similar to: make a column from the row names

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 an array some thing like this: > avglog January February March April May June July August September 60102 83397 56774 48785 49010 40572 38175 47037 51402 The class of "avglog" array. > class(avglog) [1] "array" > str(avglog) num [1:9(1d)] 60102 83397 56774 48785 49010 ... - attr(*,

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 <-

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 <-

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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 <- read.csv(file="simple.csv",head=TRUE,sep=",") > bmtrend will display something in the middle of display [

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 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

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

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

How to read xml in UTF-8 format. I have one XML file but i want to read it in UTF-8 format. How this is possible? <barplot> <cd> <name>?</name> <value1>3</value1> <value2>2.9</value2> </cd> <cd> <name>?</name>

Thanks and good point about unspecified behavior. The way it behaves now (when it doesn't ignore) is more consistent with data.frame() though so I prefer that to a "warn and ignore" behaviour: data.frame(a = 1, b = 2, 3) #> a b X3 #> 1 1 2 3 data.frame(a = 1, 2, 3) #> a X2 X3 #> 1 1 2 3 (and in general warnings make for unpleasant debugging so I prefer

I am writing a simple R program to execute a t-test repeatedly on data contained in a data frame. My data looks like this: Category Value1 Value2 1 .5 .8 1 .3 .9 . . . . . . . . . 2

This fixes Github issue #2 reported by "kupiakos". <https://github.com/libguestfs/hivex/issues/2> --- generator/generator.ml | 2 +- python/t/210-setvalue.py | 26 +++++++++++++++++--------- 2 files changed, 18 insertions(+), 10 deletions(-) diff --git a/generator/generator.ml b/generator/generator.ml index 908c5f3..02aaf12 100755 --- a/generator/generator.ml +++

Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars<-c("var1","var2","var3") I'd like my aggregate formula (its "cbind" part) to be able to use my "myvars" object. Is it

Dear R helpers I have one data as given below. date                     value1          value2             value3 30-Nov-2010           100                 40                 61 25-Nov-2010           108                 31                 88 14-Sep-2010            11                 180               56 I want the following output date                 name       amount 30-Nov-2010      value1

Hi all, I am trying to replace all uses of a value1 *inside of a given function* to use value2. My strategy was to iterate all the basic blocks, again iterating all the instructions in each basic block, and using replaceUsesOfWith(value1, value2) for every instruction. This used to work all right, but now I am finding some problems. There are instructions like this: store i32 0, i32*

Dear experts, I am trying to create a data frame from the residuals I get after having applied a linear regression to each column of a data frame, but I don't know how to create this data frame from the resulting list since the list has differing numbers of rows. So for example: age<- c(5,6,10,14,16,NA,18) value1<- c(30,70,40,50,NA,NA,NA) value2<- c(2,4,1,4,4,4,4) df<-

Dear list members, I would like to create two series of plotted mean values and error bars, yet with different locations along the x-axis. Plotting of first series using plotCI with the standard arguments goes without any problem. However I do not succeed to add the second series in the same plot, which should be horizontally shifted from the first series along the x-axis. The “add=TRUE”

Hi, I have a data fram like this: date column1 column2 column3 value1 value2 value3 1-1 A B C 10 5 2 2-1 A B D 5 2 0 3-1 A B E 17 10 7 How can I reshape it to: date column1 column2 column3 v x 1-1 A B C value1 10 1-1 A B C value2 5 1-1 A B C value3 2 2-1 A B D value1 5 2-1 A B D value2 2 2-1 A B D value3 0 3-1 A B E value1 17 3-1 A B E value2 10 3-1 A B E value3 7 Thx! Regards, Richard