similar to: How to increase decimal places

Dear All, I'm trying to read a text data file that contains several records separated by a blank line. Each record starts with a row that contains it's ID and the number of rows for the records (two columns), then the data table itself, e.g. 123 5 89.1791 1.1024 90.5735 1.1024 92.5666 1.1024 95.0725 1.1024 101.2070 1.1024 321 3 60.1601 1.1024 64.8023 1.1024 70.0593

Hello All, Is there any R function out there to turn a multi-way contingency table back to a flat file table of individual rows and attribute columns.? Thanks! marco --------------------------------- [[alternative HTML version deleted]]

Hi all, I am doing some clustering and the clustered results are presented in a "pairs" plot showing 4 clusters... I made the data points belong to 4 different clusters displaying different colors. Now I want to select the best clustered class, how can I click on the data point, and the program returns the index of that cluster(its class number, or color number)? Also, if I would

Hi, I have a string vector, say, x <- "a", "aab" and anther string vector, say, y <- "a", "aa", "aab", "aabc". Is there any R function to get the indices of y for the elements of x, that is, foo(x, y) will give me the index vector c(1, 3)? I know i can combine apply() and grep("^aab$", y) to do it. But is there any

Hello, I have two dataframes of the following structures: str(a) `data.frame': 1354896 obs. of 14 variables: $ V1 : int 0 1 2 3 4 5 6 7 8 9 ... $ V2 : int 0 0 0 0 0 0 0 0 0 0 ... $ V3 : int 74 12305 103 12337 46 57 12446 90 12097 79 ... $ V4 : num 11.8 1529.2 17.8 1579.4 6.7 ... $ V5 : int 88 11040 104 11557 56 58 11040 74 10991 81 ... $ V6 : num 15.5 1921.3 20.3

Hi everyone, I want to highlight something in a plot. So I want to write a text with a yellow background. I tried to make use of text(x,y,"hallo",bg="yellow") but that does not work. I know I am a handful. Sorry! Maja! --

Hello. In the file list.txt, I have the name of n files in data frame format. I want to make an image for each file and save the images in pdf format. To do this, I do the following: llista = scan(file="list.txt",what=list(nom="")) for (file.name in llista[[1]]){ aux=read.table(file=file.name) aux=as.matrix(aux) pdf() image(aux)

I am ashamed to be asking this question, but I couldn't find the solution anywhere. Searching for "if" and "R" is not very productive... I cannot get a simple if statement to work. I have data on college students. I want to make a string variable that has the names of the years. That is, when the year variable i is equal to 1, I want to have a variable called years

Hi there is it possible that pdfs generated using the pdf() function with default settings leads to loss of information? I was plotting copy number changes from Agilent 180k data in form of rectangles (rect()) while each rectangle represents one region of copy number change. When plotting into a pdf I noticed that some very small rectangles do not appear (even after extensive zooming) in the pdf

Dear friends, suppose i want to do the following caulation for 100 times, how to put the results of x , y and z into the same dataframe/dataset? x<-runif(1) y<-x+1 z<-x+y thanks in advance! -- Kind Regards, Zhi Jie,Zhang ,PHD Department of Epidemiology School of Public Health Fudan University Tel:86-21-54237149 [[alternative HTML version deleted]]

Hello everyone, I've a problem and dont know how to solve. This is my first posting and it would be fantastic if you could help me. I want to divide n objects in k classes and need an output with all (n+1)(n+2)/2 possibilities. For example n=4, k=3: That would be: 4 0 0 3 1 0 3 0 1 2 2 0 2 1 1 2 0 2 1 3 0 1 2 1 1 1 2 1 0 3 0 3 1 0 2 2 0 1 3

Hello everybody, I've a question about "autoregressive Regressionmodels". Let Y[1],.....,Y[n], be a time series. Given the model: Y[t] = phi[1]*Y[t-1] + phi[2]*Y[t-1] + ... + phi[p]*Y[t-p] + x_t^T*beta + u_t, where x_t=(x[1t],x[2t],....x[mt]) and beta=(beta[1],...,beta[m]) and u_t~(0,1) I want to estimate the coefficients phi and beta. Are in R any functions or packages for

Hello everybody, I'm trying to predict a linear regression model but it does not work. My Model: y = Worktime + Vacation + Illnes + Bankholidays My modelmatrix is of dimension 28x4 Then I want to make use of the function predict because there confidence.intervals are include. My idea was: mod <- lm(y~Worktime+Vacation+Illnes+Bankholidays) newdate=data.frame(x=c(324,123,0.9,0.1))

Are there ways to load a csv file without row.names by read.csv? Thanks. [[alternative HTML version deleted]]

Dear mailing list, may some one be kind to help me solve following problem. I am trying to write a code that will combine two tables "x" and "y". The first columns of both tables are unique identification for the rows. The first column of table "X" is a sub set of the first column of "Y". I need to find the matching rows in both tables by looking on their

Hi! I'd like to plot things with axes going from the highest to the lowest value, so that e.g. high values on the xaxis are plotted to the left and low values to the right. Could anyone tell me how this is done? I couldn't find anything in the documentation. Thanks a lot! Silli

Hi All, I want to use "hist" to get the relative frequency plot. But the range of ylab is greater than 1,which I think it should be less than 1 since it stands for the probability. Here is my code: x<-c(1,1,1,0,0,1,1,5,1,1,0,1,1,0,1,0,1,1,0,0,1,1,1,1,1,1,1,1,1,4,1,0,2,1,1,1 ,1,1,0,0,1,0,1,1,1,1,1,1,1,1,1)

Hello, this is probably trivial but I failed to find this particular snippet of code. What I got: my_dataframe (contains say a 40k rows and 4 columns) distances (vector with euclidean distances between a query vector and each of the rows of my_dataframe) What I do: after scaling data my_dataframe I calculate distances. order them then extract top five hits my_dataframe <-

Folks, I have got a strange behaviour when testing this: sum(x) != 1 let us set x<-c(70,134,1,5,0) and transform it in a vector of probabilities x<-x/sum(x) One expect sum(x) should be equal to 1, which is apparently the case > sum(x) [1] 1 However, when I try to test it I get: > if(sum(x) !=1) print("lost") else ("OK") [1] "lost" Which means

Dear R users, My situation: (1) I have limited workspace for my work harddisk (about 10 GiB). (2) I have a lot of data files in R workspace (*.RData) which most of them > 200 MiB. For some reason I zip some of them, for instance "filename.RData (250 MiB)" to "filename.zip (3MiB)". In this work I have a lot of more space of my harddisk. Normally, If I want to use