similar to: Problem with order() and I()

Here's an example (session info at the end). > tmpv <- c('\265g/L','Bq/L') > order(tmpv) [1] 2 1 > tmpv <- I(tmpv) > order(tmpv) Error in if (xi > xj) 1L else -1L : missing value where TRUE/FALSE needed > foov <- gsub('\265','',tmpv) > order(foov) [1] 2 1 > str(tmpv) Class 'AsIs' chr [1:2] "\265g/L"

You are right that there are no NAs in the practice data. But there are NAs in the moving average data. To see this, break your work into two separate steps, like this: tnr.ma <- ma(dat3[1:28], order=3) TNR_moving_average <- forecast(tnr.ma, h=8) I think you will find that the warning comes from the second step. Print tnr.ma and you will see some NAs. -Don -- Don MacQueen Lawrence

Hi Don, wow, you are so right. I picked that piece up from the bloggers tutorial and since I am R naive yet, I thought it was all one step moving_average = forecast(ma(tdat[1:31], order=2), h=5) Truly, I usually print and check at every step I can, as painful as it is sometimes. Great lesson for this novice usR. So the first and last values are NA in each case? Do you know why? Should I replace

Dear list-reader, by running the following script: library(zoo) sessionInfo() search() packageDescription("zoo") data(EuStockMarkets) dax <- as.zoo(EuStockMarkets[1:10, "DAX"]) daxr <- diff(log(dax)) identical(as.vector(qnorm(daxr)), qnorm(coredata(daxr))) qqnorm(coredata(daxr)) qqnorm(daxr) qqnorm() produces an error: > qqnorm(daxr) Fehler in if (xi == xj) 0L

Luca, If speed is important, you might improve performance by making d0 into a true matrix, rather than a data frame (assuming d0 is indeed a data frame at this point). Although data frames may look like matrices, they aren?t, and they have some overhead that matrices don?t. I don?t think you would be able to use the [[nm]] syntax with a matrix, but [ , nm] should work, provided the matrix has

Is there a package or a command that does window aggregation like select sum(col1) over (partition by col2, col3 order by col4 rows between unbounded preceding and current row) as sum1 from table1 ; the above is Netezza syntax, but Postgre has same capability. Stephen B [[alternative HTML version deleted]]

Thanks. I thought lm() function is for linear model, such as the correlation below: Y= aX + b On Wed, Jun 14, 2017 at 5:25 PM, MacQueen, Don <macqueen1 at llnl.gov> wrote: > Start with the lm() function; i.e., see > > ?lm > > -Don > > -- > Don MacQueen > > Lawrence Livermore National Laboratory > 7000 East Ave., L-627 > Livermore, CA 94550 >

Hello Don, thank you for your response. I appreciate your help. I am using the forecast package, originally I found it following a forecasting example on bloggers.com https://www.r-bloggers.com/time-series-analysis-using-r-forecast-package/ And subsequently located the complete pdf https://cran.r-project.org/web/packages/forecast/forecast.pdf Since I created this practice data using the

My guess would be that if you inspect the output from ma(dat3[1:28], order=3) you will find some NAs in it. And then forecast() doesn't like NAs. But I can't check, because I can't find the ma() and forecast() functions. I assume they come from some package you installed; it would be helpful to say which package. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000

Thank you for both replies Don & Rui, The very issue here is that there is a search that needs to be done within a text field and I agree with Rui later comment that regexpr might indeed be the time consuming piece of code. I might try to optimise this piece of code later on, but for the time being I am working on the following part of building a neural network to try indeed classifying some

Thanks for your replies. I tried the regression, but then got a NA value for the slope. And here is the error message: Coefficients: (1 not defined because of singularities) On Thu, Jun 15, 2017 at 12:20 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote: > Hi > > But X can be some function like - sin, cos, log, exp... > > Cheers > Petr > > > -----Original

I forgot to explain why my suggestion. The logical condition returns FALSE/TRUE that in R are coded as 0/1. So all you have to do is coerce to integer. This works because the ifelse will return a 1 or a 0 depending on the condition. Meaning exactly the same values. And is more efficient since ifelse creates both vectors, the true part and the false part, and then indexes those vectors in

Hi Rui Thank you for your suggestion, I have tested the code suggested by you against that supplied by Don in terms of timing and results are very much aligned: to populate a 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the one with ifelse employed 80 secs, hence unfortunately not really any significant time saved there. Nevertheless thank you for your contribution.

Conversion of a data frame to a matrix using as.matrix() when a column of the data frame is POSIXt and all other columns are numeric has changed in R 1.9.0 from R 1.8.1. The new behavior issues a warning message and converts to a character matrix. In R 1.8.1, such an object was converted to a numeric matrix. Here is an example. #### R 1.9.0 #### > foo <- data.frame(

What Dave said, plus here's a hint. Try this example (which uses base graphics): plot(1:5) plot(1:5, cex.lab=2) Then look at the help page for par help('par') or ?par to search for other graphics parameters (base graphics) you can use to change various things. Success will depend, as Dave indicated, on how the package author handled the plotting options in rsFit(). -Don --

I use the method, df$Time = as.POSIXct(df$Time), but it has the warning message: Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format On Thu, Dec 14, 2017 at 1:31 PM, MacQueen, Don <macqueen1 at llnl.gov> wrote: > In addition to which, I would recommend > > df <- read.table("DATAM", header = TRUE, fill = TRUE, >

Dear useRs, i want to split a matrix having 1116rows and 12 columns. i want to split that matrix into 36 small matrices each having 12 columns and 31 rows. The big matrix should be splitted row wise. which means that the first small matrix should copy values which are in first 31 rows and 12 columns of the big matrix. similarly 2nd small matrix should contain values from 32nd to 63rd row of the

Using Ulrik?s example data (and assuming I understand what is wanted), here is what I would do: ex.dat <- c("FName: fname1", "Fval: Fval1.name1", "Fval: ", "FName: fname2", "Fval: Fval2.name2", "FName: fname3") tst <- data.frame(x = ex.dat, stringsAsFactors=FALSE) sp <- strsplit(tst$x, ':', fixed=TRUE) chk <-

Um, maybe just dat1$C <- match(dat1$B, unique(dat1$B)) Indexing 1:k with numbers between 1 and k is a bit of a no-op... AFAICT, this even works without stringsAsFactors=FALSE -pd > On 11 May 2018, at 21:30 , MacQueen, Don <macqueen1 at llnl.gov> wrote: > > dat1$C <- seq(length(unique(dat1$B)))[ match( dat1$B, unique(dat1$B) )] -- Peter Dalgaard, Professor, Center for

Start with the lm() function; i.e., see ?lm -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 6/14/17, 3:40 PM, "R-help on behalf of lily li" <r-help-bounces at r-project.org on behalf of chocold12 at gmail.com> wrote: Hi R users, I have some data points (Xi, Yi), and they may follow such a