search for: webapp5

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2010 Nov 17
Give me all operator
Hello is there in R any operator that give you all the data of a matrix for example in matlab x(2,3) returns the 2ndth row and 3rdth column x(2,:) returns all the columns of the 2nd row. In R now I would like to print all the  CRagent[[i]][2]  CRagent[[:]][2] doesnot work of course. Other option is to make a loop with an index i that spans from 1:last element of CRagent[[]] but this is not
2010 Dec 01
Problem in reading Excel spreadsheets
Hi folks, Win 7 64bit R 2.12.0 32bit Problem in reading Excel spreadsheets (the text file, research_databaseI.xls, was download on Internet) > data=odbcConnectExcel(file.choose()) > sqlTables(data) TABLE_CAT TABLE_SCHEM 1 C:\\Users\\satimiswin764\\Documents\\research_databaseI <NA> 2
2010 May 28
Using a loop to define new variables
Hi, I'm a novice R user, much more used to SAS. My problem is pretty simple - basically, in a data frame, I have variables named x1,....,x10 and y1,...,y10; and I would like to create r1 = x1 / y1 etc Apologies if this is way too rudimentary - but I couldn't find any posts online which solve this exact issue. Cheers, Andre
2011 Feb 21
assign value to multiple objects with a given ls pattern
Dear R colleagues, This seems pretty straight forward but I have been banging my head on this for some time and can't seem to find a solution suppose I have something like a1<-1; a2<-2; a3<-3; a4<-4; b1<-3; b2<-4 I would like to quickly assign to objects with a certain pattern, e.g., those in ls(pattern="a") a specific value, e.g., "99", without
2010 Mar 07
Some hints for the R beginner
There is now a document called "Some hints for the R beginner" whose purpose is to get people up and running with R as quickly as possible. Direct access to it is: JRR Tolkien wrote a story (sans hobbits) called 'Leaf by Niggle' that has always resonated with me. I offer you an imperfect, incomplete
2010 Nov 17
Extract elements that starts with...
Hi I got an table that contains this colum and i wonder if there is any nice way to extract all rows that contains all dose that start with A02, so all the rows that have A020 and A021 etc etc... | A010 | | A010 | | A010 | | A020 | | A020 | | A020 | | A020 | | A020 | | A021 | | A021 | Thx for your help //Joel -- View this message in context:
2011 Jan 24
crazy loop error.
Dear R-users, This is a loop which is part of a bigger script. I managed to isolate the error in this loop and simplified it to the bare minimum and made it self-contained. a<-c(2,3,4,5,5,5,6,6,6,7) for(n in 1:10) { print(paste("n: ",n)) z1<-a[n] #make a list container ldata<-list() t=1 while(z1==a[n]) { #add dataframes to list ldata[[t]]<-paste("hello") n=n+1
2010 Aug 09
Fwd: RE: pvclust function
...eproducible code. > -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ********** [[alternative HTML version deleted]] [[alternative HTML version deleted]]
2010 Sep 16
Help for an absolutely r-noob
Hello together, I am an absolute noob in R and therefore I need help urgently. I have received a script from my tutor with plot functions in it. However, I can' manage to adapt these plots. The hole script is as follows: setwd("E:/") ##### (1) Read data ### dat <- read.table("Komfort_Tatsaechliche_ID_Versuchsreihe_1.txt", header=TRUE, sep="\t",
2010 Apr 02
plot area: secondary y-axis does not display well
Dear useRs, I'm having a slight problem with plotting on 2 axes. While the following code works alright on screen, the saved output does not turn out as desired i.e. the secondary y-axis does not display fully. Just run the code and look at image output. Suggestions please... thanks, Muhammad --- rm(list=ls()) x <- 1:100 y <- 200:300 par(mar=c(5,5,5,7)+0.1) # inner margin
2011 Jun 23
trying to import xls or xlsx files
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
2010 Apr 09
How to run Shapiro-Wilk test for each grouped variable?
I want to run Shapiro-Wilk test for each variable in my dataset, each grouped by variable groupFactor. I have these working commands: > data.n<-names(data) # put names into a vector called data.n > by(eval(parse(text=(paste("data",data.n[3],sep="$")))), data$factor, shapiro.test) #run shapiro.test but I must to change the variable number manualy. How to automate
2010 Nov 04
Problems with points in plots when importing from pdf to an SVG editor
Dear R-users When trying to import graphics from an pdf-file to a Vector graphics editor (I use Inkscape, but i've confirmed the same problem on adobe products), all points in the graphics turn out as "q"s. This example displays the beaviour: pdf(file="points are weird.pdf") plot(1:5) When importing the file to inkscape, I get five neatly arranged little
2010 Feb 24
problem with looping on sqlSave()
Dear R users, I have a follow-up question on sqlSave(). Since most of the output from the tests I use are lists, I would like to loop to export each element of the list and append it to the sheet. Here is what I do: > library(RODBC) > test <- structure(list(m = structure(c(0.090909090909091, 0.181818181818182, 0.272727272727273, 0.363636363636364, 0.454545454545455,
2010 Dec 07
increase or decrease variable by 1
many languages have shorthands for that operation like: variable += 1 or ++variable is there something like that in R ? -- View this message in context: Sent from the R help mailing list archive at
2010 Sep 22
Unique subsetting question
Hi all, I'm looking at a large data set, and I'm interested in removing rows where only one variable is duplicated. Here's an example: > presidents Qtr1 Qtr2 Qtr3 Qtr4 1945 NA 87 82 75 1946 63 50 43 32 1947 35 60 54 55 1948 36 39 NA NA 1949 69 57 57 51 1950 45 37 46 39 1951 36 24 32 23 1952 25 32 NA 32 1953 59
2011 Jul 05
How to translate string to variable inside a command in an easy way in R
I want to write a function that get 2 strings y and z and does the following R command. temp<-qq1[qq1$z==y,] for example if it get y="AMI" and z="PrimaryConditionGroup" It should do the following temp<-qq1[qq1$PrimaryConditionGroup=="AMI",] I could do it by the following function that is ugly and I wonder if there is an easier way to do it espacielly when temp
2010 Sep 06
WriteXLS problem
Hi R users: I don't know if you have had the following problem trying to export to an "xls" format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even "xlsx" package that take too long and do not finish. The data frame I try to