x(ids, na.rm = TRUE) : no non-missing arguments to max;
returning -Inf
Hope this helps,
Rui Barradas
After playing with this for a little while, I realized that the problem
with plotting the confidence limits is the addition of ylim(470, 500).
The confidence values are outside the y
nt port on each invocation
free_port(random=TRUE)
Hope this is helpful,
Dan
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27;t see the code you used to generate the curve. What kind of curve
are you trying to plot, and why are you trying to plot that curve? As
Rui suggested, it typically doesn't make sense to use line plots to
represent discrete data. If you explain what your end goal is, someone
may be able t
01 7:28:10'
> dt <- strptime(in_dt_str,'%Y-%m-%d %H:%M:%S')
if the date and time are stored in different variables then paste them
together before converting
> in_dt <- '2022-01-01'
> in_tm <- '7:28:10'
> dtm <- strptime(paste(in_dt, in_tm
t;,"très important","pas important","pas important","important","important","très important","très
important","pas important","pas important")
statut2<-c("moyennement riche","pas riche&quo
quot;date" functions to get what you want. For example
# pivot your data
data_long <- data %>% pivot_longer(starts_with("X"), names_to =
"chr_date", values_to = "LST")
# now you can use various data functions to get your month, day, and year
# for exampl
e, "1 day"), unit="secs"))
# plot event date by time grouped by date
ggplot(myDat, aes(x=tme, y = dte, group = dte)) + geom_line() +
geom_point() +
expand_limits(y=as.Date(c('20210301', '20210331'),'%Y%m%d'
close))
AAPL %>%
ggplot(aes(x = date, y = close)) +
geom_line() +
labs(title = "AAPL", y = "Closing Price", x = "") +
coord_x_date(xlim=xminmax, ylim=yminmax) +
theme_tq()
Hope this is helpful,
Dan
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and provide commented, minimal, self-contained, reproducible code.
You have the wrong format for your date
es of con
observed median lower upper
Q25 0.054000 0.054000 0.01525 0.11275
Q50 0.139275 0.139275 0.06140 0.31000
Q75 0.315000 0.315000 0.17300 0.45250
Quartiles of vac
observed median lower upper
Q25 0.01250 0.01250 0.00125 0.026000
Q50 0.02675 0.02675 0.01665 0.144575
ut in your
SAS log or listing.
Dan
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UBSCRIBE and more, see
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Works fine in Windows 10 64-bit with R-4.0.2, so I would echo Bert
Gunter's ad
for me.
df[sample(nrow(df),3), 'treated'] <- TRUE
Hope this is helpful,
Dan
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PLE
On 6/10/2020 1:20 AM, Luigi Marongiu wrote:
isoDates = as.Date(oriDates, format = "%m/%d/%y")
You need to use the format for European short dates.
isoDates = as.Date(oriDates, format = "%d/%m/%y")
Hope this is helpful,
Dan
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Po
-04-08","2020-04-09","2020-04-10"))
nc<-c(1,1,2,7,3,6,6,20,17,46,67,71,56,70,85,93,301,339,325,226,608,546,1069,1264,1340,813,608)
plot(as.Date(mydates),nc,pch=16,type="o",col="blue",ylim=c(1,1400),
xlim=c(min(as.Date(mydates)),max(as.Date(mydates
Here is one more option using the ave() function. Using Jim's data and
naming convention
fkdf$X1_change <- ave(fkdf[,'X1'], fkdf$Country, FUN=function(x)
c(0,diff(x)))
fkdf$X2_change <- ave(fkdf[,'X2'], fkdf$Country, FUN=function(x)
c(0,diff(x)))
hope t
e sense to include both
in your model.
Dan
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.
If you wish to do something else, you will need to be more specific
about what you want.
Hope this is helpful,
Dan
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h
I will leave the solution of the problem to you.
Dan
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PLEASE do read the posting guide htt
- strptime("02/20/13 00:00:00", "%m/%d/%y %H:%M:%S")
dt2 <- strptime("07/03/18 15:30:00", "%m/%d/%y %H:%M:%S")
dt <- seq(from=dt1, to=dt2, by=900)
dt[length(dt)]
There might also be some useful functions in the lubridate package.
Hope this is helpful,
D
rrences of '%'.
The pattern is more strict, and that could cause the conversion to fail
if the process that created the strings resulted in trailing spaces.
Without the '$' the conversion succeeds.
df <- data.frame(variable = c("12.6% ", "30.9%", &quo
to above
(which describes the dataMultilevelIV dataset)?
Dan
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PLEASE do read the posting gu
x27;Dx',ave(rep(1, nrow(have)), have[,1:2], FUN =
seq_along), sep='')
# cast the data into wide format
cast(cbind(have,dxnames), ClaimServiceID + ClaimID ~ dxnames,
value='DiagnosisCode')
Hope this is helpful,
Dan
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Port Townsend, WA USA
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ary(devtools)
install_github("kasperwelbers/semnet")
Hope this is helpful,
Dan
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PLE
object does not
have 3 rows, so you get an error and SlopeDiff is not created.
You need to correct these problems, and any others, so that your code
runs correctly when there are no data problems. Then you can worry
about trapping errors in the case where there are data problems.
Hope this i
median squared error", shouldn't the
final line of the function be
median((y - ypred)^2)
Dan
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(i in 1:10) lst[i] <- i
mean(lst) # does not work
The documentation for mean, ?mean, says that it is looking for a numeric
or logical vector. To convert your list to a numeric vector you could
unlist() it.
mean(unlist(lst))
Hope this is helpful,
Dan
--
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Port Townsend,
this is helpful,
Dan
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Port Townsend, WA USA
On Mon, Mar 12, 2018 at 3:49 AM, Bert Gunter wrote:
You need to re-read ?density and perhaps think again -- or do some study --
about how a (kernel) density estimate works. The points at which the
estimate is calculated are *not*
d not get the message you are getting. I
suspect you have something in your workspace that is causing the problem.
Dan
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ading
Tools/RV Tools/myfile.r"
I don't know how rscript handles the '\' character (i.e. as an escape or
not) so I changed the '\' to '/' just to be safe. And note, the program
pathname and the file being passed need to
at your real world task actually is.
Hope this is helpful,
Dan
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PLEASE do read the posting gui
On 11/7/2017 12:01 PM, Tom Backer Johnsen wrote:
Dear R-help,
I am running a Mac under Sierra, with R version 3.4.2 and RStudio 1.1.383. When
running head () or tail () on an object in a script using source (
minimal, self-contained, reproducible code.
this looks like a good place for apply()
apply(data,2,function(x) sum(x != 0, na.rm=TRUE))
Hope this is helpful,
Dan
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0:00",
"2012-04-24 06:00:00", "2012-04-24 12:00:00", "2012-04-24 18:00:00",
"2012-04-25 00:00:00", "2012-04-25 06:00:00", "2012-04-25 12:00:00",
"2012-04-25 18:00:00", "2012-04-26 00:00:00", "2012-04-26 06:00:0
y making sure the NUMLOCK key is on, hold down the alt key and
press 0228 on the numeric keypad.
Hope this is helpful,
Dan
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):1])
Hope this helps,
Dan
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PLEASE do read the posting guide http://www.R-project.org/post
o generate your random samples, and then describe
what you want, e.g. "plot sample means and standard errors estimated
from the samples," we can play along at home. Then you may get some
usable help.
You could also Google something like "R plot means and standard errors&qu
DON'T use Excel. Ever. See:
http://www.stat.uiowa.edu/~jcryer/JSMTalk2001.pdf
cheers,
Rolf Turner
Unfortunately, that link appears to be broken / does not exist anymore.
Dan
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Should that be the Rcpp package and not Rccp?
Dan
--
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Port Townsen
n?
Thanks in advance, best
Jean-Philippe
I don't know about efficiency, but it looks like you could do something
like this:
y <- t(matrix(t(dataGaus),4))
Maybe someone will come along with something better,
Dan
--
Daniel Nordlund
Port
e along with a "better" answer, but I played
with this a bit and this is what I came up with. After running your code
I ran
DF2$site <- substr(as.character(DF2$variable),1,5)
DF2$var <- substr(as.character(DF2$variable),7,10)
DF3 <- cast(DF2,year +
sion 3.3.1)
What can I do?
As far as I know, write.xls and write.table are not packages, they are
functions. There is a write.xls function in the xlsx (and also the
openxlsx) package and write.table is a built-in R function that exists
in the util package.
Dan
--
Daniel Nordlund
Port Townsen
ing the same answer."
Someone may then be able to do more than guess at what the problem is.
Dan
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d
document will help:
https://cran.r-project.org/web/packages/xlsx/xlsx.pdf
Dan
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PLEASE do read
nd
> strptime("Thu, 25 Aug 2016 6:34 PM",format="%a, %d %b %Y %I:%M %p")
[1] "2016-08-25 18:34:00 PDT"
works for me.
Hope this is helpful,
Dan
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producible exampole, I did notice one
problem in the example you did give of a directory with spaces. It
looks like you were using single quotes (') around the path/filename.
Windows requires that there be double quotes (") around any
path/filenam
need to get an XPORT format file, or have SAS available, or get
some 3rd party software that will read SAS datasets.
Dan
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ong at home" (i.e. give us a reproducible
example).
Dan
Daniel Nordlund
Port Townsend, WA
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PL
Frederic,
you need to be more clear about what you want to do. If you are
sampling 100 numbers from 1:100 with replacement, the sum will always be
greater than 50 (as has already been pointed out).
In addition, you mention trying to eliminate zeros. If you are sampling
from 1:100 there wil
t; (in descending
order). If you want the character value of line 7 to sort last, it
would need to be "06.1 (0.61)" or " 6.1 (0.61)" (notice the leading space).
Hope this is helpful,
Dan
Daniel Nordlund
Port Townsend, WA USA
__
R
ositions.
begs <- function(x) c(0,x[-length(x)])+1
Then, then use that function in your call to str_sub
str_sub(test_string,begs(ends),ends) %>% print
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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eate a pdf document and the warning is cluttering
up my pdf document. Is there anyway to alter the plot statement to avoid
the warning? If not, is there a way to get knitr not to print out the
warning to the markdown document so that it doesn't end up in the pdf
document?
Thanks,
row.names = c(NA,
-8L), class = "data.frame")
print(f)
f$dateandtimes <-
as.chron(as.POSIXct(as.character(f$date),format = "%Y-%m-%d
%H:%M:%S"))
print(f)
f$justtimes <- times(as.numeric(f$dateandtimes) %% 1)
uot;2015-10-30 00:50:00",
"2015-10-30 09:30:00", "2015-10-30 21:10:00", "2015-10-31 00:50:00",
"2015-10-31 10:30:00"), class = "factor"), value = c(88L, 17L,
80L, 28L, 23L, 39L, 82L, 79L)), .Names = c("date", "value"), ro
o convert strings to factors,
so you can modify as needed. In addition, if your files aren't as
regular as I inferred, you can increase the number of rows to read in
the first line to ensure getting the classes right.
Hope this is helpful,
Dan
--
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Bothell, WA USA
___
x27;, believeNRows=FALSE, colQuote=NULL)
Hope this is helpful,
Dan
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PLEASE do read the posting guide http
equal to
zero!
On Jul 13, 2015 5:31 PM, "Daniel Nordlund" mailto:djnordl...@frontier.com>> wrote:
On 7/13/2015 3:01 PM, Lida Zeighami wrote:
Hi there,
I have a matrix which its elements are 0, 1,2,NA
I want to remove the columns which the colsums are equ
orld
problem. There is no error checking, and for large samples it may not
scale well.
Hope this is helpful,
Dan
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m contain identical letters 1 and 3, AND ~50% again contain identical
letters 2 and 3 (except in this example as it is not possible from the choices).
Can multiple probability weightings be combined in such a manner?
Ben,
If I correctly understand your requirements, you can't do what you
lowing:
gaussianKernel <- function(u) exp(-u^2/2)/(2*pi)^.5
densityFunction <- function(x, df, ker, h){
difference = t(t(df) - x)/h
W = sum(apply(difference, 1, ker)) / (nrow(df)*h)
}
If you are wanting to do density estimation for real world work, I would
get help from someone i
;- rnorm(100,200,50)
# create deciles for binning
decile <- as.numeric(cut(rt, quantile(rt,0:10)/10),include.lowest=TRUE))
# collect into a dataframe (not really necessary)
df <- data.frame(rt=rt, decile=decile)
#compute the bin means
aggregate(rt,list(decile),mean,data=df)
This sho
On 5/16/2015 1:19 PM, Daniel Nordlund wrote:
On 5/16/2015 12:32 PM, li li wrote:
Hi all,
I wrote the following code and have two questions:
(1) As you can see, I would like different colors for different
types. It does not come out that way in the graph from this code.
Anyone know how to
over the colors chosen. I created a vector, color, specifying
the colors and the order they will be applied.
2. the scales parameter is what you are looking for. The parameter is
looking for a list which consists of name=value pairs. You can rotate
the labels by a specified number of deg
index=3)
Error in boot.ci(results, type = "bca", index = 3) :
object 'results' not found
A reproducible example means that when I run your code on my machine, I
get the same results / warnings / errors that you get. I got something
different.
Dan
--
Daniel Nordl
e help says of the formula
argument
formula
formula for restricted MIDAS regression or midas_r object. Formula must
include fmls function
your formula does not include the fmls() function, it uses mls(). So I
think your problem may have to do with how you are calling the midas
assuming you want to sample without replacement.
Generalizing it to other data structures is left as an exercise for the
reader.
replicate(100,mean(sample(yourdata,30, replace=FALSE)))
hope this is helpful,
Dan
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-- To UNSUBSCRIBE and more, see
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and provide commented, minimal, self-contained, reproducible code.
--
Daniel Nordlund
Bothell, WA USA
_
AS you could use PROC FCMP to
turn this into a function.
If you are interested contact me offline and I will send you a PROC FCMP
implementation.
Dan
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-04 2011-11-04
Pradip K. Muhuri, PhD
SAMHSA/CBHSQ
1 Choke Cherry Road, Room 2-1071
Rockville, MD 20857
Tel: 240-276-1070
Fax: 240-276-1260
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Daniel Nordlund
Sent: Sunday, November 09, 2
rg/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
I am not familiar with the mutate() function from dplyr, but you can get
your wanted results as follows:
data2 <- within(data1, oidflag <- apply(data1[,-1], 1, max, na.rm=TRUE))
Hope this is h
,list(Population), mean))
with(your_data_frame,aggregate(R,list(Population), var))
hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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PLEASE do read the posting guide
, p=probs, replace=TRUE)
}
new_sample(x)
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Dan Abner
> Sent: Monday, June 23, 2014 3:19 PM
> To: Greg Snow
> Cc: r-help@r-p
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Daniel Nordlund
> Sent: Thursday, May 01, 2014 1:10 AM
> To: r-help@r-project.org
> Subject: Re: [R] A combinatorial assignment problem
>
>
r,m,max_iter=120) {
n <- 0
cmb <- combn(r,m)
repeat {
n <- n+1
tbl <- table(map<-cmb[,sample(1:choose(r,m),k)])
if(min(tbl) == max(tbl)-1) break
if(n > max_iter) break
}
return(t(map))
}
a <- assignment(10,7,3)
Dan
Daniel Nordlund
freezes R on my Win 7 Pro x64 box using either 64-bit R-3.0.3 or R-3.1.0. You
might try switching to shell() instead of system()
> command <- paste(aa, fnm)
> shell(command)
However, it all depends on what programs you are trying to run and what
behavior you expect.
Dan
Daniel
sp3 = structure(c(1L, 1L, 1L, 1L), .Label = "mean", class =
> "factor")), .Names = c("watershed",
> "year", "sp1", "sp2", "sp3"), class = "data.frame", row.names = c(NA,
> -4L))
>
> Any s
as cut-n-paste from your email) and got a plot
of the data with regression line without any error message. What version of R
and ISwR are you running? Have you tried closing R and starting a clean
session?
Since the example worked for me, there is not much else I can help with.
Dan
Dan
e of 1.5 hours:
>
> > as.POSIXct("2010-04-04 03:00:00")- as.POSIXct("2010-04-04 02:30:00")
> Time difference of 1.5 hours
>
>
>
>
>
>
> Any suggestions?
> Thanks,
> David Fox.
>
Daylight savings time change in Australia?
Dan
Da
read up on
?sample
Dan
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Alaios
> Sent: Monday, January 20, 2014 10:12 PM
> To: Alaios; R-help@r-project.org
> Subject
rted data starts
> from January 2, 2000.
>
> What have I done wrong?
>
You need to specify an appropriate value for the origin parameter. It looks
like as.Date in the zoo package (which masks the as.Date in base) defaults to
the Unix epoch value, origin='1970-01-01'. Yo
#check for excess of 'a's or 'b's if(suma - sumb > 2) p <- .2
if(sumb - suma > 2) p <- .8
}
}
If this doesn't do what you want, then write back to R-help describing what
your ultimate goal is (i.e., what this method of randomization is supp
oc CPORT was used to write the file,
read.xport will not be able to read it. read.xport only reads files written
using the XPORT engine.
If you don't have access to SAS you will need to get whomever created the file
to re-create it.
Dan
Daniel Nordlund
Bothell, WA USA
> -Original
cko
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > __
> > > R-help@r-project.org mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> >
Yeah, I tried building the package and got essentially the same warnings and
decided that further assistance required someone above my pay grade. :-)
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
The R high performance computing sig might be useful for some of these
questions.
https://stat.ethz.ch/mailman/listinfo/r-sig-hpc
Dan
Daniel Nordlund
Bothell, WA USA
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behal
aniel Nordlund
Bothell, WA USA
_
From: Shang Zuofeng [mailto:zuofengsh...@gmail.com]
Sent: Monday, August 19, 2013 2:16 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] How to rebuild an R package that has been removed from CRAN?
Thanks, Dan!
The package is "ass
age, why it was removed from CRAN, and a host of other
things. If you want more detailed help, you need to provide the "at a minimum"
info requested in the posting guide. It would also help if you told us what
package you are trying to install.
Dan
Daniel Nordlund
Bothell, WA
, then it is even
easier. I stored the following data in a file, temp.txt,
X1 X2 X3
age race stat
12 2 35
17 6 55
and read it like this
dat1<- read.table("c:/tmp/temp.txt", header=TRUE, skip=1)
and got this
> dat1
age race stat
1 122 35
2 176 55
>
Hope
age on my system
that I tried.
So if you can't read the data one way, try another. You could install and load
the sos package and runthe following function
findFn('xls')
and you will get all sorts of suggestions.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
_
with). Try the following (and notice the space at the end of "Infected ".
RECinf<-subset(REC2, INFECTION=="Infected ")
David's suggestion worked because you did include a space there.
Dan
Daniel Nordlund
Bothell, WA USA
_
pecies c present at site 1 on date 3, but it is not present in your desired
data. It is not at all clear (nor is it deducible) how you get from your
example data to your desired data. If you clarify your requirements, maybe
someone will be able to help.
Dan
Daniel Nordlund
Bothell, WA USA
Thank you in advance for your help?
>
>
The data appear to be tab delimited with the decimal point being a comma (',').
So, try read.csv2()
heisenberg <- read.csv2(file="comprice.csv", header=TRUE, sep="\t")
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
thod actually used.
No guessing necessary, as the R-help is quite explicit
Details
The currently available RNG kinds are given below. kind is partially matched to
this list. The default is "Mersenne-Twister".
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
ot;\\begin{figure}\n")
cat("\\includegraphics[width = ", wid, ", height = ", hei, "]{proj1-fig1}\n")
cat("\\end{figure}\n")
@
If I have got any of this wrong, I am sure someone will come along and correct
me.
Hope this is helpful,
Dan
Daniel Nordl
The statement
print((731:1095)[-SEQ])
does not remove the values of SEQ (i.e. 731 732 733 734 735 736) from the
printed sequence, but instead uses SEQ to index into the vector created by
731:1095. But the vector 731:1095 has length 365, so no elements are removed
because the sm
flow going on.
> >
> > The internal L'Ecuyer RNG is a modification of the original one (and I'm
> not
> > familiar with it), but they seem to relax that restriction.
> >
> > Hana
I apologize if this message is posted twice. I am having email problems.
I t
22
[3,] 31 23 21
> matrix(sample(m),nrow=3)
[,1] [,2] [,3]
[1,] 21 23 1
[2,] 31 33 23
[3,]23 22
> matrix(sample(m),nrow=3)
[,1] [,2] [,3]
[1,] 312 21
[2,] 331 22
[3,] 23 233
>
Hope this is helpful,
Dan
Daniel Nordlund
Bot
work:
b <- a[-nrow(a),]
If you haven't already read the manual, "An Introduction to R", that ships with
every copy of R, then now is the time.
hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
R-help@r-project.org ma
value
1/3 ends up being slightly smaller that one-third.
> print(1/3, digits=20)
[1] 0.1483
>
> b <- 1000^(1/3)
> print(b, digits=20)
[1] 9.9982236
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
__
;ll be playing with your code (which is a model of readability, and a
> lesson to me on same, BTW) and keep you posted on my progress.
>
> Warmly, Andrew
>
Andrew,
R-help is not really the venue for discussing SAS programming and how the SAS
data step reads fixed width files. If yo
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