Hi All,
I am trying to start R Server to run some Java Script in my local machine
using the library, 'Rook'. I use Windows 7.
And my codes are following,
> library(Rook)
> myD3dir <- 'D:\\STUDIES\\Java script\\d3-master'
> s <- Rhttpd$new
> s$start(quiet=TRUE)
and i get the following error for
Hi,
I have written a function "harvest" and I would like to run the function for
each value in a vector c(1:1000). The function returns 4 list objects
(obj_1, obj_3, obj_3, obj_4) using the following code at the end of the
function: return(list(obj_1 = obj_1, obj_2 = obj_2, obj_3 = obj_3, obj_4 =
I tried adding my jars to classpath using
#!/usr/bin/Rscript
library(rJava)
#.jclassPath()
.jinit()
.jaddClassPath(dir("target\\mavenLib", full.names=TRUE ))
.jclassPath()
but got
in .jnew("com/algoTree/ClientElasticSearch/ElasticSearchLoader") :
java.lang.ClassNotFoundException
then I t
On Nov 26, 2012, at 11:10 PM, David Winsemius wrote:
On Nov 26, 2012, at 3:05 PM, Aki Hoji wrote:
Hi all,
I have this microarray large microarray data set (ALL) from which
I would like to subset or extract a set of data based on a factor
($mol.biol).I looked up some example of sub
Works for me. Your description is kind of thin for providing a helpful
response, though.
Did you install as administrator? (I have found it works best to not do that)
Did you open the RGui program or R at the command line?
What messages did it display when you started the program (before you type
On Nov 26, 2012, at 3:05 PM, Aki Hoji wrote:
Hi all,
I have this microarray large microarray data set (ALL) from which I
would like to subset or extract a set of data based on a factor
($mol.biol).I looked up some example of subsetting in, picked up
two commands and tried both bu
Looks like
format.summary.difftime <- function(sd, ...) structure(sd$string,
names=rownames(sd))
does the job.
any reason not to use it?
On Mon, Nov 26, 2012 at 7:36 PM, William Dunlap wrote:
>> why do I see NULLs?!
>
> because
>
>> > ... format.difftime does a reasonable job (except that it does
Brent,
Try setting the cex parameter to make the numbers smaller.
Also try modifying the e.size (ellipse size) parameter.
Then, to make much better looking graphs, see if you can install Rgraphviz and
try using the fa.rgraph function or use the graphviz=TRUE parameter.
fa.diagram was a hack to
Hi,
I have installed R 2.15.2 on windows 7.
http://cran.cnr.berkeley.edu/
I tried to run some simple graph code:
http://www.harding.edu/fmccown/r/
But, no graphs are presented or poped up.
Any help will be appreciated.
Thanks
[[altern
Brent,
No, cex doesn't work (as you have discovered). That is a bug.
I will work on it.
In the meantime, try Rgraphviz called from fa.rgaph or
use the output from fa.graph which produces a dot file for processing with any
graphic package (including graphviz) which handles the dot language.
I am trying to call a r file named es.r which have lotes of R functions.
These R functions are internally calling java functions by using ".jnew()"
and ".jcall()". I have added
necessary jar's to the Classpath and I am able to run es.r from command
prompt .
But when I tried to call it
Hello,
When I run the following glm model:
modelresult=glm(CID~WS+SS+DV+DS, data=kimu, family=binomial)
I get the following warning messages:
1: glm.fit: algorithm did not converge
2: glm.fit: fitted probabilities numerically 0 or 1 occurred
What I am trying to do is model my response variable
In this case, instead of naming the column 'c', it names it 'args[1]' as a
string, not a variable.
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Thanks Jim and Rui.
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__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
Yeah, I meant 3 instead of 5.
This was just an example, it's not what Im really doing.
I am using a 'arules' package for data mining, and I have to pass and
'arg[]' element and use it as the new column name of a data.frame.
It's a bit complicated, so I used this example, and I would like to use
'
My problem is relatively straight forward, but I cannot seem to find a way to
make it work.
I have a RCBD with repeated measurements over time. I have created a fit
using the gamm4 package. My model is:
fit4a <- gamm4(Rate ~ s(Time,by=trt,bs="cr")+trt,data=qual.11.dat,
random=~(1
Hi all,
I have this microarray large microarray data set (ALL) from which I would like
to subset or extract a set of data based on a factor ($mol.biol).I looked
up some example of subsetting in, picked up two commands and tried both but I
got error messages as follows
> testset <- sub
Any idea on the reason why the instruction
p1x<-c(2, 1, 0, 0, 0, 7, 1, 0, 2, 0, 5, 0, 0, 0, 0, 0, 10, 1, 3, 0, 2, 0, 0,
0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 12, 0, 0,
1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 4, 0, 1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 1,
0, 1, 0, 0, 0, 3, 0, 0, 1,
HI,
Just a modification of Rui's solution:
x1<-within(x,TYPE<-as.character(TYPE))
group<-cumsum(c(TRUE,x1$TYPE[-1]!=x1$TYPE[-length(x1$TYPE)]))
res<-as.data.frame(do.call(rbind,lapply(split(x1,group),function(x)
c(C1=min(x[,1]),C2=max(x[,2]),TYPE=x[,3][1]))),stringsAsFactors=FALSE)
res[,-3]<-sap
Thank you Paul,
It works.
Now I will study for repeating the same code for different days.
Regards,
Vincent
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The question says that there is an experiement to investigate the effect on
breathing rate when doing different types of exercise with wearing more
clothes or less clothes (factor A, coded 1,2).
The duration of exercise was 10min, 20min, 30min (factor B, coded 1,2,3).
Could you give m anymore hel
On Mon, Nov 26, 2012 at 5:33 AM, Florian Weiler wrote:
> Dear all,
>
> I am using the book "Generalized Linera Models and Extension" by Hardin and
> Hilbe (second edition, 2007) at the moment. The authors suggest that
> instead of OLS models, "the log link is generally used for response data
> th
Dear Bill
Wow! Thank you so much for your rapid reply - you are such a kind person,
thank you!
I'll try fa.rgraph - thanks
Thanks
Best wishes
Brent
-Original Message-
From: William R Revelle [mailto:reve...@northwestern.edu]
Sent: Tuesday, 27 November 2012 5:54 p.m.
To: Brent Caldwell
C
Dear R help
I have conducted a fa() analysis, and I want to use fa.diagram to assess the
extent to which the 11 latent factors predict the 37 items in a psychological
battery. However, the display on the screen has very large font size for the
coefficients of the relationship between the 11 f
On Monday, November 26, 2012, Sam Steingold wrote:
[snip]
>
> there is precisely one country for each id.
> i.e., unique(country) is the same as country[1].
> thanks a lot for the suggestion!
>
> > R> result <- f[, list(min=min(delay), max=max(delay),
> > count=.N,country=country[1L]), by="share.i
> why do I see NULLs?!
because
> > ... format.difftime does a reasonable job (except that it does not copy
> > the input names to its output).
Replace your call of the form
format(difftimeObject)
with
structure(format(difftimeObject), names=names(difftimeObject))
to work around this.
Bill
Thanks, Arun! This pretty much does what I was looking for. Looks like I
should get more familiar with the apply functions, it seems like they can
solve a lot of these problems!
Best,
Steve
On Sun, Nov 25, 2012 at 8:09 PM, arun wrote:
> HI Steve,
>
> You could try this:
> ylist<-lapply(y,functi
Thanks a lot - almost there!
--8<---cut here---start->8---
format.summary.difftime <- function(sd, ...) {
t <- matrix(sd$string)
rownames(t) <- rownames(sd)
print(t)
format(as.table(t))
}
print.summary.difftime <- function (sd, ...) {
print(format(sd),
Hello,
Also gives an error in R 2.15.2 on Windows 7. I'd change "when the
dimnames of the table are not integers " to not numeric as that's what
the code for plot.table tests. And it seems to come from seq.int, since
with table value of 2, seq.int produces a vector of length 2 but the
table l
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a "table" object for printing. Hence,
write
a format.summary.difftime if you want objects of class "summary.difftime" (which
I assume summary.difftime produces) to be formatted
Hi,
> * Steve Lianoglou [2012-11-26 17:32:21
> -0500]:
>
>> --8<---cut here---start->8---
>>> f <- data.frame(id=rep(1:3,4),country=rep(6:8,4),delay=1:12)
>>> f
>>id country delay
>> 1 1 6 1
>> 2 2 7 2
>> 3 3 8 3
>> 4
Hi,
On Mon, Nov 26, 2012 at 4:57 PM, Sam Steingold wrote:
[snip]
>> Could you please copy paste the output of `(head(infl, 20))` as
>> well as an approximation of what the result is that you want.
Don't know how "dput" got clipped in your reply from the quoted text I
wrote, but I actually asked
> * David Winsemius [2012-11-26 08:46:35 -0800]:
>
> On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
>
>> summary(infl), where infl$delay is a difftime vector, prints
>>
>> ...
>>
>>delay
>> string:c("492.00 ms", "18.08 min", "1.77 hrs", "8.20 hrs", "8.13 hrs",
>> "6.98 days")
>> secs :c("
hi Steve,
> * Steve Lianoglou [2012-11-26 16:08:59
> -0500]:
> On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold wrote:
>>> * Steve Lianoglou [2012-11-19 13:30:03
>>> -0800]:
>>>
>>> For instance, if you want the min and max of `delay` within each group
>>> defined by `share.id`, and let's assum
On Mon, Nov 26, 2012 at 2:41 PM, Ludo Pagie wrote:
> Hi all,
>
> there appears to be something strange with the plotting of tables of 1
> dimension; if I attempt to make a plot of a table of characters with only
> 1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
> 'x' and
On Mon, Nov 26, 2012 at 10:38 AM, wrote:
> Goodmorning,
> I'moneafazrtrbhoquhasa variablefactorcomtwoNivesCandH.Queoaatravésdlinear
> regressionrelationshipetrehaSaerifthe
> variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat showsthe
> two curvesHandCLIRand
> Selecionar tudo
> Th
On Mon, Nov 26, 2012 at 4:46 PM, David Winsemius wrote:
>
> On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
>
>> this overcomes the summary generation, but not printing:
>>
>> --8<---cut here---start->8---
>> summary.difftime <- function (v, ...) {
>> s <- su
Hi Sam,
On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold wrote:
> Hi,
>
>> * Steve Lianoglou [2012-11-19 13:30:03
>> -0800]:
>>
>> For instance, if you want the min and max of `delay` within each group
>> defined by `share.id`, and let's assume `infl` is a data.frame, you
>> can do something like
Dear All,
I could use a bit of help here, this function is hard to figure out (for me at
least) I have the following so far:
PKindex<-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,28,25,22,18,14,11))
Dose<-200
Tinf <-0.5
defun<- function(time, y, parms) {
dCpdt <- -parms["kel"]
Hi,
> * Steve Lianoglou [2012-11-19 13:30:03
> -0800]:
>
> For instance, if you want the min and max of `delay` within each group
> defined by `share.id`, and let's assume `infl` is a data.frame, you
> can do something like so:
>
> R> as.data.table(infl)
> R> setkey(infl, share.id)
> R> result <
On Nov 26, 2012, at 1:47 PM, Raffaello Vardavas wrote:
> Thank you Marc,
>
> I will study the material you sent and follow up on this at R-SIG-DB - should
> I still have problems.
>
> I'm using a Mac OS X 10.6.8
>
> Thanks.
> Raff.
In that case, if you use RODBC, you will want to get the O
Hi,
Just a modification:
x1<-gsub("[$]"," ",x)
read.table(text=x1,sep="",header=FALSE,stringsAsFactors=FALSE)
# V1 V2 V3
#1 qwer 12 qwre
#2 ewrtr 7789 ewwe
A.K.
- Original Message -
From: David L Carlson
To: 'Bert Gunter' ; 'Sarah Goslee'
Cc: 'r-help'
Sent: Monday, Nove
Thank you Marc,
I will study the material you sent and follow up on this at R-SIG-DB - should I
still have problems.
I'm using a Mac OS X 10.6.8
Thanks.
Raff.
> Subject: Re: [R] remote connection to an Oracle database - using RODBC -
> RMySQL..?
> From: marc_schwa...@me.com
> Date: Wed,
Another approach is to use gsub() followed by read.table():
> x <- c("qwer$$12$$qwre", "ewrtr$7789$ewwe")
> x <- gsub("\\$+", "\\$", x)
> x <- read.table(text=x, sep="$", stringsAsFactors=FALSE)
> x
V1 V2 V3
1 qwer 12 qwre
2 ewrtr 7789 ewwe
Hello,
You had a typo in the lines() instruction, the parenthesis didn't close
after c(y...).
Anyway, I'm not sure I understand but to have horizontal lines, just
reverse the roles of x and y. (And change pch = "_" to pch = "|").
plot(y, x, type="n",axes=F, xlab="PR(95% CI)",ylab=" ")
points
Try this code that uses "segments" to draw in the bars
# put data into a nicer form
y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17)
ym <- matrix(y
, ncol = 3
, byrow = TRUE
, dimnames = list(NULL, c("Val", "Lower", "Upper"))
)
ym <- cbind(ym, x = 1:4) # add the x
I have a timeseries with some missing data points that I need smoothed
using a Savitzky-Golay. Right now I am using the sgolayfilt function
in the signal package. It fails if I have NA values in the timeseries.
If I replace NA with NULL sgolayfilt does not fail but it drops the
NULL values which is
Sarah:
You may not agree, but the following avoids the IMHO ugly rbind() and
separate conversion to numeric by using scan():
## first, as before..
x <- c("qwer$$12$$qwre", "ewrtr$7789$ewwe")
x <- strsplit(x, "\\$+")
## Then, instead ...
x.convert <- data.frame(scan(textConnection(unlist(x)),wh
Hi everyone,
I have tried to produce a table with reclassification (continuous NRI, p
value) using the reclassification() function in the PredictABEL package. My
problem is that I can only get the results printed on the screen, rather
than stored in an object (as when doing a regular regression)
Dear All:
I would any appreciate any help with this plot I am struggling with.
I have 4 estimates (95% CIs) I want to plot. I want the CI lines to be
horizontal on each plotted point. I was trying to tweak some old codes (was
for a vertical CI lines) into horizontal but not much dice.
Many th
Yes, I'm sorry ... i was checking for another column so i made a mistake.
Thank you people ;)
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To whom It May Concern,
I am working on a dimensional reduction problem using Smoothly Clipped
Asolute Deviation (SCAD) Penalty according to "Variable Selection via
Nonconcave Penalized Likelihood and its Oracle Properties, J.Fan and R. Li,
JASA, Dec 2001". I found an R package named *"ncvreg*" wh
The message is referring to the format of your input data. The package needs
a matrix or vector format.
I suggest using this:
xm<- as.matrix(x)
return.modwt<-modwt(xm, filter="la8", n.levels=5, boundary="periodic",
fast=TRUE)
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does anybody have a suggestion as to how to use R to fit some date to a
cosine function and then have some output statistics that will evaluate the
fit?
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I noticed the following problem with cacheSweave.
If I want to print the result of a list object, with cache=TRUE option, if I
just use summary(x), the output would not appear in the tex file. If I use
print(summary(x)) instead, the output would appear.
With cache=FALSE option, however, summary(
In the example below, I don't see any "$$$" separators. Are you sure this is
supposed to be the separate or just a single dollar sign? If this is the case,
you don't specify what is to happen when multiple separators appear next to
each other. From your example, it appears that two separators
Hello Petr Savicky, hello all,
I have a situation similar to the previous one,
I need to group a data.frame in a specific way,
col1 col2 score
2873 3192 319
4268 4451 183
5389 5534 145
6622 10622 4000
12631 17853 5222
20408 20615 207
21595 21838 243
23121 2313918
the out put
Hi,
You could do this:
Lines<-"qwer$$12$$qwre
ewrtr$7789$ewwe
"
res<-unlist(strsplit(Lines,split="\\$|\n"))
as.data.frame(matrix(res[res!=""],nrow=2,byrow=TRUE),stringsAsFactors=FALSE)
V1 V2 V3
#1 qwer 12 qwre
#2 ewrtr 7789 ewwe
A.K.
- Original Message -
From: ar
try this: (provide sample data next time)
> # create some test data
> x <- data.frame(LST = runif(1000), NDVI = runif(1000, 0, 1))
> head(x,10) # show some data
LST NDVI
1 0.86542839 0.95129647
2 0.88910058 0.75971649
3 0.44086718 0.86532140
4 0.99879370 0.05511501
5 0.02401
-- Forwarded message --
From: Catarina Maia
Date: 2012/11/22
Subject: prediction problem
To: r-help@r-project.org
Hello,
I am using the mda package and in particular the fda routine to
classify/predict in terms of color to a set of 20 samples for which i don´t
know the color.
try this:
> x <- read.table(text = "C1 C2 TYPE
+ 10 20 A
+ 33 44 B
+ 66 80 A
+ 111 140 B
+ 144 220 B
+ 300 340 A
+ 380 449 A
+ 455 500 B
+ 510 520 A
+ 540 580 B", header = TRUE, as.is = TRUE)
> # mark s
Hello,
Try the following.
x <- read.table(text="
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
", header = TRUE)
x
fun <- function(x){
mn <- which.min(x$C1)
mx <- which.max(x$C2)
c(C1 = x$C1[mn], C2 = x$C2[mx], TYPE = x$TY
EcoFranc,
di you remember to use the argument
header=TRUE
?
If not, the variable name is interpreted as data and will flip you into
factor rather than numeric.
On Mon, Nov 26, 2012 at 12:15 PM, jim holtman wrote:
> Also when reading a CSV file, if you do not want characters columns
> converted
Also when reading a CSV file, if you do not want characters columns
converted to factors, use 'as.is = TRUE' as one of the parameters. If
you have a column that is a factor, to convert it to numeric you have
to do the following:
as.numeric(as.character(factorColumn))
notice that you have to conv
Lorenzo,
I'd suggest posting such questions to the R-sig-geo list, which seems
more suitable.
The book "Applied Spatial Data Analysis with R" written by Roger
Bivand, Edzer Pebesma and Virgilio Gómez-Rubio has exactly what you want.
http://www.amazon.com/Applied-Spatial-Data-Analysis-
Robert -- it may be that some of your factor variables cannot be
coerced to numeric because they have non-numeric elements.
This is why we ask for a reproducible example when posting. I
suspect just a few lines from the data frame would tell the story.
On 26-Nov-12, at 7:59 AM, EcoFranc wrot
Hello,
You're wrong, unique doesn't count unique returns a vector without
repetitions.
unique(a) # includes 1 and 4
[1] 1 2 3 6 7 8 4
length(unique(a))
[1] 7
Hope this helps,
Rui Barradas
Em 26-11-2012 14:55, Hard Core escreveu:
this is wrong because with the command "unique" it counts th
On Nov 26, 2012, at 3:53 AM, R_Antony wrote:
> Hi,
> Here im not able to connect with MS-SQLSERVER database with *.R-File.
>
> Previously i was able to do in R in different machine and configuration was
> R Ver-2.11.1
> Package Installed : RODBC_1.3-2
>
> And now where im doinig now there the
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
this overcomes the summary generation, but not printing:
--8<---cut here---start->8---
summary.difftime <- function (v, ...) {
s <- summary(as.numeric(v), ...)
r <- as.data.frame(sapply(s,difftime2string),s
Hi,
Imagine the column is named XX.
Type:
> nrow(table(banca_impresa$XX))
and you'll get how many different categories there are in that column.
(If you type
> table(banca_impresa$XX)
you'll get the frequencies).
José
José Iparraguirre
Chief Economist
Age UK
T 020 303 31482
E jose.iparra
Most likely there's something amiss in your csv file: R won't convert
numeric data to factors unless there are non-numeric characters
included.
First check your csv file for errors. If that doesn't solve your
problem, please provide a reproducible example.
http://stackoverflow.com/questions/596326
> x <- c(1, 2, 3, 1, 6, 7, 8, 3, 4, 4)
> length(unique(x))
[1] 7
On Mon, Nov 26, 2012 at 5:20 AM, Hard Core wrote:
> Hello,
> Suppose that i have a dataframe
> a <- read.dta("banca_impresa.dta")
>
> i have a column with 17900 obs like
>
> 1
> 2
> 3
> 1
> 6
> 7
> 8
> 3
> 4
> 4
>
> and i want to kn
You state that you want "$$$" as the separator, but your example has
"$$" and "$" so I'm assuming an indeterminate number of consecutive
$.
x <- c("qwer$$12$$qwre", "ewrtr$7789$ewwe")
x <- strsplit(x, "\\$+")
x <- do.call(rbind, x)
x <- data.frame(x, stringsAsFactors = FALSE)
x$X2 <- as.numeric(x
Hi,
maybe somebody would be kind enough to help a bloody and unprofessional
beginner like me (and I hope I did not overlook the answer to my question on
the website). I've imported a csv data frame into R, but I can't run a
regression because R interprets 4 out of 5 variables as factors (rather th
My apologies.
I still do not understand the difference; good luck.
Hard Core wrote
> this is wrong because with the command "unique" it counts the only values
> that are unique ..
>
> in my column, for instance, 1 and 4 are not unique so the formula doesn't
> work in my case
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Hey
The code need some corrections and I would kindly ask for help.
Say:
I have a table with two columns:
col1=LST and col2=NDVI
i would like to sort all data by NDVI.
in reality the NDVI ranges between 0 and 1 (although some values might be
minus also).
I want to sort by NDVI values and then ma
I am following instructions online for cluster analysis using the mclust
package, and keep getting errors.
http://www.statmethods.net/advstats/cluster.html
These are the instructions (there is no sample dataset unfortunately):
# Model Based Clustering
library(mclust)
fit <- Mclust(mydata)
plot(fit
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ). With more than one value I don't get
error
It seems that your sqlTables also give no results.
So there must be something wrong with the ODBC connect.
I didn't use odbcConnectAccess, but made an ODBC connection, and then used
ODBCconnect to connect to our database.
Maybe you can define a user DSN and try it this way?
Bart
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That did it!
Thanks so much as always.
I emailed the question to you because I think you are an R expert based
on all the suggestions, feedback and codes I have received from you in
the past.
Yes, I do look for answers in the open
forum but when it comes to a question for which the ans
HI Marcel,
Sorry, that was a mistake.
I guess this should be the one:
df$condition2<-ifelse(df$condition1<=1,runif(length(df$condition1),0,1),df$condition1)
df$condition2
#[1] 0.5207776 0.5227388 0.5196520 0.6552820 2.000 2.000 2.000
#[8] 2.000 3.000 3.000 3.000 3.0
length(unique(a))
Hard Core wrote
> Hello,
> Suppose that i have a dataframe
> a <- read.dta("banca_impresa.dta")
>
> i have a column with 17900 obs like
>
> 1
> 2
> 3
> 1
> 6
> 7
> 8
> 3
> 4
> 4
>
> and i want to know the number of the different values so in this case it
> would be 7
> Ho
this is wrong because with the command "unique" it counts the only values
that are unique ..
in my column, for instance, 1 and 4 are not unique so the formula doesn't
work in my case
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Hey again!
I finally, after some work done before, had time to apply the code.
The sorting of the table did not work well or maybe something was
misunderstood.
I have a table with 973 rows and 1329 col (ascii/text file). I want to sort
the table that all columns are one under each other so that
Hi
I have set of records seperated by a separator say "$$$" i want to get the
values in a dataframe.
eq
qwer$$12$$qwre
ewrtr$7789$ewwe
I want the output as\
V1 V2V3
qwer 12 qwre
ewrtr 7789 ewwwe
Please help me
-
Thanks in Advance
Arun
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View this message in con
Hello members,
I have this data frame with 3 columns,
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
Here, the rows 4 , 5 has type "B" and sim
Goodmorning,
I'moneafazrtrbhoquhasa
variablefactorcomtwoNivesCandH.Queoaatravésdlinear
regressionrelationshipetrehaSaerifthe
variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat
showsthe two curvesHandCLIRand
Selecionar tudo
Thank you
Ana C. Rocha Rua
_
I am trying to call a r file named es.r which have lotes of R functions.
These R functions are internally calling java functions by using ".jnew()"
and ".jcall()". I have added
necessary jar's to the Classpath and I am able to run es.r from command
prompt .
But when I tried to call it
Hello,
Suppose that i have a dataframe
a <- read.dta("banca_impresa.dta")
i have a column with 17900 obs like
1
2
3
1
6
7
8
3
4
4
and i want to know the number of the different values so in this case it
would be 7
How can i do?
Thank you
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You should read your error messages more carefully, especially: could not
find function!
take a look at loadworkbook and loadWorkbook.
Bart
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Hi,
Here im not able to connect with MS-SQLSERVER database with *.R-File.
Previously i was able to do in R in different machine and configuration was
R Ver-2.11.1
Package Installed : RODBC_1.3-2
And now where im doinig now there the same code im trying to execute but its
not connecting.
configur
(from R-help Digest, Vol 117, Issue 26)
Date: Sun, 25 Nov 2012 07:49:05 -0800 (PST)
From: billycorg
To:r-help@r-project.org
Subject: [R] creation of an high frequency series
Message-ID:<1353858545067-4650744.p...@n4.nabble.com>
Content-Type: text/plain; charset=us-ascii
Hi R Users!
I would li
On Nov 26, 2012, at 7:37 AM, "ONKELINX, Thierry"
wrote:
> Dear all,
>
> I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase).
> I keep getting an error about the channel when using sqlQuery(). However,
> sqlTables() does not complain about the channel and lists all tabl
this overcomes the summary generation, but not printing:
--8<---cut here---start->8---
summary.difftime <- function (v, ...) {
s <- summary(as.numeric(v), ...)
r <- as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) <- c("string")
r
First a statistical issue: The survfit routine will produce predicted survival curves for
any requested combination of the covariates in the original model. This is not the same
thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True
adjustment requires a populati
John Kane
Kingston ON Canada
> -Original Message-
> From: landronim...@gmail.com
> Sent: Sun, 25 Nov 2012 15:02:16 +0100
> To: jrkrid...@inbox.com
> Subject: Re: [R] RExcel, ROOo and LibreOffice inquiry
>
> On Sun, Nov 25, 2012 at 2:38 PM, John Kane wrote:
>> Can you supply a link for
Dear all,
I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase). I
keep getting an error about the channel when using sqlQuery(). However,
sqlTables() does not complain about the channel and lists all tables in the
database. If I try sqlFetch(), then R crashes.
I'm happy
Not unless we have more information. Please read the Posting Guide to see how
to make it easier for people to answer your question.
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Oritteropus
Sent: Thursday, November 2
Webinar signup:
Advances in Gradient Boosting: the Power of Post-Processing
December 14, 10-11 a.m., PST
Webinar Registration:
http://2.salford-systems.com/gradientboosting-and-post-processing/
Course Outline:
*Gradient Boosting and Post-Processing:
o What is missing from Gradien
Dear all,
I am using the book "Generalized Linera Models and Extension" by Hardin and
Hilbe (second edition, 2007) at the moment. The authors suggest that
instead of OLS models, "the log link is generally used for response data
that take only positive values on the continuous scale". Of course the
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