Hi Dan
For devices png, pdf, postscript and ? others the pointsize argument
controls the font size which is modified by cex
For lattice there are the settings in trellis.par.get()
trellis.par.get()$fontsize
$text
[1] 12
$points
[1] 8
which you can set and there is no real need to change font
Hi
Following Pauls reply you may like to increase the panel space available by
using useOuterStrips which makes the smaller proportions easier to see and
compare
If you also wanted to change colours auto.key is the better option. Colours
are a bit garish
See
?xyplot
and
panel.barchart
library(
Works great, thank you Paul!
On Wed, Jul 27, 2016 at 5:24 PM, Paul Murrell
wrote:
> Hi
>
> Try this ...
>
> barchart(Class~Freq|Sex + Age, Titan,
> groups=Survived,
> panel = titanpanel,
> stack=TRUE, layout=c(4,1),
> key=simpleKey(title="Survived", text=level
> On Jul 27, 2016, at 12:02 PM, Jeff Newmiller wrote:
>
> An alternative (more compact, not necessarily faster, because apply is still
> a for loop inside):
>
> f <- function( m, nx, ny ) {
> # redefine the dimensions of my
> a <- array( m
> , dim = c( ny
>, n
What if it is being used on a platform other than Windows? This problem is more
challenging than you seem to think it is.
I would suggest including a kind of "par" argument to your function that lets
the user change your defaults.
--
Sent from my phone. Please excuse my brevity.
On July 27, 2
In general, some assembly is required. This stuff doesn't quit fit in a one
page email. You may want to start playing around with model.matrix and qr() to
see the light.
One tricky bit is how much you know already. There is some risk for me of
having to rewrite a linear algebra textbook...
I'l
Hi
Try this ...
barchart(Class~Freq|Sex + Age, Titan,
groups=Survived,
panel = titanpanel,
stack=TRUE, layout=c(4,1),
key=simpleKey(title="Survived", text=levels(Titan$Survived),
rectangles=TRUE, points=FALSE, columns=2))
Paul
On 28/07
I have constructed a barchart that requires a panel call, but the panel
reduces the facsimiles of bars in the legend to small colored circles. You
can see this behavior in the following example:
Titan <- as.data.frame(Titanic)
titanpanel <- function(x,y,...){
panel.barchart(x,y,...)
}
barchart(C
Hi All,
I am putting together a package that (among other things) draws some nice
graphs for users. I place some explanatory text on figs using "text" and
"mtext". But the size of the text depends on the Windows display settings:
Smaller (100%), medium (125%) or larger (150%) (In Windows 7... Contr
Hi all,
I have four matrix tmp_a, tmp_b, tmp_c and tmp_d whose dimensions are
shown as below.
I want to take one row from each of these matrices and then put the four
selected rows into a matrix.
I want to count the number of such matrices for which the vector of row sum
is less than or equal to (
Hi Peter,
Thank you for your good answer. I am sorry for the late reply.
*An ortogonalized model matrix generates a decomposition of the model space
into orthogonal subspaces corresponding to the terms of the model.
Projections onto each of the subspaces are easily worked out. E.g., for a
two-w
Sorry if my earlier remarks were unfair to City but their request reminded
me of some problems that I had with using R, similar software and the
obstacles placed in my way by an IT Department many years ago. If it is of
any help I have R installed on a Desktop and two laptops (running 64-bit
Window
Thanks, this works. I thought the code should put in the for loop, but it
turns out an extra line.
On Wed, Jul 27, 2016 at 12:50 PM, wrote:
> Hello,
>
> Try
>
>
> chars = c('A','B','C','D')
> matrix1 = matrix(nrow = length(1:100), ncol = length(1:5)*length(chars))
> k = 0
> for(i in 1:length(1:5
> On Jul 27, 2016, at 8:20 AM, Stefan Kruger wrote:
>
> Hi -
>
> I'm new to R.
>
> In other functional languages I'm familiar with you can often seed a call
> to reduce() with a custom accumulator. Here's an example in Elixir:
>
> map = %{"one" => [1, 1], "three" => [3], "two" => [2, 2]}
> ma
> On Jul 27, 2016, at 5:03 AM, Ismail SEZEN wrote:
>
>
>>> The time column is factor. I want to convert each time to the following
>>> format:
>>>
>>> 20111211003000
>>>
>>> 20111211013000
>>>
>>> 2011121102
>>>
>>> (Year)(month)(date)(hr)(min)(sec)
>>
>>> newTimeDt <- strptime(Time, f
Correction to my code. I created a "doc" variable because I was thinking of
doing something faster, but I never did the change. grep needed to work on
the original source "dat" to be used for counting.
Fixed:
combs = structure(list(V1 = c(65L, 77L, 55L, 23L, 34L), V2 = c(23L, 34L,
34L, 77L, 65L
If I understood the request this is the same programming task as counting
words in a document and counting character sequences in a string or
matching bytes in byte arrays (though you don't want to go down that far)
You can do something like what follows. There are also vectorized greps
in strin
> On Jul 27, 2016, at 8:56 AM, Anamika Chaudhuri wrote:
>
>
> Y<-matrix(1:40,ncol=2)
> Y1<-Y/60 # estimates of p
>
> #print(Y1)
>
> sigma2<-
> matrix(c(var(Y1[,1]),cov(Y1[,1],Y1[,2]),cov(Y1[,1],Y1[,2]),var(Y1[,2])),2,2)
> #print(sigma2)
>
> rho<-sigma2[1,2]/sqrt(sigma2[1,1]*sigma2[2,2])
> #
An alternative (more compact, not necessarily faster, because apply is still a
for loop inside):
f <- function( m, nx, ny ) {
# redefine the dimensions of my
a <- array( m
, dim = c( ny
, nrow( m ) %/% ny
, ncol( m ) %/% nx )
)
On 07/27/2016 11:49 AM, lily li wrote:
If replace 1:5 to char2 = c('east','west','south','north','central'),
how to put on column names with the original structure? Thanks again.
Have you tried to actually "replace 1:5 with char2", literally?
You need to try things before you ask. That's the o
I see. But please understand that initializing the values is not the
same as setting the colnames. How providing this almost-impossible-
to-read initialization code helps with respect to your question which
is about setting the colnames?
I know people often asked you to show the code in your prev
Hello,
Try
chars = c('A','B','C','D')
matrix1 = matrix(nrow = length(1:100), ncol = length(1:5)*length(chars))
k = 0
for(i in 1:length(1:5)){ # or more simply just for(i in 1:5)
for(j in 1:length(chars)){
k = k+1
matrix1[,k] = k
}
}
matrix1
tmp <- expand.grid(chars, 1:5, stringsAsFac
If replace 1:5 to char2 = c('east','west','south','north','central'), how
to put on column names with the original structure? Thanks again.
On Wed, Jul 27, 2016 at 12:40 PM, Hervé Pagès wrote:
> Hi,
>
>
> On 07/27/2016 11:17 AM, lily li wrote:
>
>> Hi all,
>>
>> I want to ask that how to create
Thanks. I shorten a more complex matrix to this example, but use the
original structure of the code. The original matrix has all characters
instead of 1:5.
On Wed, Jul 27, 2016 at 12:42 PM, Hervé Pagès wrote:
> On 07/27/2016 11:40 AM, Hervé Pagès wrote:
>
>> Hi,
>>
>> On 07/27/2016 11:17 AM, li
On 07/27/2016 11:40 AM, Hervé Pagès wrote:
Hi,
On 07/27/2016 11:17 AM, lily li wrote:
Hi all,
I want to ask that how to create column names for a matrix. For example,
the matrix below, the column names should be: 1-A, 1-B, 1-C, 1-D, 2-A,
2-B,
2-C, 2-D, 3-A, etc. Thanks for your help.
chars =
Hi,
On 07/27/2016 11:17 AM, lily li wrote:
Hi all,
I want to ask that how to create column names for a matrix. For example,
the matrix below, the column names should be: 1-A, 1-B, 1-C, 1-D, 2-A, 2-B,
2-C, 2-D, 3-A, etc. Thanks for your help.
chars = c('A','B','C','D')
matrix1 = matrix(nrow = l
Hi,
It is not a just 79 triplets. As I said, there are 79 codes. I am making
triplets out of that 79 codes and matching the triplets in the list.
Please find the dput of the data below.
> dput(head(newd,10))
structure(list(uniq_id = c("1", "2", "3", "4", "5", "6", "7",
"8", "9", "10"), hi = c("1
Hi all,
I want to ask that how to create column names for a matrix. For example,
the matrix below, the column names should be: 1-A, 1-B, 1-C, 1-D, 2-A, 2-B,
2-C, 2-D, 3-A, etc. Thanks for your help.
chars = c('A','B','C','D')
matrix1 = matrix(nrow = length(1:100), ncol = length(1:5)*length(chars)
Hi Luis,
What features do you want? What did you try? What was missing?
I've not used VSCode, but for Atom check out
https://atom.io/packages/repl, https://atom.io/packages/language-r,
https://atom.io/packages/autocomplete-r, and possibly
https://atom.io/packages/hydrogen
Best,
Ista
On Wed, Jul
You said you had 79 triplets and 8000 records.
When I compared 100 triplets to 1 records it took 86 seconds.
So obviously there is something you're not telling us about the format
of your data.
If you use dput() to provide actual examples, you will get better
results than if we on Rhelp have
Hi,
Thanks for the solution. But I am afraid that after running this code still
it takes more time. It has been an hour and still it is executing. I
understand the delay because each triplet has to compare almost 9000
elements.
Regards,
Sri
On Wed, Jul 27, 2016 at 9:02 PM, Sarah Goslee
wrote:
If you have a simple list of vectors (call it lst), use
lengths = sapply(lst, length)
In general, you may want to look at functions lapply and sapply which
apply a function over a list, in this case the function length().
Peter
On Wed, Jul 27, 2016 at 8:20 AM, Stefan Kruger wrote:
> Hi -
>
> I
Hi David,
Thanks for your response. rbind doesnot seem to work.
Here is a reproducible example
Y<-matrix(1:40,ncol=2)
Y1<-Y/60 # estimates of p
#print(Y1)
sigma2<-
matrix(c(var(Y1[,1]),cov(Y1[,1],Y1[,2]),cov(Y1[,1],Y1[,2]),var(Y1[,2])),2,2)
#print(sigma2)
rho<-sigma2[1,2]/sqrt(sigma2[1,1]*sigm
Hi -
I'm new to R.
In other functional languages I'm familiar with you can often seed a call
to reduce() with a custom accumulator. Here's an example in Elixir:
map = %{"one" => [1, 1], "three" => [3], "two" => [2, 2]}
map |> Enum.reduce(%{}, fn ({k,v}, acc) -> Map.update(acc, k,
Enum.count(v),
I've been keen to use either VSCode or Atom and have not had much success. Any
suggestions?
Regards,
- Luis
---
This communication may contain confidential and/or privileged information. If
you are not the intended recipient (or have re
\n<<
Hi,
I am trying to set up a dynamic panel data model using the pgmm function from
the plm package. The formula that I want to estimate is something like this:
[y(t) - y(t-1)] ~ [x(t) - x(t-1)] + [x(t-1) - y(t-1)]
where the lagged value of the dependent variable should be instrumen
Hello, guys!
I tried to do that:
> teste = fuzzy_partition( varnames = c( "a", "b" ), FUN = fuzzy_trapezoid,
> corners = c( 0, 1, 2, 3), corners = c(4, 5, 6, 7))
Error in FUN(i, ...) :
argumento formal "corners" corresponde a múltiplos argumentos especificados
So, how can I use fuzzy_partiti
Hi Marc,
I have something very close now, and think to leave it like that.
Thank you so much for your help to solve the problem.
Cheers,
asarr
On Mon, Jul 25, 2016 at 6:03 PM, Marc Schwartz wrote:
> Hi,
>
> First, I noted again that you still have:
>
>xaxp=c(181,2005,1)
>
> in the first
Hi Jon,
The versions are: raster_2.5-2 sp_1.2-3
On Wed, Jul 27, 2016 at 9:31 AM, Jon Skoien
wrote:
> Hi Lily,
>
> You can ignore the first three lines, they are just for information.
> Which version of raster do you have? I see that the attempt to read the
> crs-variable was not wrapped in a t
This should be faster. It uses apply() across the blocks.
> ilon <- seq(1,8,nx)
> ilat <- seq(1,4,ny)
> cells <- as.matrix(expand.grid(ilat, ilon))
> blocks <- apply(cells, 1, function(x) tst[x[1]:(x[1]+1), x[2]:(x[2]+1)])
> block.means <- colMeans(blocks)
> tst_2x2 <- matrix(block.means, 2, 4)
>
Hi,
With the caveat that I speak for myself only, a few comments:
1. Given that Citi appears to have been running R on Windows 7 for some period
of time, I am not sure that they are looking for reasons to deny users access
to R. They are simply looking for some indication that R will run on Win
Hi,
It's really a good idea to use dput() or some other reproducible way
to provide data. I had to guess as to what your data looked like.
It appears that order doesn't matter?
Given than, here's one approach:
combs <- structure(list(V1 = c(65L, 77L, 55L, 23L, 34L), V2 = c(23L, 34L,
34L, 77L, 6
Hi Lily,
You can ignore the first three lines, they are just for information.
Which version of raster do you have? I see that the attempt to read the
crs-variable was not wrapped in a try-call in earlier versions.
Upgrading might help.
Best,
Jon
On 7/27/2016 5:13 PM, lily li wrote:
> Hi Jon,
Hi,
Apologizes for the less information.
Basically, myCombos is a matrix with 3 variables which is a triplet that is
a combination of 79 codes. There are around 3lakh combination as such and
it looks like below.
V1 V2 V3
65 23 77
77 34 65
55 34 23
23 77 34
34 65 55
Each triplet will compare in
When I first graduated some 50 years ago I worked in the Department of
Finance. On small piece of my work involved getting CPI data from the CSO
and doing some calculations. A specific person in the CSO (Central
Statistics Office) usually supplied this information and this fact was
always recorded
Not entirely sure I understand, but match() is already vectorized, so you
should be able to lose the supply(). This would speed things up a lot.
Please re-read ?match *carefully* .
Bert
On Jul 27, 2016 6:15 AM, "sri vathsan" wrote:
Hi,
I created list of 3 combination numbers (mycombos, around
A reproducible example makes your problem easier to understand.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in the stat
Hi,
I created list of 3 combination numbers (mycombos, around 3 lakh
combinations) and counting the occurrence of those combination in another
list. This comparision list (mylist) is having around 8000 records.I am
using the following code.
myCounts <- sapply(1:nrow(myCombos), FUN=function(i) {
Dear Achim Zeileis, dear John Fox,
Thank you for your time! Both worked well.
lrtest(Restrict, Full)
#Df LogLik Df Chisq Pr(>Chisq)
1 27 -882.00
2 28 -866.39 1 31.212 2.313e-08 ***
anova(Restrict, Full)
Resid. df Resid. Dev TestDf LR stat. Pr
>> The time column is factor. I want to convert each time to the following
>> format:
>>
>> 20111211003000
>>
>> 20111211013000
>>
>> 2011121102
>>
>> (Year)(month)(date)(hr)(min)(sec)
>
>> newTimeDt <- strptime(Time, format="%d%b%y:%H:%M")
>> newTimeDt
> [1] "2011-12-27 00:30:00 PST" "20
Dear Faradj Koliev,
There is an anova() method for "polr" objects that computes LR chisquare tests
for nested models, so a short answer to your question is anova(Full,
Restricted).
The question, however, seems to reflect some misunderstandings. First aov()
fits linear analysis-of-variance mode
Hi all,
I need to aggregate some matrix data (1440x720) to a lower dimension (720x360)
for lots of years and variables
I can do double for loop, but that will be slow. Anybody know a quicker way?
here an example with a smaller matrix size:
tst=matrix(1:(8*4),ncol=8,nrow=4)
tst_2x2=matrix(NA,nc
On Wed, 27 Jul 2016, Faradj Koliev wrote:
Dear all,
A quick question: Let?s say I have a full and a restricted model that looks something like this:
Full<- polr(Y ~ X1+X2+X3+X4, data=data, Hess = TRUE, method="logistic?) # ordered logistic regression
Restricted<- polr(Y ~ X1+X2+X3, data=da
Dear all,
A quick question: Let’s say I have a full and a restricted model that looks
something like this:
Full<- polr(Y ~ X1+X2+X3+X4, data=data, Hess = TRUE, method="logistic”) #
ordered logistic regression
Restricted<- polr(Y ~ X1+X2+X3, data=data, Hess = TRUE, method="logistic”) #
or
I think you get the error because you passed the object from nc_open(),
you should rather pass the filename of the netCDF, so:
pre = brick("sample_precip_daily.nc")
From the test-file, this gives you a RasterBrick-object with 5*9 pixels
and 7305 layers.
Then you can extract data.frames of each
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