similar to: How to modify a column of a matrix

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Dear all, I am a newbie in R and could not find help on this problem. I am trying to plot an histogram with probabilities in the y axis. This is the code I am using: #TLC uniform n=30 mi=1; mx=6 nrep=1000 xbar=rep(0,nrep) for (i in 1:nrep) {xbar[i]=mean(runif(n,min=mi,max=mx))} hist(xbar,prob=TRUE,breaks="Sturges",xlim=c(1,6),main=paste("n =",n), xlab="Média",

Hi, Does anyone know if there is a function to find the maximum likelihood estimates of glm using Newton Raphson metodology instead of using IWLS. Thanks Valeska Andreozzi -------------------------------------------------------- Department of Epidemiology and Quantitative Methods FIOCRUZ - National School of Public Health Tel: (55) 21 2598 2872 Rio de Janeiro - Brazil

I am recoding some data. Many values that should be 1.5 are recorded as 1-2. Some example data and my solution is below. I am curious about better approaches or any other suggestions. Thanks! # example input data myData <- read.table(textConnection("id, v1, v2, v3 a,1,2,3 b,1-2,,3-4 c,,3,4"),header=TRUE,sep=",") closeAllConnections() # the first column is IDs so remove

Dear R users: Is there a built-in and simple way to insert new columns after other columns in a dataframe? I.e. currently I have: V1 V2 V3 V4 [1,] [2,] Etc. But I want V1 V5 V2 V3 V4 [1,] [2,] Etc. Can this be done in one line? Jon Minton [[alternative HTML version deleted]]

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Hi all .., I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. When I extract one dataframe and apply my code it works very well as follows :- First suppose this is my first dataframe in the list > OneDF <- Mylist[[1]] > OneDF ID Pdate

Folks: I consider my reply below rather clumsy: One has to keep track of index numbers other than that which is inserted and must separately change column names. Is there as "essentially better" way to do this, either via base R or via an R package. I leave it to you to define "essentially better." Thanks. Cheers, Bert On Mon, Aug 1, 2011 at 10:17 AM, Bert Gunter

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Hi, I'm trying to concatenate values from two columns in a data frame. For example, I have the following data.frame: C1 C2 C3 C4 C5 A B *F C* Q G H *I J* T K D *R S* E P L *M N* O I'd like to concatenate text from columns C3 and C4, to yield either a list or vector, like so: NewCol FC IJ RS MN Is this feasible in R? Thanks!

> On Apr 15, 2018, at 4:08 AM, Allaisone 1 <Allaisone1 at hotmail.com> wrote: > > > Hi all .., > > > I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. > > > When I extract one dataframe and apply my code it works very well as

Hi All I have a datafram which looks like this : CustomerID DietType 1 a 1 c 1 b 2 f 2 a 3 j 4 c 4 c 4 f And I would like to reshape this so I can

Hi, Here's my problem... I have a data frame with three columns containing strings. The first columns is a simple character. I want to get the index of that character in the second column and use it to extract the item from the third column. I can do this using a scalar method. But I'm not finding a vector method. An example is below. col1 col2 col3 'L'

Dear R-Users, I have a dataframe (''forexdata'') of daily returns from the foreign exchange market for three currencies - British Pound (bp), Canadian Dollar(cd), Deustche Mark (dm) vis-a-vis the US Dollar and the Date Of Trade(yymmdd). For some dates the returns are missing (recorded as zero) as there were no trades in that currency for that date. My task is to substitute the

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<-

Hi, I am trying to get faster search results on our webmail client(Roundcube). Besides using Lucene for FTS are there other options? Would having all mails indexed give fast results? Currently the time it takes to search 25,000mails is 4mins. If indexed how much faster are we looking at? Really appreciate if someone could advise about this. Thanks Kevin

Hi All, I want to change debug information of an llvm instruction so that the modified debug info is subsequently passed to executable binary. So if I use "addr2line" utility on the binary, it will return my modified debug information. I've tried to change by using the following code snippet: MDNode *N = Inst->getMetadata("dbg"); DebugLoc Loc =

Hi Allaisone, If you want a data frame as the output you will have to put up with a few NA values unless each Customer has the same number of diet types: a1df<-read.table(text="CustomerID DietType 1 a 1 c 1 b 2 f 2 a 3 j 4

I do have some data of dim 100*3 (i.e 100 rows and 3 columns ) stored in a txt file. I want to read the data into R, Perform the same operation in each row and store the result in a forth column( that I should create). I do not know how I can create a forth column to store the result of the operation for each row? -- View this message in context:

Hello, I have a data frame with many rows, and I want to create a column with a name only at every 12th row, starting from 97 to 278. Thanks in advance!