similar to: creating a function

Is there a library that is able for example to 1. merge 2 dataframes by row eg.: rbind(dataframe1, dataframe2):data.frame 2. delete a column from a dataframe del(dataframe, colname) or del(dataframe, colindex= 1):data.frame? 3. Select lines from a dataframe by a specific function ? select(dataframe, func=small(x){x<1}, colindex=3): data.frame? 4. converting all double columns of a data.frame

Hi: I very recently started experimenting with R and am occasionally running into very basic problems that I can't seem to solve. If there is an R-newbies forum that is more appropriate for these kinds of questions, please direct me to it. I'd like to automatically add vectors to a dataframe. I am able to build command strings that would do what I want, but R is not executing them. A

Dear Folks-- I'm trying to make a function that takes the columns I select from a data frame and then uses a for loop to print some information about each one, starting with the column name. I succeed in returning the column name, but nothing else I have tried using the variable colName, containing the name of the column, to refer to the column itself has worked. Below I show my

hi, i have a list of data.frame that has same structure. i would like to know a efficient way of rbind-ing it. right now, i write: n = length(temp) # 'temp' is a list of data.frames temp2 = data.frame() for (i in 1:n) temp2 = rbind( temp2, temp[[i]]) return(temp2) but this is not an efficient way since we keeping overwriting temp2. i wonder if there's faster way. thanks --

Hi, Try either: set.seed(28) stats1<- as.data.frame(matrix(rnorm(5*10000),ncol=5)) pdf(paste("test",1,".pdf",sep="")) par(mfrow=c(2,1)) lst1<- lapply(names(stats1),function(i) {hist(stats1[,i],100,col="lightblue",main=paste0("Histogram of ",i),xlab=i );qqnorm(stats1[,i])}) dev.off() #or

Hello, I am having a problem with write.fwf in Windows. I wrote a code to ingest a number of text files with weather data in them, process them, and then output a text file with two parts: 1) a set of column names, 2) the processed data table. I wrote and tested the program on my Mac, and it worked fine. However, on the windows machine, where I intend the work to be done, when I run the

Hi, I often run into this problem: I have a data.frame with one column containing entries that are not unique. What I then want is a subset of the data.frame in which the entries in that column have become the 'unique' of the original column. Normally I program around it by taking the unique of the column and making a new data.frame with it and filling the rest of the data. (By the way,

All, Given a data frame and a list containing factor definitions for certain columns, how can I apply those definitions from the list, rather than doing it the standard way, as noted below. I'm lost in the world of do.call, assign, paste, and can't find my way through. For example: #set up df y <- data.frame(colOne = c(1,2,3), colTwo =

How does one remove a column from a data frame when the name of the column to remove is stored in a variable? For Example: colname <- "LOT" newdf <- subset(olddf,select = - colname) The above statement will give an error, but thats what I'm trying to accomplish. If I had used: newdf <- subset(olddf,select = - LOT) then it would have worked, but as I said the column

Hello everyone, A recent (in 2.5 I suspect) change in R is giving me trouble. I want to apply a function (tolower) to all the columns of a data.frame and get a data.frame in return. Currently, on a data.frame, both apply (for arrays) and lapply (for lists) work, but each returns its native class (resp. matrix and list): apply(mydat,2,tolower) # gives a matrix lapply(mydat,tolower) # gives

Si, ese es el problema, lo que implica por ejemplo que no pueda ordenar el data.frame por grados. Javier Rubén Marcuzzi De: JCMld Enviado: miércoles, 1 de junio de 2016 4:28 Para: 'Javier Marcuzzi'; R-help-es en r-project.org Asunto: RE: [R-es] data.frame colname igraph Creo que la conversión a data.frame está devolviendo un data frame con una sola columna llamada

Hi I want to extract columns from a data frame using a vector with the desired column names. This short example uses the select argument in the subset function to accomplish what I am trying to do. Is there a better solution? #names of desired columns colnames <- c("col1","col3") #my data data <-

Every time I have to prefix a dataframe column inside the indexing brackets with the dataframe name, e.g. df[df$colname==value,] -- I am wondering, why isn't there an R scoping rule that search starts with the dataframe names, as if we'd said with(df, df[colname==value,]) -- wouldn't that be a reasonable default to prepend to the name search path? Cheers, Alexy

Hi! All, I have 2 correlation matrices of 4000x4000 both with same row names and column names say cor1 and cor2. I have extracted some information from 1st matrix cor1 which is something like this: rowname colname cor1_value a b 0.8 b a 0.8 c f 0.62 d k 0.59 - - -- -

Hello alltogether, how is it possible to assign the colnames of a data.frame to a function called by apply, e.g. for labeling a plot? Example: I want to plot several qqnorm-plots side by side and there should be a maintitle for each qqnorm-plot which is identical to the respective colname. I checked, but the column which is processed by the function called by apply does not contain a colname

I am running R version 2.15.1 in Windows XP I am having problems with a function I'm trying to create to: 1. subset a data.frame based on function arguments (colname & parmname) 2. rename the PARMVALUE column in the data.frame based on function argument (xvar) 3. generate charts plotvar <- function(parentdf,colname, parmname,xvar,yvar ){ subdf <-

summary() for a factor prints: ColName SNDK : 72 VXX : 36 MWW : 30 ACI : 28 FRO : 28 (Other):1801 it would have been much more useful if it additionally printed frequency stats as if by summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x) -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://jihadwatch.org

Dear R Users on a self-written function for calculating maximum likelihood probability (plz check function code at the bottom of this message), one value, wden, suddenly jump to zero. detail info as following: w[11]=2.14 lnw =2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182 0.178 0.179... w[11]=2.14 lnw=2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182

Hello to Rusers! I am puzzled with R and I really do not know where to look in for my problem. I am moving from SAS and I have difficulties in translating SAS to R world. I hope I will get some hints or pointers so I can study from there on. I would like to do something like this. In SAS I can write a macro as example bellow, which is afcourse a silly one but shows what I don't know how

Hi, I would like to extract columns from a dataframe using a vector of desired column names. The following working example uses the select argument in the subset function to accomplish what I am trying to do. Is there a better solution? Thanks. #my data data <- data.frame("col1"=c(1,2,3),"col2"=c("A","B","C"),"col3"=c(4,5,6))