similar to: if confusion

Why does: (shape/scale) * (1:365/scale)^(shape - 1) return a vector of numbers but calling a function hasard(1:365,shape,scale) defined like: hazard <- function(x,shape,scale) { return (shape/scale) * (x/scale)^(shape - 1) } Only return a single value? It is like x becomes a single value passed as an argument. Thank you. Kevin

Hi, is there a simpler way to count the number of elements in a vector that are not NA than this: countN <- function (v) { return (Reduce(function (x, y) x + y, ifelse(is.na(v), 0, 1))) } ? - Godmar

Dear all, I'm a relative new user of R and I have a problem with merging a collection of matrices. All matrices in this collection have the same dimension (532 rows and 532 columns), but can differ in the row and columns names. I'd like to merge these matrices in such a way that the resulting matrix only contains unique row- and column-names and that in case of a match between the row or

Dear R People: I am looking at the ctime attribute of two different files. It contains the year, month, day, time of creation and time zone. Is there a way to determine the difference between the ctimes of two files, please? I looked in the chron package and nothing seemed to work. Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and

Dear R People: I have several vectors, sa1, sa2,...sa27 of varying lengths. I want to produce one vector xener[1:27] which has the minimum of each sa[i]. I'm trying to set up a loop and use the assign statement, but here are my results: > for(i in 1:27) { + xener[i] <- min(assign(paste("sa",i,sep=""))) + } Error in assign(paste("sa", i, sep =

I've created a short program to print a table of learning curve factors. However, I cannot figure out how to format the table to: 1) Get rid of the [1]s in the first column and replace it with the values of N. 2) Line up the first row with the factors (decimal fractions). Thanks for any help. The complete program and output is as follows: > Lc<-seq(0.70,0.95,0.05) #Specify learning

I'm looking for something along the lines of which ( table ( x ) == max ( table ( x ) ) ) to find the most common level of one factor by several other factors. For instance, I've got > X <- data.frame ( + x = factor ( sample ( c ( "A" , "B" , "C" , "D" ) , 20 , r = T ) ) + , z1 = factor ( sample ( c ( "Before" ,

I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a "blank sheet". I would like to "plot" a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]]

Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat<-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4,

Given a Date object or simply a year is there an R function to tell me if the it is a leap year or not? I was hoping for something like 'is.leapyear'. I probably can build my own function (year divisible by 4 etc.) but I would rather use an existing function if it is available. Thank you. Kevin

I'm trying to install the Hmisc package, however it doesn't appear in the list after "utils:::menuInstallPkgs()". Any help? -- David

Hi I'm new to R and would like to implement a SAS-like macro variable in R. What I'd like to do is take the simple R code below and change the "=TEF" to different letters to refer to different companies' data for download. # DOWNLOADS FILES FROM YAHOO INTERNET

I have two dataframes, the first column of each dataframe is a unique id number (the rest of the columns are data variables). I would like to figure out how many times each id number appears in each dataframe. So far I can use: length( match (dataframeA$unique.id[1], dataframeB$unique.id) ) but this only works on each row of dataframe A one-at-a-time. I would like to do this for all of

I noticed that when I fit a linear model using 'lm' there is an attribute called "factors" that is added to the "term". It doesn't seem to appear for 'model.matrix', just 'lm'. I have been unable to find where it gets constructed or what it means? It looks like a two dimensional array that I may be able to use so I would just like to get some

I had a question about the basic power functions in R. For example from the R console I enter: -1 ^ 2 [1] -1 but also -1^3 [1] -1 -0.1^2 [1] -0.01 Normally pow(-1, 2) return either -Infinity or NaN. Has R taken over the math functions? If so I would think that -1^2 is 1 not -1 and -0.1^2 is 0.01 not -0.01. Thank you. Kevin

Sometimes I write code like this: > qf.a <- subset(qf, pubid %in% c(104, 106, 107, 108)) > qf.b <- subset(qf, !pubid %in% c(104, 106, 107, 108)) and I get a little worried that maybe I've remembered the precedence rules wrong, so I change it to > qf.a <- subset(qf, pubid %in% c(104, 106, 107, 108)) > qf.b <- subset(qf, !(pubid %in% c(104, 106, 107, 108))) and pretty

I am trying to upgrade to R 2.14 from R 2.13.1 I have compied all the libraries from the 'library' directory in my existing installation (2.13.1) to the installed R 2.14. Now I want to uninstall the old installation (R 2.13.1) and I get the error: Internal Error: Cannot find utCompiledCode record for this version of the uninstaller. Any ideas? Kevin [[alternative HTML

Given a dataframe like > data ID Type N 1 45900 A 1 2 45900 B 2 3 45900 C 3 4 45900 D 4 5 45900 E 5 6 45900 F 6 7 45900 I 7 8 49270 A 1 9 49270 B 2 10 49270 E 3 18 46550 A 1 19 46550 B 2 20 46550 C 3 21 46550 D 4 22 46550 E 5 23 46550 F 6 24 46550 I 7 > containing an identifier (ID), a

I am running R 2.9.2 and creating a PDF that I am trying to open with Adobe Reader 9.2 but when I try to open it the reader responds with "There was an error opening this document. The file is damaged and cannot be repaired.: I am using the R command(s): pdf(file="cat.pdf", title="Historical Sales By Category") for(j in 1:length(master)) { d <-

This may be a begining question. If so, please bear with me. If I have some data that based on the historgram and other plots it "looks" like a beta distribution. Is there a function or functions within R to help me determine the model parameters for such a distirbution? Similarily for other "common" distirbutions, Poisson(lambda), Chi-Square(degrees of freedom, chi-square