You can change ex <- data.frame("gIN" = gIN, "bc1" = bc1, "bc2"=bc2,
"group" = group)
to
ex <- data.frame("gIN" = c(gIN,gIN), "bcX" = c(bc1,bc2), "group" =
c(group,group))
On Wed, Jun 3, 2015 at 2:27 PM, hedelhusk [via R] <
ml-node+s789695n4708145...@n4.nabble.com> wrote:
> Hello,
>
> I would
Dear Sir,
I have converted the data into numeric as the function read.excel read the
data file as a character (as it was said by David Meyer, maintainer of the
package e1071) and after that run the function svm. Then another error is
found there. The detail description of the code and error is giv
Hi valerio,
Any chance of letting me know how "it didn't work out"? For example
your code and the resulting error message.
Jim
On Wed, Jun 3, 2015 at 10:32 PM, valerio orfano wrote:
> Hi Jim and thanx. Unfortunately it didn’t work out!!
>
> any other help?
>
> rgds valerio
> On 03 Jun 2015, at
Hi Pijush,
As before, please keep the messages on the mailing list. In your
example above you have not defined "y", so the function is probably
complaining about being passed a NULL for the dependent variable. I
forgot to add in my previous message that I imported your data by
exporting it from XLS
> On 03 Jun 2015, at 22:05 , Erica Cseko Nolasco wrote:
>
> I tried using TRUE instead T, but it gave me the same error. (Script on the
> bottom)
> I really want a paired test and it seems to give me as we can see on the
> following matrix.
> No idea what is happening
>
I get the impression
I found the gather function from the tidyr package, which worked nicely:
gather(ex,bcX,value, bc1:bc2)
gIN group bcX value
1 A_1 A bc1 1219.79
2 A_2 A bc1 1486.84
3 A_3 A bc1 1255.80
4 A_4 A bc1 941.87
5 B_1 B bc1 588.19
6 B_2 B bc1
And I think this will do it too.
library(reshape2)
melt(ex, id.vars= c("gIN", "group"),
variable.name = "bc",
value.name = "value",
na.rm = FALSE)
John Kane
Kingston ON Canada
> -Original Message-
> From: jonsle...@gmail.com
> Sent: Wed, 3 Jun 2015 18:25:02 -0
Hello,
I would like to ask for some advice in reformatting a data frame such as
the following one:
gIN <- c("A_1","A_2","A_3","A_4","B_1","B_2")
bc1 <- c(1219.79, 1486.84, 1255.80, 941.87, 588.19, 304.02)
bc2 <- c(319.79, 186.84, 125.80, 94.87, 1008.19, 314.02)
group <- c("A","A","A","A","B","B"
I tried using TRUE instead T, but it gave me the same error. (Script on the
bottom)
I really want a paired test and it seems to give me as we can see on the
following matrix.
No idea what is happening
> pairwise.t.test(x = data$tss,data$pa,p.adjust.method =
'bonferroni',paired=TRUE)
Error in compl
> On 03 Jun 2015, at 19:14 , Erica Cseko Nolasco wrote:
>
> Thanks Jim,
>
> I removed the corresponding cases and tried again. What I discovery now is
> if I run the function without paired = T (pairwise.t.test(x =
> data$tss,data$pa) is works. However, if I add the paired = T
> (pairwise.t.tes
You can use
labels=as.expression(lapply(1:9, function(i)bquote(hat(lambda)[.(i)])))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jun 3, 2015 at 12:39 PM, Frank S. wrote:
> Hi everyone, I have allocated greek letters with subscript numbers on
> x-axis of my R plot this way:
> plot(10
Just to be on the safe side, what is T? It is recommended to use TRUE in
case you set T to something else. I think this is very unlikely to solve
the problem but it is worth trying.
On 03/06/2015 18:14, Erica Cseko Nolasco wrote:
Thanks Jim,
I removed the corresponding cases and tried again.
Hi everyone, I have allocated greek letters with subscript numbers on x-axis of
my R plot this way:
plot(10:18, 30:38, lwd=2, xlim=c(10,18), ylim=c(30,38), xaxt="n",
xlab="weights", ylab="value")
points(10:18, 30:38, pch=19)
axis(1, at=10:18, labels=c(expression(hat(lambda)[1]),
expression(ha
On Jun 3, 2015, at 9:39 AM, John Wasige wrote:
> Hello community,
>
> Could somebody help on how I can
> add
> a legend to my heatmap plot. I need to know what values from the data do
> colours ( "red","yellow", "yellowgreen","lightblue4") represent from the
> data. The script is here below
On 6/3/2015 11:26 AM, Boris Steipe wrote:
If letters 1 and 2 must be equal with p=0.5, and 1 and 3 must be equal with
p=0.5, then letter 1 must be the same as either 2 or 3. Therefore:
Choose a letter.
Make a pair of (letter, (not letter)).
Reverse the pair with p = 0.5
Concatenate your letter
Valerio,
Actually htmlize works pretty well. I used it for a long time. If you want to
get the most help from this forum you need to provide a reproducible example
showing htmlize not working for you.
Nevertheless, today I would recommend going with RMarkdown. It may be overkill
for your ex
If letters 1 and 2 must be equal with p=0.5, and 1 and 3 must be equal with
p=0.5, then letter 1 must be the same as either 2 or 3. Therefore:
Choose a letter.
Make a pair of (letter, (not letter)).
Reverse the pair with p = 0.5
Concatenate your letter and the pair.
Is that what you need?
B.
Or try the brand-new way:
http://www.r-pkg.org/search.html?q=html
--Ista
On Wed, Jun 3, 2015 at 1:06 PM, MacQueen, Don wrote:
> When it's as simple as in Boris's example, just use cat() statements.
>
> Otherwise, go to CRAN, find the packages page ("Table of available
> packages, sorted by name
Thanks Jim,
I removed the corresponding cases and tried again. What I discovery now is
if I run the function without paired = T (pairwise.t.test(x =
data$tss,data$pa) is works. However, if I add the paired = T
(pairwise.t.test(x = data$tss,data$pa,paired = T) is gives me the error ' Error
in compl
When it's as simple as in Boris's example, just use cat() statements.
Otherwise, go to CRAN, find the packages page ("Table of available
packages, sorted by name"), and search for "html"
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-4
On Jun 3, 2015, at 3:30 AM, Jeremy Clark wrote:
> The coding I've settled on to save file without clipping is:
What exactly was "clipping". You earlier complained about "jaggies". There was
no restriction of the plotted lines to the plot area in the example you earlier
presented. That's what I
Hello community,
Could somebody help on how I can
add
a legend to my heatmap plot. I need to know what values from the data do
colours ( "red","yellow", "yellowgreen","lightblue4") represent from the
data. The script is here below and attached is the test dataset. Thanks for
your help.
Hohn
Hi Boris and thanx a lot
Which library should i be using to create html in R?
rgds valerio
On 03 Jun 2015, at 18:08, Boris Steipe wrote:
> ... as in: the png exists in a directory that is accessible to the server?
>
> That would be as simple as creating a HTML document with the following
> co
Hello community,
Could somebody help on how I can a legend to my heatmap plot. I need to
know what values from the data do colours ( "red","yellow",
"yellowgreen","lightblue4") represent from the data. The script is here
below and attached is the test dataset. Thanks for your help.
Hohn
###
On 03/06/2015 11:56 AM, Thierry Onkelinx wrote:
> Dear all,
>
> I'm a bit puzzled by the difference in an object when created in R 32-bit
> and R 64-bit.
>
> Consider the code below. test.rda is available at
> https://drive.google.com/file/d/0BzBrlGSuB9n-NFBWeC1TR093Sms/view?usp=sharing
>
> # Ru
... as in: the png exists in a directory that is accessible to the server?
That would be as simple as creating a HTML document with the following contents:
http://www.w3.org/TR/html4/strict.dtd";>
B.
On Jun 3, 2015, at 11:32 AM, valerio orfano wrote:
> Hi All
>
> Is there any chanc
Dear all,
I'm a bit puzzled by the difference in an object when created in R 32-bit
and R 64-bit.
Consider the code below. test.rda is available at
https://drive.google.com/file/d/0BzBrlGSuB9n-NFBWeC1TR093Sms/view?usp=sharing
# Run in R 3.2.0 Windows 32-bit, lme4 1.1-8
library(lme4)
load("test.r
Thank you! It worked.
Hanna
2015-06-03 3:49 GMT-04:00, Duncan Mackay :
> Hi
> If you use the lattice package and not resort to other packages read the ?
> xyplot carefully
> There are several ways to put a key in the graph including auto.key if you
> want something simple ie default
> The examp
svm() needs a matrix as input but read.xlsx() produces a data frame.
B.
On Jun 3, 2015, at 5:01 AM, Pijush Das wrote:
> Dear Sir,
>
>
> I am working with SVM recently, facing a problem which is given below.
> If you able to solve the problem, please send me the code.
>
>
> Thank you very
Hi All
Is there any chance in R to convert a png and/or pdf file into an html?
Any example?
I’ve tried htmlize but doesn’t work out!
Rgds valerio
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo
Hi
there is probably easy solution by gsub but it is not my cup of tea so I used
strsplit
spl<-data.frame(t(matrix(unlist(strsplit(as.character(my_dat$instrument),
"_")),2)))
res<-data.frame(spl, my_dat$mtm_value)
> res
X1 X2 my_dat.mtm_value
1 EQ 0 23
2 EQ 1
Hello,
Try the following.
tmp <- strsplit(as.character(my_dat$instrument), "_")
tmp <- t(as.data.frame(tmp))
tmp <- data.frame(instrument = tmp[,1], sr_no = as.integer(tmp[, 2]),
my_dat$mtm_value)
result <- tmp[order(tmp[, 1], tmp[, 2]), ]
rm(tmp)
rownames(result) <- NULL
result
Hope this
First of all, the as.Date() function converted it to the Date class (not
"Date,format..", and I'd suggest the terminology is important here).
> class('2015-4-1')
[1] "character"
> class( as.Date('2015-4-1') )
[1] "Date"
Second, the format() function always converts to character class.
> n <- 3
Dear R forum
I have a data (actually its a big data and I am only giving part of my
interest) as
my_dat = data.frame(instrument = c("EQ_0", "EQ_1", "EQ_10", "EQ_100", "EQ_2",
"EQ_20", "IRS_0", "IRS_1", "IRS_10", "IRS_100", "IRS_2", "IRS_20"), mtm_value =
c(23, 63, 8, 44, 68, 11, 83, 56, 73, 92
I'm running into a road block here and I can't figure out what I'm doing wrong.
I need to skip over the link if the text equals postseason. The text is in
the second li in the xpaths below in my code. I tried li[not(.,"postseason")]
as I thought that is what I needed to exclude the postseason
Dear Sir,
I am working with SVM recently, facing a problem which is given below.
If you able to solve the problem, please send me the code.
Thank you very much.
>library(e1071)
>library(openxlsx)
> List<- read.xlsx(file.choose(), sheet = 1, colNames=TRUE,rowNames = TRUE)
> Data<- read.xlsx(fi
Another trick if Bill is correct is to sort the data first, by whatever
variable the color is intended to represent.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 6/2/15, 9:06 AM, "William Dunlap" wrote:
>Points later in
Hi,
On Wed, Jun 3, 2015 at 8:26 AM, Thomas Chesney
wrote:
> I'd like to add some columns into the middle of this:
>
> vec <- c(1, -1)
> lst <- lapply(numeric(n-1), function(x) vec)
> combos <- as.matrix(expand.grid(lst))
> colnames(combos) <- NULL
This isn't quite reproducible, since n is undefi
Hi Antonello,
I had never heard of Rpubs until now so I went and had a look. It seems to
have allowed me to register but I don't seem to be able to do anything. I tried
goign to the support page (bottom of intro page) and go a message "This site is
closed" so it may be they are down or having a
Hi Jim and thanx. Unfortunately it didn’t work out!!
any other help?
rgds valerio
On 03 Jun 2015, at 02:29, Jim Lemon wrote:
> Hi valerio,
> This is a guess, but try running your code with "htmlize" (prettyR).
> Change the "pdf" call to:
>
> png("C:/Users/Administrator/Desktop/treemap1.png")
>
I'd like to add some columns into the middle of this:
vec <- c(1, -1)
lst <- lapply(numeric(n-1), function(x) vec)
combos <- as.matrix(expand.grid(lst))
colnames(combos) <- NULL
The way I have used is:
combos2 <- matrix(NA, nrow=64, ncol=9)
combos2[,1] <- 0
combos2[,2] <- combos[,1]
combos2[,3]
I don't think there is any reason to get rid of that message unless you have a
presentation problem, that is, you are including that output in a paper.
All that is, AFAIK, is a notice that ggplot() is using the default binning
rule. You can change the number of the bins if you need more gran
Hi,
it seems not possible to susbset a svydesign object (DBI svydesign) and use
variables to make the subset expression
uff<-c("14","15")
for (i in 1:length(uff))
{
subpnad<-subset(pnad, uf==uff[i] & v0302=='4')
}
the error is the following
Error in sqliteSendQuery(con, statement, bind.data)
The coding I've settled on to save file without clipping is:
library(gridExtra)
gt <- ggplot_gtable(ggplot_build(q3))
gt$layout$clip[gt$layout$name=="panel"] <- "off"
gt4 <- arrangeGrob(gt)
ggsave <- ggplot2::ggsave; body(ggsave) <- body(ggplot2::ggsave)[-2]
## from Baptiste
ggsave("gt.pdf", plot
Dear All - I'm trying to integrate the following function via nlsLM - and I
think the problem is something to do with passing the correct arguments
(error is given below) - any help gratefully appreciated - many thanks.
library(minpack.lm)
dOM <- function(x, mu = 0, sigma = 1, log = FALSE)
{
Hi, I do not know whether someone had the same problem and if s/he can help
me with some advice.
I'm using RPubs from RStudio to publish tutorials and exercises.
On June 2, I've published a Markdown post in RPubs with RStudio.
So, I'm sure the system can connect to Rpubs without any issues.
Some
Hi Pijush,
First, please keep the messages on the help list.
As far as I can determine, the "x" in your code above is a matrix of
character strings. When you transpose (t) the initial data frame, all
of the numbers are coerced to character mode as the columns must all
be of the same mode. The firs
Hi
If you use the lattice package and not resort to other packages read the ?
xyplot carefully
There are several ways to put a key in the graph including auto.key if you
want something simple ie default
The examples at the bottom give some ideas
If you want something special you can use key = lis
Dear Glenn,
Your code contains a typo: it has bindwidth instead of binwidth. Fixing
that will no longer show the message. AFAIK, the message is always
displayed when binwidth is not set. You can use suppressMessages() to hide
them.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bos
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