Cross-posted on Stack Overflow:
http://stackoverflow.com/q/11567745/271616
--
Joshua Ulrich | FOSS Trading: www.fosstrading.com
On Thu, Jul 19, 2012 at 12:23 PM, cursethiscure
wrote:
> I am working with xts dependent data, and my code is as follows (the problem
> is explained throughout):
>
>
hello,
thank you for your answer!
yes, now it is working!
marion
2011/9/19 Duncan Murdoch
> On 11-09-19 7:30 AM, Marion Wenty wrote:
>
>> Hello,
>>
>> could someone help me with this problem?:
>>
>> I would like to create a latex-script inside of a character vector in
>> order
>> to being able t
On 11-09-19 7:30 AM, Marion Wenty wrote:
Hello,
could someone help me with this problem?:
I would like to create a latex-script inside of a character vector in order
to being able to compilate it within latex in the end.
if i try the following commands:
l1<- "Hello world"
latexscript<- paste(
Hello,
could someone help me with this problem?:
I would like to create a latex-script inside of a character vector in order
to being able to compilate it within latex in the end.
if i try the following commands:
l1 <- "Hello world"
latexscript <- paste("\c",l1,"\c")
... I get an error message
This is the qqmath example from the lattice package. I added the scales to the
example. I would like to switch the axis and not sure how? Meaning I would like
the "height" on the x-axis and the probability on the y-axis. Will you show me
the correct syntax for this switch thanks.
qqmath(~ hei
On Nov 12, 2010, at 12:50 AM, sachinthaka.abeyward...@allianz.com.au
wrote:
Hi R,
In the following code my x-axis is formatted in month format. Which Im
happy with. The y-axis is what I want to re-format with something
else.
Did you mean x-axis?
My
question is, is it possible just to
Hi:
Do you mean to leave the x-axis as is and redo the y? If so, use graphical
parameter yaxt:
x <- y <- 1:10
plot(x, y, yaxt = 'n')
axis(2, at = ..., lab = ..., ...)# fill in the blanks
In case it matters, xaxt = 'n' suppresses the x-axis labels but not the y
labels.
HTH,
Dennis
On Thu,
Hi R,
In the following code my x-axis is formatted in month format. Which Im
happy with. The y-axis is what I want to re-format with something else. My
question is, is it possible just to switch off the xaxis in plot function
(see below). If not how do you get the months to show up as FEB-,
M
AM
To: r-help@r-project.org
Subject: [R] switching elements of a vector
Hi,
I would like to receive help for the following matter:
If I'm dealing with a numeric vectors containing increasing elements.
i.e.
a<-c(1,2,2,2,2,3,3,3,4,4,4,5,5,6,7,7,7)
There exist an efficient way to o
it works as well!
--
View this message in context:
http://r.789695.n4.nabble.com/switching-elements-of-a-vector-tp2228373p2228485.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
Thank you !
Great answers...now it seems very easy...
As usual...a the obviousness of a solution depends on how you face the
problem...
Thank you for help me in find the good approaches...
--
View this message in context:
http://r.789695.n4.nabble.com/switching-elements-of-a-vector-tp2228373p2
Try this:
> which(diff(a) > 0) + 1
[1] 2 6 9 12 14 15
On Mon, May 24, 2010 at 6:02 AM, speretti wrote:
>
> Hi,
>
> I would like to receive help for the following matter:
>
> If I'm dealing with a numeric vectors containing increasing elements.
> i.e.
>
> a<-c(1,2,2,2,2,3,3,3,4,4,4,5,5,6,7,7,
Hi,
is this what you need?
b <- c(NA, a[1:length(a)-1]) # shift values of a one step to the right
which(a-b == 1)
On Monday 24 May 2010 12:02:55 pm speretti wrote:
> Hi,
>
> I would like to receive help for the following matter:
>
> If I'm dealing with a numeric vectors containing increasin
On 5/24/2010 6:02 AM, speretti wrote:
> Hi,
>
> I would like to receive help for the following matter:
>
> If I'm dealing with a numeric vectors containing increasing elements.
> i.e.
>
> a<-c(1,2,2,2,2,3,3,3,4,4,4,5,5,6,7,7,7)
>
> There exist an efficient way to obtain an vector that indicate
Hi,
I would like to receive help for the following matter:
If I'm dealing with a numeric vectors containing increasing elements.
i.e.
a<-c(1,2,2,2,2,3,3,3,4,4,4,5,5,6,7,7,7)
There exist an efficient way to obtain an vector that indicates the position
of the changing element of "a"?
In this ca
Assuming that you are using the xts package, try this:
data(sample_matrix)
sample.xts <- as.xts(sample_matrix)
open <- as.vector(sample.xts[,1])
month <- as.Date(time(sample.xts))
plot(open, month, type="l")
-Peter Ehlers
JSmaga wrote:
Basically it works, but I use the xts format and the
Basically it works, but I use the xts format and the axis are messed up.
plot(c(time(ser)) ~ c(ser), type = "l")
Error in as.POSIXct.default(x) :
do not know how to convert 'x' to class "POSIXlt"
I don't know why do you have any idea?
Besides, why do you use the c(...) function in your code
Try this using the builtin ts series, Nile:
year <- c(time(Nile))
Nile. <- c(Nile)
plot(year ~ Nile., type = "l")
On Thu, Feb 11, 2010 at 10:19 PM, JSmaga wrote:
>
> Hi guys,
>
> I would like to know if it is possible to switch the X Y axis when plotting
> a time series.
>
> Precisely, what I w
Hi guys,
I would like to know if it is possible to switch the X Y axis when plotting
a time series.
Precisely, what I would like to do is having the dates on the Y axis, and
the numbers on the X axis.
I know it is pretty uncommon but it would help me a lot!
Thanks,
Jeremie
--
View this messa
--
View this message in context:
http://n4.nabble.com/Switching-tp1478064p1478064.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posti
Henrique Dallazuanna wrote:
> Try this:
>
> ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
>
>
Also,
ix <- 2*rep(1:(length(X)/2-1), each = 2) + 2:1
X[ix]
or
ix <- seq_along(X) + c(1,-1)
X[ix]
> On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous <
> [EMAIL PROTECTED]> wrote:
>
>
>> All,
>>
Thanks Henrique, Marc, and Gabor!
On 6/27/08 10:17 AM, "Henrique Dallazuanna" <[EMAIL PROTECTED]> wrote:
> Try this:
>
> ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
>
> On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous <[EMAIL PROTECTED]>
> wrote:
>>
>> All,
>>
>> I have a long vector
on 06/27/2008 09:17 AM Marc Schwartz wrote:
on 06/27/2008 09:11 AM David Afshartous wrote:
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without
writing a
loop.
This code works:
X = c(8, 10, 6, 3, 20,
Try this:
ave(X, rep(1:(length(X)/2), each = 2), FUN=rev)
On Fri, Jun 27, 2008 at 11:11 AM, David Afshartous <
[EMAIL PROTECTED]> wrote:
>
> All,
>
> I have a long vector that contains an even number of entries. I'd like to
> switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writ
on 06/27/2008 09:11 AM David Afshartous wrote:
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writing a
loop.
This code works:
X = c(8, 10, 6, 3, 20, 1)
index = c(2,1,4,3,6,5)
X[index]
But for a
You can do it without and index like this:
c(matrix(X, 2)[2:1,])
or if you need the index for some purpose apart from this:
c(matrix(seq_along(X), 2)[2:1,])
On Fri, Jun 27, 2008 at 10:11 AM, David Afshartous
<[EMAIL PROTECTED]> wrote:
>
> All,
>
> I have a long vector that contains an even numb
All,
I have a long vector that contains an even number of entries. I'd like to
switch the 1st and 2nd entry, the 3rd and 4th, and so on, without writing a
loop.
This code works:
X = c(8, 10, 6, 3, 20, 1)
index = c(2,1,4,3,6,5)
X[index]
But for a long list is there a way to generate the index?
: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bob Green
Sent: 13 June 2008 12:14
To: r-help@r-project.org
Subject: [R] Switching the order of legend boxes in a lattice bar graph
I suspect there is a simple solution to this problem, but have been
unable to find it. Below is some code t
I suspect there is a simple solution to this problem, but have been
unable to find it. Below is some code that I have run to create 3
lattice graphs. I have been asked to change the legend so that the
'No' and dark blue are above "Y" and light blue in the legend to
mirror the stacked bars in
29 matches
Mail list logo
