similar to: Can't understand syntax

Dear R Community, I am new to R, and have a question that I suspect may be quite simple but is proving a formidable roadblock for me. I have a large data set that includes water-quality measurements collected over many 24-hour periods. The date and time of sample collection are in a combined Date/Time field in the format yyyy-mm-dd hh:mm:ss. I need to be able to subset the data for analysis of

Hi I got the following error using as.xts Error in xts(x, order.by = order.by, frequency = frequency, ...) : NROW(x) must match length(order.by) Here is how the data looks like > d1 <- read.csv(file.path(dataDir,"AppendixA-FishCountsTable-2009.csv"), as.is=T) > d1[1:3,] dive_id date time species count size site depth level TRANSECT VIS_M 1 62 10/12/2009

Hola a todos, Me gustaria pedir vuestra ayuda a encontrar el error que no consigo encontrar en este archivo. He revisado todo mil veces y probado y no doy con ello.Adjunto el archivo con Google drive porque es muy grande. ? monicap_50.csv <https://docs.google.com/file/d/0B8o2KrPEgG7ATlBMc19lTVk1d3M/edit?usp=drive_web> ? Este es el script, y lo que no entiendo que pasa es que tengo 592044

Hola!! Estoy intentando ejecutar un script com horas, al principio ejecute estos comandos DBx$Date<-strptime(DBx$Date, "%d-%m-%Y") ###Monicap use ; other use Y DBx$Year<-as.POSIXlt(DBx$Date)$year+1900 if(filename!="monicap_50.csv") {DBx$Time<-paste(DBx$Time, ":00", sep="")} Pero me daba el error de que mi base de datos tenia las

Hola Daniel, si perdona di a responder directamente y no me di cuenta. Ya se donde esta el error, pero queria preguntar si a puede ser que mi R funcione mal o algo porque esta manhana ejecute mi script y funciono perfectamente y ahora volvi a ejecutarlo y me volvio a dar el mismo problema de ayer , despues de reiniciar y demas el tinnR y el R, no se si me vacila o es que tengo algo mal en

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

Why does the subset not work in the which() version below? Thank you v1 <- subset(t1, version_1==as.character("100-1") | version_1==as.character("100-2")) a<-c("100-1", "100-2") v1 <- subset(t1, which(a==as.character(version_1)) != 0) [[alternative HTML version deleted]]

Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to include enough detail. I have a question about survival curves I have been battling off and on for a few months. No one local seems to be able to help, so I turn here. The issue seems to either be how R calculates Kaplan-Meier Plots, or something with the underlying statistic itself that I am misunderstanding. Basically,

sir i could not find the plant breeding libraray in Rgui3.0.0 [[alternative HTML version deleted]]

Short question: Is it possible to use statistical tests, like the Augmented Dickey-Fuller test, in functions with for-loops? If not, are there any alternative ways to scale measures? Detailed explanation: I am working with time-series, and I want to flag curves that are not stationary and which display pulses, trends, or level shifts. >df DATE ID VALUE2012-03-06 1

Hello Everyone I have the data long format and I want to draw the contour plot with it x1 x2 x3 0 1 2 0 2 1 0 3 5 1 1 4 1 2 2 1 3 3 when I am using contour(x1,x2,x3,col=heat.colors) or fill.contour its giving me an error that increasing x and y expected So please tell me what is the right function to draw contour when the data is not ordered and you cant order

guRus! I have a function f = exp(x^2-y+(1/z)) Also, x can take values from 1 to 37, y from 2 to 20 and Z from -13 to 51. How can I find the maximum of f using any of the optimization functions please? Is there a way to store the possible values of x, y and Z in a single variable like in a List or in a multi-dimensional array? Thanks for your help Raghu [[alternative HTML version deleted]]

Hi, I have some controller code like this: def mailing_lists @ml = MailingList.find_by_identifier(@params[:id]) @messages = @ml.messages @headers["Content-Type"] = "application/atom+xml" @response.headers[''Last-Modified''] = @messages.last.created_on.httpdateunless @ messages.empty? minTime =

Dear All, I'd like to know if it is possible to create animations with R. To be specific, I attach a code I am using for my research to plot some analytical results in 3D using the lattice package. It is not necessary to go through the code. Simply, it plots some 3D density profiles at two different times selected by the user. I wonder if it is possible to use the data generated for different

Dear All, I am trying to use the do.call command to avoid tedious loops in R when I need to call repetitively a function. I am experiencing a problem when the arguments of a function are given with a vector (I need to express then this way to be able later on to integrate them numerically with the adapt package). Consider the small script: remove(list=ls()) #library(adapt) # first a list of the

Hi all, I am trying to write a new function based on an existing function in "stats". This function requires to call a c program "C_BinDist". The new function couldn't work without load this c script. How can I find this "C_BinDist" function and load it to make my new function work through? Thanks, Dan [[alternative HTML version deleted]]

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

Hi all, I still have problems with the predict function by setting up the values on which I want to predict ie: original df: p1 (193 obs) variates y x1 x2 rm(list=ls()) x1<-rnorm(193) x2<-runif(193,-5,5) y<-rnorm(193)+x1+x2 p1<-as.data.frame(cbind(y,x1,x2)) p1 y x1 x2 1 -0.6056448 -0.1113607 -0.5859728 2 -4.2841793 -1.0432688 -3.3116807 ...... 192

Suppose plotx <- "someName" modx <- "otherName" plotxRange <- c(10,20) modxVals <- c(1,2,3) It often happens I want to create a dataframe or object with plotx or modx as the variable names. But can't understand syntax to do that. I can get this done in 2 steps, creating the data frame and then assigning names, as in newdf <- data.frame( c(1, 2, 3, 4),

Dear R-helpers, I am stuck on an error in R: When I run my code (below), I get this error back: Error in names(x) <- value : 'names' attribute must be the same length as the vector Then when I use traceback(), R gives me back this in return: `colnames<-`(`*tmp*`, value = c(""Item", "Color" ,"Number", "Size")) I'm not exactly