Eduardo M. A. M. Mendes gmail.com> writes:
>
> Hello
>
> I am using xyplot (lattice) to plot a xts variable. There are 20 variables
within the xts variable (index by a
> datetime vector sampled every 1 minute). The results are very nice and help
me to understand what is going
> on with the
Hi
Just an addition for the future :
If you wanted to have different main title plots to the same device then
par.settings can be used for each plot
xyplot(1 ~1,
par.settings = list(par.main.text = 0.85,
par.sub.text = 0.85)
cex=0.85 did the job for me.
many thanks
Ed
On May 10, 2014, at 10:13 PM, David Winsemius wrote:
>
> On May 10, 2014, at 6:06 PM, Eduardo M. A. M. Mendes wrote:
>
>> library(zoo);
>> library(lattice);
>>
>> a=matrix(runif(100),25,4);
>> b=zoo(a,seq(1,25));
>> names(b)=c("How to change the f
On May 10, 2014, at 6:06 PM, Eduardo M. A. M. Mendes wrote:
> library(zoo);
> library(lattice);
>
> a=matrix(runif(100),25,4);
> b=zoo(a,seq(1,25));
> names(b)=c("How to change the font size?","2","3","4");
> xyplot(b)
>
xyplot(b, strip=strip.custom( par.strip.text=list(cex=.5)))
--
David.
>
library(zoo);
library(lattice);
a=matrix(runif(100),25,4);
b=zoo(a,seq(1,25));
names(b)=c("How to change the font size?","2","3","4");
xyplot(b)
Ed
On May 10, 2014, at 9:52 PM, David Winsemius wrote:
>
> On May 10, 2014, at 5:47 PM, Eduardo M. A. M. Mendes wrote:
>
>> Yes, you are right. I
On May 10, 2014, at 5:47 PM, Eduardo M. A. M. Mendes wrote:
> Yes, you are right. I meant panels, xyplot generates 22 panels, which one
> with y-axis, x-axis and a title. I need to change the font size of title in
> the panels.
>
It appears to me that the answer is probably in :
?strip.d
Yes, you are right. I meant panels, xyplot generates 22 panels, which one
with y-axis, x-axis and a title. I need to change the font size of title in
the panels.
Ed
On May 10, 2014, at 8:04 PM, David Winsemius wrote:
>
> On May 10, 2014, at 3:30 PM, Eduardo M. A. M. Mendes wrote:
>
>
On May 10, 2014, at 3:30 PM, Eduardo M. A. M. Mendes wrote:
> Hello
>
> Many thanks.
>
> par.main.text is the main title (on top of all plots). I need to change the
> each of the titles in the, say, subplots.
>
One gets the response the fills ones needs when those needs are clearly
express
Hello
Many thanks.
par.main.text is the main title (on top of all plots). I need to change the
each of the titles in the, say, subplots.
Ed
On May 10, 2014, at 7:05 PM, David Winsemius wrote:
>
> On May 10, 2014, at 2:26 PM, Eduardo M. A. M. Mendes wrote:
>
>> Hello
>>
>> I am using xyp
On May 10, 2014, at 2:26 PM, Eduardo M. A. M. Mendes wrote:
> Hello
>
> I am using xyplot (lattice) to plot a xts variable. There are 20 variables
> within the xts variable (index by a datetime vector sampled every 1 minute).
> The results are very nice and help me to understand what is goin
Hello
I am using xyplot (lattice) to plot a xts variable. There are 20 variables
within the xts variable (index by a datetime vector sampled every 1 minute).
The results are very nice and help me to understand what is going on with the
data. However since the names (labels of each variable) w
Hello,
I'm a novice R user. I'd like to estimate the linear trends (b) and their
statistical significance (p-value) of quite a few univariate time series.
Because several time series show autocorrelation and heteroskedasticity,
the ordinary least squares method lm(), as I understand it, isn't the
And I misplaced the closing parenthesis in my reponse - it goes before
the .Names attribute as well.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, May 10, 2014 at 11:36 AM, William Dunlap wrote:
> You made the same error as the first time - a misplaced parenthesis.
> Make the last 2 lines
You made the same error as the first time - a misplaced parenthesis.
Make the last 2 lines
.Names = c("row", "stimulation", "positivity", "group", "copy")),
row.names=c(NA, 120L), class="data.frame")
instead of
.Names = c("row", "stimulation", "positivity", "group", "copy"),
row.names =
Hi,
---
2767.493803,4796.33016,12292.93705,3864.657567,9380.673835,14886.44683,8457.88646,26050.47191))),#
.Names = c("row", "stimulation", "positivity", "group", "copy"),
row.names = c(NA, -120L), class = "data.frame")
str(my.data)
'data.fram
Hello,
I have tried to apply this approach -- which worked for this example
-- to a larger dataset. Although I have not received error messages,
the variable I have set (my.data) is empty.
Do you have any tip about this?
Many thanks
Luigi
# code::
my.data<-structure(list(
column_1 = 1:120,
col
On May 10, 2014, at 3:54 AM, Duncan Murdoch wrote:
> On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
>> On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham wrote:
>>> Beware of the is.* functions:
>>>
>>> * is.object() does not test the usual definition of objects
>>> * is.vector() does not test th
> d <- data.frame(id=1:4, no.contacts=c(2,3,3,1))
> d$contacts_list <- list(3:4, c(1,3,4), c(4,2,1), 1 )
If you store that information in a longer format, with each row being
an edge to the relationship graph, it can make further processing
easier:
d2 <- with(d,
data.frame(id=rep(id, vappl
One answer to your question might be ?get. A way better answer is to never use
assign to create xx1 etc in the first place but to store those results in a
list that you can simply index later. If you were to create a reproducible
example [1] by defining mydata you would be more likely to get an
Hi,
Try:
dat <- data.frame(id=1:4, contacts_list=I(list(3:4,c(1,3,4), c(4,2,1), 1)),
`number of contacts`=c(2,3,3,1),check.names=FALSE)
attr(dat$contacts_list,"class") <- NULL #if needed
dat
A.K.
Dear Group,
I have data like the following
id contacts_list number of contacts
Thanks, modifying the predict function to allow for new levels was what was
required
Brian
--
View this message in context:
http://r.789695.n4.nabble.com/Extracting-the-names-of-coefficients-of-random-effects-tp4689109p4690298.html
Sent from the R help mailing list archive at Nabble.com.
On 10/05/2014, 7:46 AM, Ragia Ibrahim wrote:
Dear Group,
I have data like the following
id contacts_list number of contacts
---
1 3 4 2
2 1 3 43
34 2 1
Dear Group,
I have data like the following
id contacts_list number of contacts
---
1 3 4 2
2 1 3 43
34 2 1 3
411
-
On 10 May 2014, at 12:54 , Duncan Murdoch wrote:
> On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
>> On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham wrote:
>>> Beware of the is.* functions:
>>>
>>> * is.object() does not test the usual definition of objects
>>> * is.vector() does not test the
Thankyou very much arun. Its always nice to hear from you.
Eliza
> Date: Sat, 10 May 2014 03:55:29 -0700
> From: smartpink...@yahoo.com
> Subject: Re: [R] adding rows
> To: r-help@r-project.org
> CC: ruipbarra...@sapo.pt; eliza_bo...@hotmail.com
>
>
>
> HI,
> If you want to try other ways:
>
>
HI,
If you want to try other ways:
fun1 <- function(mat, rowN) {
dm <- dim(mat)[1]
rowN1 <- rowN - 1
indx <- rep(1:rowN, dm - rowN1) + rep(seq(0, dm - rowN), each = rowN)
indx1 <- (seq_along(indx)-1)%/%rowN+1
as.vector(tapply(indx, list(indx1), FUN = function(i) sum(mat[i, ])
On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham wrote:
Beware of the is.* functions:
* is.object() does not test the usual definition of objects
* is.vector() does not test the usual definition of vectors
* is.numeric() does not work the same way
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham wrote:
> Beware of the is.* functions:
>
> * is.object() does not test the usual definition of objects
> * is.vector() does not test the usual definition of vectors
> * is.numeric() does not work the same way as is.character() or is.integer()
> * is.
NOELIA LEGAL hotmail.com> writes:
>
> Please I need your help.I'm intrested to know if there is any R-package
for fit a Pearson III extreme value
> distribution to data.Thanks a lot
> Noelia
> [[alternative HTML version deleted]]
>
>
As it wa
I'd like to create several tables in which my data are stored.
I did the follow:
x<-mydata
splitted<-split(x[,1],x[,2])
#Create my function for the tables
fxdata<-function(){
y.row=c("0","1","2","3","4","5")
yy=c(rep(0,6))
w=data.frame(A3=yy,A2=yy,A1=yy,A0=yy)
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