similar to: Re :

On Sat, May 1, 2010 at 11:02 AM, Mohan L <l.mohanphy@gmail.com> wrote: > Hi All, > > I am new to R and the mailing list. > > I have a data file (.xls) format with little bit large (16 column,35000 > rows ) data file. I am trying to read this file for calculation. I have > converted the file into .csv format and read like that : > > > bmtrend <-

Dear All, I have data some thisng like this : > data <- read.csv(file='ipsample.csv',sep=',' , header=TRUE) > data State Jan Feb Mar Apr May Jun 1 AAA 1 1 0 2 2 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 0 0 0 1 2 1 4 DDD 5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111

Dear All, I have data some thing like this: State Jan Feb Mar Apr May Jun AAA 1 1 0 2 2 0 BBB 1298 1195 1212 1244 1158 845 CCC 0 0 0 1 2 1 DDD 5 11 17 15 10 9 EEE 18 28 27 23 23 16 FFF 68 152 184 135 111 86 I want to sum all the column(Jan, Feb, Mar ...) and have to merge the total at last row. like this: StateJanFebMarAprMayJunAAA110220BBB12981195121212441158845CCC 000121DDD51117

Hi all, I have data like this: >sample <- read.csv(file="sample.csv",sep=",",header=TRUE) > sample stdate Domain sex age Login 1 01/11/09 xxx FeMale 25 2 2 01/11/09 xxx FeMale 35 4 3 01/11/09 xxx Male 18 30 4 01/11/09 xxx Male 31 3 5 02/11/09 xxx Male 32 11 6 02/11/09 xxx Male 31 1 7 02/11/09

Dear All, > avglog 01/11/09 02/11/09 03/11/09 04/11/09 9.750000 4.500000 4.500000 8.666667 > avglog1 <- data.frame(avglog) > avglog1 avglog 01/11/09 9.750000 02/11/09 4.500000 03/11/09 4.500000 04/11/09 8.666667 The first column isnt a column, It's the row names. I makeing a column from the row names by using the following > value1$Day <- rownames(value1) >

Dear All, I have two data like this : $cat main.csv name,id,memory,storage mohan,1,100.20,1.10 ram,1,200,100 kumar,1,400,50 xxx,1,100,40 aaa,1,800,45 mount,1,200,80 > main <- read.csv(file='main.csv',sep=',' , header=TRUE) > main name id memory storage 1 mohan 1 100.2 10 2 ram 1 200.0 100 3 kumar 1 400.0 50 4 xxx 1

Hi All, I have the data like this : >sample <- read.csv(file="sample.csv",sep=",",header=TRUE) > sample stdate Domain sex age Login 1 01/11/09 xxx FeMale 25 2 2 01/11/09 xxx FeMale 35 4 3 01/11/09 xxx Male 18 30 4 01/11/09 xxx Male 31 3 5 02/11/09 xxx Male 32 11 6 02/11/09 xxx Male 31 1 7 02/11/09

Dear All, I have data some thing like this: > sample State Jan Feb Mar A 1 1 1 B 1298 1293 1294 C 0 0 0 D 5 5 5 E 18 18 18 I need to multiply "Jan" column *1000 and divided by the same number, like this : > data[,-(1:2)] * ((data[,2] *1000)/data[ ,2]) I doing some thing

Dear All, I have .csv file it looks like this : rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t' , header=FALSE) > head(rawdata,n=5) Tenant Domain Owner Current State 1 \\N ROOT admin Running 2 \\N ROOT admin Stopped 3 \\N ROOT admin Running 4 \\N

Dear All, I am executing my R script from PHP using the shell_exec. $return = shell_exec("/usr/bin/R --slave --args ".$Domain.",".$Gender." < /var/www/html/trends/newTrend.R"); In newTrend.R, I am plotting graph and save in pdf file using the below code. It will create a pdf file with two line. pdf(file="/var/www/html/trends/bmtrend.pdf", height=7.5,

Hi all, I have data size of : > dim(sample) [1] 35943 17 The first column is "stdate" - is date ( 01/11/2009 00:00:00,02/11/2009 00:00:00,02/11/2009 00:00:00 etc... ) Login is 13th column - is numbers (12,0,1 erc...) The below operation return the following error. > sample1 <- read.csv(file="sample1.csv",sep=",",header=TRUE) > avglog <-

Yes. I tried that already. Not straightforward. data <- read.csv("20_newsgroups.csv",fill=TRUE,as.is=T,header=F, quote="", sep=",", encoding="UTF-8") This line does read it haphazardly. The emails in the column are split into multiple columns and there are several columns with just ?NA?. Totally 202 columns. And then I removed columns with NA?s and

You might want to try some of the suggestions mentioned in this post: https://stackoverflow.com/q/17414776/2140956 Jean On Thu, Aug 10, 2017 at 7:59 AM, <Mohan.Radhakrishnan at cognizant.com> wrote: > Hi, > > Reading http://ssc.wisc.edu/~ahanna/20_newsgroups.csv after downloading > it using > > data <- read.csv("20_newsgroups.csv",header=TRUE) > >

Hi, Reading http://ssc.wisc.edu/~ahanna/20_newsgroups.csv after downloading it using data <- read.csv("20_newsgroups.csv",header=TRUE) throws this. Warning message: In scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : EOF within quoted string So, for example, the first line in the file is this. This column contains only such text. Is there a way read it ?

Dear All, I have a data frame with 5 column and 201 row data. I want to add one more column between column 1 and 2 with value of 1. So the new column has to be the second column filled with 1. Any help will be appreciated. Thanks for your time. Thanks & Rg Mohan L

Dear All, I have a data frame like this: name ip Bsent Breceived a 1 0.00 0.00 a 2 1.43 19.83 a 1 0.00 0.00 a 2 1.00 1.00 b 1 0.00 2.00 b 3 0.00 2.00 b 2 2.00 0.00 b 2 2.00 0.00 b 1 24.40 22.72 c

Hi All, I have txt file like : $ cat data.txt US 10 UK 12 Ind 4 Germany 14 France 8 > rawdata <- read.table(file='data.txt',sep='\t' , header=FALSE) > rawdata V1 V2 1 US 10 2 UK 12 3 Ind 4 4 Germany 14 5 France 8 I want to draw pie chart for the above data. How to split rawdata into : con <-

Hi everyone I am trying to use grep in a for loop to compare a string value. It works if I use the actual index value but when I use the for loop index, it doesn't work. Any suggestions plz. Here is the code: data <- read.table(file="Sigmoid.csv", head=FALSE, sep=","); c1 <- data$V1 c2 <- data$V2 c3 <- data$V3 c1data <- data.frame(c1); c2data <-

Dear All, I have a data frame with State and 12 Months as column. I want to convert all the 12 month column from factor to numeric. any help will be greatly appreciated. > str(data) 'data.frame': 33 obs. of 9 variables: $ State : Factor w/ 33 levels "Andaman and Nicobar Islands",..: 1 2 3 4 5 6 7 8 9 10 ... $ April : Factor w/ 21 levels

Dear All, I have a data frame "data" with Jan , Feb as column. I using like this data$Jan to take Jan column. I have the Jan in another variable var1 <- "Jan" data$var1 I need to replace the variable value there . Any help will be greatly appreciated. Thanks & Rg Mohan L [[alternative HTML version deleted]]