod(a)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be
Hi
you need the tensor library:
> library(tensor)
> z <- array(runif(27),rep(3,3))
> w <- runif(3)
> w[1]* z[,,1] + w[2]*z[,,2] + w[3]*z[,,3]
[,1] [,2] [,3]
[1,] 1.2700333 1.1920113 0.8015904
[2,] 0.5175217 0.7808569 0.6306053
[3,] 0.8386015 0.6143882 0.6382314
> tens
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