similar to: Updating a Data Frame

Hi, I am trying to execute the following SQL statement using two data frames: tab1, tab2 : Two Tables Select tab1.*, tab2.*, tab1.tobiiTime - tab2.ruiTime as timeDiff, IFNULL(n-m, -9999999) as alwaysIncrement FROM tab1 LEFT OUTER JOIN tab2 On tab1.data1 - tab2.mouseX = 0 And tab1.data2 - tab2.mouseY = 0 I am trying to do the following in R:- *#Getting error here:* data

Tu fichero tiene los decimales como puntos y no como comas como tu le indicas. Te dejo un ejemplo #--------------------------------------------------------------------------------------------------------------------- setwd(dir="c:/Users/usuario/Desktop/") library(outliers) filename<-"timediff.csv" time<-read.csv(filename, sep=";",header=TRUE,dec=".")

Hola Marta, si observas cualquier artículo de psicología esto es una práctica típica. Te digo psicología porque creo que tus datos son de ese tipo. Tienes bibiligrafía de artículos cientificos en las que se quitan valores siguiendo los criterios que te he dicho solo has de buscar bibliografía del campo en el que te mueces. Osea que nos quitarlos por que sí, aunque en cada area de ciencia tienes

I have the following data (a subset of the real data set): structure(list(Date = structure(c(1209625080, 1209641460, 1209652500, 1209676800, 1209682860, 1209692100, 1209706980, 1209722580, 1209726300, 1209739620, 1209762780, 1209765720, 1209770520, 1209791040, 1209812580, 1209829920, 1209837180, 1209848160, 1209854640, 1209859440, 1209870780, 1209887760, 1209901080, 1209921660, 1209929280,

Hi. I searched the list and didn't found nothing similar to this. I simplified my example like below: #I need calculate correlation (for example) between 2 columns classified by a third one at a data.frame, like below: #number of rows nr = 10 #the third column is to enforce that I need correlation on two variables only dataf =

Dear all, In order to ease the transition from SPSS to R for some of my colleagues, I am trying to create a function which would show the variables and their labels (if those exist), using function "label" in package Hmisc. A toy example would be this: my.data <- data.frame(age=c(24,35,28), gender=c("Male", "Female", "Male")) require(Hmisc)

Hola! Quiero encontrar los outliers de una columna de datos que son diferencias de tiempo, aunque esta en formato numero no en horas. El paquete que encontre que me podria servir es el outliers e hice este script: library(outliers) filename<-"timediff.csv" time<-read.csv(filename, sep=";",header=TRUE,dec=",") chisq.out.test(time$TimeDiff) Pero me da error

Hello; I am having a problems with the interpretation of models using ordered or unordered predictors. I am running models in lmer but I will try to give a simplified example data set using lm. Both in the example and in my real data set I use a predictor variable referring to 3 consecutive days of an experiment. It is a factor, and I thought it would be more correct to consider it ordered. Below

Hello, I have two problems in automating multiple glm(s) operations. The data file is tab delimited file with headers and two columns. like "ABC" "EFG" 1 2 2 3 3 4 dat <- read.table("FILENAME", header=TRUE, sep="\t", na.strings="NA", dec=".", strip.white=TRUE) dataf <- read.table("FILENAME", header=FALSE,

Hello All, I am trying to write a function which would operate on columns of a dataframe specified in parameters passed to that function. f = function(dataf, col1 = "column1", col2 = "column2") { dataf$col1 = dataf$col2 # just as an example } The above, of course, does not work as intended. In some languages one can force evaluation of a variable, and then

Hi, I like the formatting and the appearance of lattice plots. But I have not succeeded in gettting the right format in my plots with the lattice package in one of my applications. In?the code shown below, I start by constructing a general data frame and show my attempts with the lattice package commands. After that, I use the graphics package and show the kind of plot that I want to get. I would

Hi, I was trying to use the :select parameter to select a calculated row, but it appears that the parameter is not picked up by the paginate method. I want to add the following to the select statement: timediff(now(), created_on) as age So I tried: @story_pages, @stories = paginate :stories, {:per_page => 10, :include => ''user'', :select =>

Dear All I am reposting because I my problem is real issue and I have been working on this. I know this might be simple to those who know it ! Anyway I need help ! Let me clear my point. I have huge number of datapoints plotted using either base plot function or xyplot in lattice (I have preference to use lattice). name xvar p 1 M1 1 0.107983837 2 M2 11

Hi, Here, i have a matrix like this MyMatrix <- *DATETIME HEADER1 HEADER2* 1/1/2010 0:10 197.1947 100.0859 1/1/2010 0:20 203.8811 100.1013 1/1/2010 0:30 206.564 100.0433 1/1/2010 0:40 207.9563 99.9393 i want to get the time difference in minutes between two date. TimeDiff <- MyMatrix[1,1] - MyMatrix[2,1] TimeDiff

Hi, I am writing a function that writes to a csv file for every call. However, for the subsequent calls, I want to append the data to the existing csv file without appending the column names again. I tried searching in the previous posts, but I am stuck with different errors. Here is what I am doing (dataF is a data-frame):- outputFilePath <- paste(getwd(), "/",

In R, I'm imported a data frame of 2,321,123 by 4 called "dataF". I converted the data frame "dataF" to a matrix dataM <- as.matrix(dataF) Does R have an efficient routine to treat the special elements that contain "inf" in them. For example, can you separate the rows that have "inf" elements from the matrix into a separate matrix without

Dear list, ? would anybody be able to tell me why the statement ? Tripstatistics=aggregate(TripsData[2:3],by=list(Trip=Tripmatch),FUN="mean") ? seems to work well with TripsData 1 but not with TripsData 2 ? ? With TripsData 2 it yields ? Error in FUN(X[[1L]], ...) : arguments must have same length I can't see a difference in the two data sets. Could someone shed light on the error

Hi, 1.) The "median" function does not work well. Please refer to the data below (same data is attached as txt-delimited). This is what I try to do in R: median ( dataf [2:9] ) I get warning: "needs numeric data" 2.) BUT if apply the median to a single vector: median ( dataf [,2]] ) then it works: 3.) How come the "median"

Dear R users, Here is my problem: # example data name <- c(paste ("M", 1:1000, sep = "")) xvar <- seq(1, 10000, 10) set.seed(134) p <- rnorm(1000, 0.15,0.05) dataf <- data.frame(name,xvar, p) plot (dataf$xvar,p) abline(h=0.05) # I can know which observation number is less than 0.05 which (dataf$p < 0.05) [1] 12 20 80 269 272 338 366 368 397 403 432 453

Hi listers, I want to make lattice plots xyplots with the indication of legends inside each panel with only the points and the lines actually ploted inside each given panel according to the group(ing) factor. The code below shows what I have achieved so far and I hope will make clear what I want to have. It seems to me that my solution is a very "dirty hack" and there certainly is