,",
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 a
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])
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 =
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
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,
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
>
Niloofar.Javanrouh javanrouh_n at 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 9:43 AM, Niloofar.Javanrouh
javanrou...@yahoo.com 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
On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
javanrou...@yahoo.com 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 10:16 PM, David Winsemius dwinsem...@comcast.net wrote:
On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
javanrou...@yahoo.com wrote:
hello,
i want to differentiate of L with respect to b
when:
L= k*ln
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,]
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])}),function(x) apply(x,2
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 jfca...@gmail.com 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
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.
On Fri, Jun 1, 2012 at 5:23 PM, David Winsemius dwinsem...@comcast.net 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
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
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
80251
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 -
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 =
?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
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
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()
z -
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 -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.
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
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
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
This
?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
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 like to
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 * v_j
?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:
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 in R?
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