similar to: Trying to learn how to write a function... can't define a variable??

library(xlsReadWrite) mydata<-read.xls("file path", header=TRUE) however if I change xls to csv it works just fine. Any ideas what I'm doing wrong? I have have also using the package gdata with the exact same error. Below is the error that pops up. Error in findPerl(verbose = verbose) : perl executable not found. Use perl= argument to specify the correct path. Error in

Can someone verify for me if the for loop below is really calculating the nonzero min for each row of a matrix? I have a bug somewhere in the is section of code. My first guess is how I am find the the nonzero min of each row of my matrix. The overall idea is to make sure I am investing all of my money, i.e. new.set is a set of indicator variables for each stock for a particular portfolio, i.e.

General problem: I have 20 projects that can be invested in and I need to decide which combinations meet a certain set of standards. The total possible combinations comes out to 2^20. However I know for a fact that the number of projects must be greater than 5 and less than 13. So far the the code below is the best I can come up with for iteratively creating a set to check against my set of

I wish to create a matrix of all possible percentages with two decimal place percision. I then want each row to sum to 100%. I started with the code below with the intent to then subset the data based on the row sum. This works great for 2 or 3 columns, but if I try 4 or more columns the number of rows become to large. I would like to find a way to break it down into some kind of for loop, so

I am just learning to use qplot and can't get the fill/colour to work. Below is the R code for a scatter plot and bar graph. library(ggplot2) x<-c(1,2,3,4,5,6,7) y<-c(1,2,3,2,5,6,3) qplot(x,y, main="Scatter Plot Test", xlab="X Label Test", ylab="Y Label Test", colour="blue")z<-c("van", "van", "van",

I have 4 columns and 56 rows of made up data that I want to plot as a series of bar graphs. The idea is to create one bar graph for each of the 4 columns using a for loop. I tried the following command in RStudio and when I type x in the console I get just the 4th graph instead of all four graphs. I did not define what x is before hand. I was not sure what it would be. Any suggestions on how you

I have several thousand rows of shipment data imported into R as a data frame, with two columns of particular interest, col 1 is the entry date, and col 2 is the tracking number (colname is REQ.NR). Tracking numbers should be unique but on occassion aren't because they get entered more than once. This creates two or more rows of with the same tracking number but different dates. I wrote a for

Hi, I want to know if it's possible to pass a vector v=c(x,y,...) to a function f(x,y,...) so that each vector element corresponds to a formal argument of the function. For python programmers: f(*v). Specifically, what I'm trying to achieve is: given a list of coordinates l=list(c(x1,y1,z1), c(x2,y2,z2),...) I would like to obtain the corresponding elements in some array A (3-dim in this

Hi! Yesterday I accidentally discovered this: > a <- LETTERS[1:5] > a [1] "A" "B" "C" "D" "E" > > a[1] <- factor(a[1]) > a [1] "1" "B" "C" "D" "E" BUT: > b <- factor(LETTERS[1:5]) > b [1] A B C D E Levels: A B C D E > b[1] <- factor(b[1]) > b [1] A B C D E

Hello Everyone, My name is Thomas and I have been using R for one week. I recently found your site and have been able to search the archives of posts. This has given me some great information that has allowed me to craft an initial design to an inquiry I would like to make into the breakdown of McNemar's test. I have read an intro to R manual and the posting guides and hope I am not violating

Good day, For lack of a better solution (or perhaps I am ignorant to something more elegant), I have been bootstrapping panel data by hand so to speak and I would like to know if there is a way to define multiple variables in a loop using the loop variable. I found a post (here: https://stat.ethz.ch/pipermail/r-help/2002-October/026305.html ) that discussed naming multiple variables but it

Hello, I would like to take a data set of doctors, and recategorize them into numbers. For example, John Doe would be assigned as 1 Jane Doe would be assigned as 2 Does anyone know how to do this in R? It seems quite easy, yet I cannot find the code. Thank you, Michael Leitson Research Intern MSAS Candidate Michael.Leitson@wellstar.org

Following the beginning tutorial, I typed help.start() and was given a choice of browsers. I picked chrome but got back that chrome is not found. I cannot seem to change or get rid of it now. I looked at /etc/R/Renviron but don't see anywhere to set browser (R_BROWSER = ${R_BROWSER ...etc) - nothing about chrome. I tried starting as "R_BROWSER= R" - same thing. How do I change the

I took some data from an online poll about which R GUI people used most and I am messing around with it to learn how to use qplot. Specifically I am making a horizontal bar graph and I have two questions. 1. The categories are ordered in rather strange way at least to me. It is not alphabetical or ascending/descending order of votes cast so i had to manually state the order I wanted which is

General goal: Write R code to find the inverse matrix of an nxn positive definite symmetric matrix. Use solve() to verify your code works. Started with a 3x3 matrix example to build the code, but something dosen't seem to be working. I just don't know where I am going wrong. ##Example matrix I found online A<-c(4,1,-1,1,2,1,-1,1,2) m<-matrix(A,nrow=3,ncol=3) ##Caculate the eigen

Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the "subset" in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs <- dead==FALSE & recTreat==FALSE which

Hello, This is a beginner question. I am trying to loop through numbered variables with "apply" to calculate weighted means. My data is "data", the variables are "var1" to "var100", the weight is "weight". The command works using sapply(paste('data$var', 1:100, sep=''), function(x) weighted.mean(eval(parse(text=x)), data$weight))

I'm used to run that on Matlab >> for indice=1:23, V=vector(1:indice) end that give me 23 vectors as output. Why the same command on R " for ( i in 23:0 ) { V <- vector [ 1 : i ] } " returns one only vector? How can I obtain the 23 vectors that I need ? Thank you. -- View this message in context:

I have columns of data in Excel 2007, A2:A196, B2:B196...ET2:ET196 that I would like to place into arrays in R. I have been trying to write a macro that would automatically create all of my arrays for me with a array names coming from the cells A1, B1, etc. I can manually create an array using RExcel<Put R Var< Array, but I don't want to manually create all 150 arrays this way. Thanks

Dear friends, Being very new to this, I was wondering if I could get some pointers and guidance to interpreting the results of performing a linear regression with ordinal predictors in R. Here is a simple, toy example: y <- c(-0.11, -0.49, -1.10, 0.08, 0.31, -1.21, -0.05, -0.40, -0.01, -0.12, 0.55, 1.34, 1.00, -0.31, -0.73, -1.68, 0.38, 1.22, -1.11, -0.20) x <-