Yes, I should have put
> library(foreach)
> library(zoo)
at the top.
On 11/06/2016 05:20 PM, Duncan Murdoch wrote:
> On 06/11/2016 5:02 PM, Jim Lemon wrote:
>> hi James,
>> I think you have to have a starting date ("origin") for as.Date to
>> convert numbers to dates.
>
> That's true with the
Add a
simplify = FALSE
to the call to replicate, and you'll get a list.
replicate(5, matrix(rnorm(4), 2, 2), simplify = FALSE)
Peter
On Wed, Nov 9, 2016 at 10:45 AM, Rui Barradas wrote:
> Hello,
>
> I also thought of replicate() but it creates an 2x2x5 array, not a list.
> Maybe it's all the
Thanks - it made me realize that I had reversed the column and row selection
rowMeans(x[seq(1:dim(x)[1],by=4),-1])
Jim
On Thu, Nov 10, 2016 at 8:30 AM, Uwe Ligges
wrote:
>
>
> On 09.11.2016 22:06, Jim Lemon wrote:
>>
>> Hi Milu,
>> Perhaps this will help:
>>
>> apply(as.matrix(x[-1,seq(1:dim(x
On 09.11.2016 22:06, Jim Lemon wrote:
Hi Milu,
Perhaps this will help:
apply(as.matrix(x[-1,seq(1:dim(x)[1],by=4)]),1,mean)
More efficient than apply(..., 1, mean): rowMeans()
Best,
Uwe Ligges
Jim
On Thu, Nov 10, 2016 at 4:00 AM, Miluji Sb wrote:
Thanks a lot for your quick reply.
Hi Milu,
Perhaps this will help:
apply(as.matrix(x[-1,seq(1:dim(x)[1],by=4)]),1,mean)
Jim
On Thu, Nov 10, 2016 at 4:00 AM, Miluji Sb wrote:
> Thanks a lot for your quick reply. I made a mistake in the question, I
> meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
>
> Sin
Thanks, David.
It does read without error using readBin. And using file.info as you
suggest seem to be a step in the right direction (the length of foo agrees
with 'wc -l' on the file). I just get different errors on subsequent
attempts to use it.
(I should have mentioned in the beginning, whatev
Hi Don,
Since there is not a current R package implementation of the blowfish algorithm
for encyrpt/decrypt, as far as I can tell, your only options via R would be to
call an external script/program that provides that functionality from within R,
or consider integrating one of the compiled lang
Hello,
I also thought of replicate() but it creates an 2x2x5 array, not a list.
Maybe it's all the same for the OP.
Rui Barradas
Em 09-11-2016 18:41, Marc Schwartz escreveu:
On Nov 9, 2016, at 12:32 PM, Evan Cooch wrote:
So, its easy enough to create a random matrix, using something like (
Not answering the question, but if you have the same dimensions of the
matrices everywhere, it is much more efficient to sample all numbers in
a row and put stuff into an array:
array(rnorm(5*4), dim=c(2,2,5)))
Best,
Uwe Ligges
On 09.11.2016 19:51, Marc Schwartz wrote:
On Nov 9, 2016, a
> On Nov 9, 2016, at 10:35 AM, MacQueen, Don wrote:
>
> Bob,
>
> Thanks for responding. I've tried the functions in bcrypt, and get, for
> example,
>
>> infl <- 'path.to.the.encrypted.file'
>> junk <- 'the.password'
>> foo <- readLines(infl)
> Warning message:
> In readLines(infl) :
> incompl
>>
>> Hi,
>>
>> See ?replicate
>>
>> Example:
>>
>> ## Create a list of 5 2x2 matrices
>
>
> Sorry, correction on my reply.
>
> I copied the wrong output, It should be:
>
> > replicate(5, matrix(rnorm(4), 2, 2), simplify = FALSE)
>
Perfect does the trick. Thanks also for the lapply solutions fro
> On Nov 9, 2016, at 12:41 PM, Marc Schwartz wrote:
>
>
>> On Nov 9, 2016, at 12:32 PM, Evan Cooch wrote:
>>
>> So, its easy enough to create a random matrix, using something like (say)
>>
>> matrix(rnorm(4),2,2)
>>
>> which generates a (2x2) matrix with random N(0,1) in each cell.
>>
>> B
Hello,
Try using ?lapply, its return value is a list. Note however that the
*apply functions are loops in disguise. There's nothing wrong with your
solution to the problem, but it's true that most R users find lapply
more elegant, like you say.
hold <- lapply(1:5, function(x) matrix(rnorm(4)
> On Nov 9, 2016, at 12:32 PM, Evan Cooch wrote:
>
> So, its easy enough to create a random matrix, using something like (say)
>
> matrix(rnorm(4),2,2)
>
> which generates a (2x2) matrix with random N(0,1) in each cell.
>
> But, what I need to be able to do is create a 'list' of such random m
So, its easy enough to create a random matrix, using something like (say)
matrix(rnorm(4),2,2)
which generates a (2x2) matrix with random N(0,1) in each cell.
But, what I need to be able to do is create a 'list' of such random
matrices, where the length of the list (i.e., the number of said ra
Bob,
Thanks for responding. I've tried the functions in bcrypt, and get, for
example,
> infl <- 'path.to.the.encrypted.file'
> junk <- 'the.password'
> foo <- readLines(infl)
Warning message:
In readLines(infl) :
incomplete final line found on 'path.to.the.encrypted.file'
> tmp <- checkpw(junk,
Thanks, Marc,
Unfortunately, I didn't choose the encryption method. My (only) need is to be
able to decrypt an existing file that is maintained by others. I am able to use
system() on a simple perl script to do the job, but I'm hoping for an R-only
solution, primarily for portability.
-Don
--
> On Nov 9, 2016, at 9:24 AM, David Winsemius wrote:
>
>
>> On Nov 9, 2016, at 9:00 AM, Miluji Sb wrote:
>>
>> Thanks a lot for your quick reply. I made a mistake in the question, I
>> meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
>
> Then try matrix( sapply( split(
I was assuming you had the data in my earlier reply. You could also look at the
covmat= argument in princomp():
# Create some random data
> set.seed(42)
> dat <- mapply(rnorm, n=rep(100, 5), mean=(1:5)*5, sd=1:5)
# Construct covariance and correlation matrices
> dat.cov <- cov(dat)
> dat.cor <- c
> On Nov 9, 2016, at 9:00 AM, Miluji Sb wrote:
>
> Thanks a lot for your quick reply. I made a mistake in the question, I
> meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
Then try matrix( sapply( split( x , 0:(dim(x)[2]/4-1)*4+1] ), mean)
Again assuming that dim(x)[1]
Better is...
rowMeans(x[0:((dim(x)[1]-1)/4)*4+1, ])
On Wed, Nov 9, 2016 at 9:00 AM, Miluji Sb wrote:
> Thanks a lot for your quick reply. I made a mistake in the question, I
> meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
>
> Sincerely,
>
> Milu
>
> On Wed, Nov 9, 2016
rowMeans(x[0:(dim(x)[1]/4-1)*4+1, ])
On Wed, Nov 9, 2016 at 9:00 AM, Miluji Sb wrote:
> Thanks a lot for your quick reply. I made a mistake in the question, I
> meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
>
> Sincerely,
>
> Milu
>
> On Wed, Nov 9, 2016 at 5:45 PM, Dal
Have you read the manual page for prcomp()?
?prcomp
Look at the examples, particularly the use of the scale.= argument.
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help
Well, it seems you can't -- prcomp() seems to want the data matrix.
But it would be trivial using svd() -- or possibly even eigen() -- if
you understand the underlying linear algebra.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking th
Thanks a lot for your quick reply. I made a mistake in the question, I
meant to ask every 4 (or 12) rows not columns. Apologies. Thanks again!
Sincerely,
Milu
On Wed, Nov 9, 2016 at 5:45 PM, Dalthorp, Daniel wrote:
> Hi Milu,
> The following should work for an array x (provided dim(x)[2] is di
Hi Milu,
The following should work for an array x (provided dim(x)[2] is divisible
by 4):
colMeans(x[,0:(dim(x)[2]/4-1)*4+1])
-Dan
On Wed, Nov 9, 2016 at 8:29 AM, Miluji Sb wrote:
> Dear all,
>
> I have a dataset with hundreds of columns, I am only providing only 12
> columns. Is it possible t
Dear all,
I have a dataset with hundreds of columns, I am only providing only 12
columns. Is it possible to take the mean of every four (or 12) columns
(value601, value602, value603, value604 etc.in this case) and repeat for
the hundreds of columns? Thank you.
Sincerely,
Milu
structure(list(va
Please learn to use the R ecosystem -- or web search (e.g.
"differential equations in R") -- both of which would have led you
to:
https://cran.r-project.org/web/views/DifferentialEquations.html
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and stic
Dear all;
I can not get legend position at the top when using the following code. It
does give at the right.
Your help is greatly appreciated.
p3 <- ggplot(a,
aes(x = SUPER.PATHWAY, y = SI)) +
#theme_classic()
theme_classic(legend.position="top", axis.text=element_text
Dear R users,
I am trying to do a Principal Components Analysis using the prcomp() function
based on the correlation matrix. How can I determine to calculate PCA on a
correlation or covariance matrix using prcomp()?
Thanks in advance.
[[alternative HTML version deleted]]
The speed enhancement using your f2() is remarkable when compared to the
apply() method I implemented. In the larger context of my actual problem, this
essentially now solves the big computational hog and I can do some real work in
a meaningful timeframe as a result.
Thank you for all the sugge
Hello every1,
Could you please suggest a good package (if any) for solving PDEs using FEMs
approach in R ?
I'm building a model where I need to solve Poisson, heat transport eq ( linear
+ heatsource in 2D or 3D in cylindrical Coords ) and combine the results
through a semi analytical model.
There are many good R tutorials on the web. Some are listed here:
https://www.rstudio.com/online-learning/#R
But you can search around for others.
If you are unable or unwilling to put in the time and effort to learn
R, there's not much we can do.
Cheers,
Bert
Bert Gunter
"The trouble with ha
Geez, I must be too excited. I meant:
stringsAsFactors=FALSE
Jim
On Wed, Nov 9, 2016 at 7:44 PM, Jim Lemon wrote:
> Hi Henry,
> You are certainly starting from the beginning. first, when you import
> the data from a CSV file, remember to add:
>
> read.csv(...,stringsAsFactors=TRUE)
>
> There wi
Hi Henry,
You are certainly starting from the beginning. first, when you import
the data from a CSV file, remember to add:
read.csv(...,stringsAsFactors=TRUE)
There will doubtless be other problems, but you have to start somewhere.
Jim
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