similar to: Subsets without NA

I'm trying to plot observations against observation dates and getting julian dates along the x-axis: library(date) Week<-as.date(c("05/02/03","05/09/03","05/16/03","05/23/03","05/30/03","06/0 7/03","06/14/03")) Leafminers<-c(0,2,5,10,4,6,5) Diglyphus<-c(0,0,4,5,7,3,1) LeafDig<-cbind(Week,Leafminers,Diglyphus)

I have two problems where I've come up with some code that will do the analysis that I want, but it looks pretty clumsy. In the first case, I calculate the variance on five different columns for each of 14 clusters and get them into one matrix. I get the job done, but I would have thought that it could be done in one or two lines, not six, and be generalized so that it didn't matter how

Is there a way of extracting mean distance or dissimilarity for a given strata from a 'dist' or 'dissimilarity' object, e.g. extract mean distances for each species in Anderson's iris data? data(iris) iris.dist<-dist(iris[,1:4]) then what? Mikkel Grum, PhD Genetic Diversity Scientist International Plant Genetic Resources Institute (IPGRI) Sub-Saharan Africa Group ***

I'm trying to plot four different sorghum types on a plot using a different colour/symbol combination for each sorghum type (TYPBOTA in the script below). What am I doing wrong?: plot(xx$LATITUDE,xx$SFD1, points(xx$SFD1,xx$LATITUDE, col=1:4[codes(xx$TYPBOTA)],pch=c(4,3,1,2)[codes(xx$TYPBOTA)]) grid() I get the following error message Error in 1:4[codes(xx$TYPBOTA)] : NA/NaN argument

Can anyone tell me what's wrong with this command?? xx<-subset(x,LOCAL.NAME==c("Chifumbata","Chikota"),select=c(13,16,19,23,26,2 9,30,33:48)) Warning message: longer object length is not a multiple of shorter object length in: LOCAL.NAME == c("Chifumbata", "Chikota") both of the following commands work fine

Any idea why clusters five and six fall off the scale in the following? library(cluster) library(maptree) data(flower) dfl2<-daisy(flower,type=list(asymm=c(1,3),ordratio=7)) hdfl2<-hclust(dfl2) prune.dfl2<-prune.clust(hdfl2,k=6) plot(prune.dfl2) Of course the following will display all six clusters, but without a scale: draw.tree(prune.dfl2) Best wishes, Mikkel Mikkel Grum

This message is in MIME format. Since your mail reader does not understand this format, some or all of this message may not be legible. ------_=_NextPart_000_01C35A53.75780090 Content-Type: text/plain; charset="iso-8859-1" Anyone have a clue why hclust() and agnes() produce different results in the example below when both use method="average"?? I'm not able to reproduce

First question: I have an Excel file with both character and numeric variables that I want to import to a data.frame, so I've saved the data to a tab-delimited text file. read.table : * Won't identify the mode of my character variables, as some of them contain only numbers and none of them are surrounded by quotes, * Won't allow me to specify the mode of my variables, * Doesn't

Is there any function in R like is.not.found(x, y) meaning if you can't find object x, then use object y?? Mikkel Grum

I run a batch file with the following command in Windows XP: C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore <C:\users\me\file.R> C:\users\me\file.out 2>&1 Is there any way to get only the output of R in file.out, without getting all the code from file.R too? Any help greatly appreciated, Mikkel

Hi Can anyone explain why a date becomes numeric when you loop over a series of dates? > dt <- Sys.Date() > dt [1] "2017-11-14" > class(dt) [1] "Date" > dts <- dt - 1:0 > class(dts) [1] "Date" > > for (i in dts) { + print(i) + print(class(i)) + print(as.Date(i, "1970-01-01")) + print(class(as.Date(i,

I want to do something like df[df$b %in% df2[i, 2], ] where df$b is a numeric vector and df2[i, 2] is a factor with levels like "7:9, 12:14". For example: a <- c(paste("A", 1:10, sep = ""), paste("B", 1:10, sep = "")) b <- 1:20 df <- as.data.frame(cbind(a, b)) df$b <- as.numeric(levels(df$b))[as.integer(df$b)] f <-

What am I doing wrong, or is the \r that I'm getting in the example below a bug? > a <- (1:10) > b <- (LETTERS[1:10]) > df <- as.data.frame(cbind(a, b)) > > df a b 1 1 A 2 2 B 3 3 C 4 4 D 5 5 E 6 6 F 7 7 G 8 8 H 9 9 I 10 10 J > library(RSQLite) > drv <- dbDriver("SQLite") > con <- dbConnect(drv, dbname = "Test")

What am I doing wrong, or is the \r that I'm getting in the example below a bug? > a <- (1:10) > b <- (LETTERS[1:10]) > df <- as.data.frame(cbind(a, b)) > > df a b 1 1 A 2 2 B 3 3 C 4 4 D 5 5 E 6 6 F 7 7 G 8 8 H 9 9 I 10 10 J > library(RSQLite) > drv <- dbDriver("SQLite") > con <- dbConnect(drv, dbname = "Test")

I have a situation where I'm filling out a dataframe from a database. Sometimes the database query doesn't get anything, so I end up trying to place NULL in the dataframe like below. > temp <- NULL > xmat <- as.data.frame(matrix(NA, 2, 3)) > xmat[1, 2:3] <- temp Error in if (m < n * p && (n * p)%%m) stop(gettextf("replacement has %d items, need %d",

Dear useRs, I'm writing regular survey reports using Sweave. Each report has several sections along the lines of: \section*{Disease X} <<MapX,fig=TRUE,echo=FALSE>>= image(vectorx,vectory,matrixz) @ Notes with or without Sexpr{a}. \vfill \pagebreak \section*{Disease Y} <<MapY,fig=TRUE,echo=FALSE>>= ...etc. Often one or more of the diseases is not observed (all

Is the following expected behaviour for a date used in an ifelse function? > date <- Sys.Date() > date [1] "2007-12-30" > ifelse(TRUE, date-1, date) [1] 13876 > ifelse(FALSE, date-1, date) [1] 13877 > ifelse(TRUE, as.character(date-1), date) [1] "2007-12-29" > if (TRUE) {date} [1] "2007-12-30" It would seem more natural to me if a date produced

How do I get output with "\color{blue}", i.e. with only one backslash??? > "\color{blue}" [1] "color{blue}" Warning messages: 1: '\c' is an unrecognized escape in a character string 2: unrecognized escape removed from "\color{blue}" > "\\color{blue}" [1] "\\color{blue}" > Any help greatly appreciated. Best regards,

Dear R-helpers, When I use sqlQuery in the package RODBC, I cannot break the line, but have to write the entire SQL Query on the same line. Is this expected behaviour? It is definitely workable, but makes the queries a slightly difficult to read and edit. I'm using R 1.9.1 and RODBC 1.0-4 on Windows Server 2003 and querying a Sybase database. Best wishes, Mikkel

Is there any way to pause R for a gvien time period, i.e. without the need for user intervention? I've got a loop that I want to have work as hard as it can as long as there is data coming in, but when there is no new data, I would like it to pause before checking again. Something like in the following construction: Set up environment: load libraries, establish database connections, etc.