search for: doye

Can anybody help me write a code on the following data example, which fills out all NA values by using a linear interpolation with the two closest values? Doy is day of year (%j). Code example: yr<-c(rep(2000,14)) doy<-c(16:29) dat<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6) ta<-cbind(yr,doy,dat) ta yr doy dat [1,] 2000 16 3.2 [2,] 2000 17 NA [3,] 2000 18 NA

Dear All I am quite new to R and would appreciate some help fitting 95% confidence intervals to a nls function. I have the data DOY CET 90 5.9 91 8 92 8.4 93 7.7 95 6.6 96 6.8 97 7.1 98 9.7 99 12.3 100 12.8 102 11 103 9.3 104 9.8 105 9.9 107 7.7 110 6.2 111 5.9 112 5.9 113 3.4 114 3.5 116 3.3 117 5.4 118 6.3 119 9.7 120 11.2 121 7.3 124 7.8 etc I am trying to use some code that has been

I have a large dataset looking like this (as an example): doy<-c(178,179,180,181,182,183,184,185,186,187,188) s1<-c(0 , 0, 2.4 , 0 , 3.34 , 0 , 5.34 , 0 , 0 , 0 , 6.9) s2<-c(0 , 9.72, 0, 10.56 , 2.67 , 0 , 6.45 ,0 , 0 , 9, 3.6) dat<-cbind(doy,s1,s2) dat I need to make a barplot where the two time series s1 and s2 are plottet beside each other for each doy. How can I do that?

Hello, helpeRs, I have a vector of numbers from 1-365 (days of the year) that I would like to convert to a date. There are no NA's and no missing values. I did not insert leading zero's for numbers less than 100. Using the syntax: dat$doy.1 <- as.numeric(format(dat$doy, "%j" )) I get the following error message: Error in prettyNum(.Internal(format(x, trim, digits,

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966

Dear All, I'm having problems getting text to appear on a second y axis....because this can only be done within a plot statement and my attempts to add the label as text didn't work because the "locator" refers to the plot area.... Any thoughts greatly appreciated. Thanks in advance, Sion >par(mai=c(1,1,1,1.5)) >plot(time[DOY<230],Hmeas[DOY <

Problem: I keep getting: "Error in (index - 9):index : argument of length 0" when I run this code in an automated loop, but when I step through it by hand, I have no problems. It keep crashing when i<-10. I cannot figure this out to save my life! Please help and THANK YOU in advance! variables: FM100 is a matrix of columns: year, day (all years set to 366 days), region 1, region

On 2011-01-27 14:22, Rasti Matus wrote: > I have a few questions regarding machine learning packages in R, > e.g. earth package ... > > 1) How could I force the algorithm to generate knots only with > values < 90 for variable doy? Not possible in the current implementation. In principle it would be possible to extend earth so the user-defined "allowed" function

Hi, I have a few questions regarding machine learning packages in R, e.g. earth package and its forward pass and prunning pass, and hinge functions selection in the example below. 1) How could I force the algorithm to generate knots in hinge function only with values < 90 for e.g. variable doy? 2) Is it possible to restrict the algorithm to generate maximum number of terms per a selected

Hi. I'm analysing some fish biological traits with a gam in mgcv. After several tries, I got this model log(tle) = sexcolor + s(doy, bs = "cc", by = sexcolor) +log(tl) sexcolor is a factor with 4 levels doy is "day of year", which is modeled as a smoother tl is "total length of the fish" The summary of this models is (only parametric coefficientes): Parametric

Dear helpeRs- I'm using a for loop to create a series of models. I'm trying to assign a name to each model created, using the loop index. The loop gets stuck at the name of the model, giving the error "target of assignment expands to non-language object". The linear model runs without error; only the name is problematic. Here is the current loop syntax. The use of dat

About time series graphs, I need help to move on: A time series of data directly from a data logger comes in the dat format created below: year<-c(rep(2005,10)) doy<-c(rep(173,5),rep(174,5)) time<-c(15,30,45,100,115,15,30,45,100,115) dat1<-c(0.022128,0.035036,0.051632,0.071916,0.081136,0.07837,0.083902,0. 126314,0.080214,0.117094) dat2<-

Buenas, He probado a intentar graficar mas de 700.000 puntos y es una locura. No doy con una libreria que consiga graficar grandes volumenes de datos. ?Existe alguna en R? Gracias Jes?s [[alternative HTML version deleted]]

Hola a todos, les consulto algo fácil, pero no me doy cuenta. Estoy siguiendo un ejemplo de un libro, pero no llego al resultado con R. Lo escribo el problema en Word para usar el editor de ecuaciones, porque es tan sencillo que mi error es increíble. El código en R son solo 3 renglones. Javier ------------ próxima parte ------------ Se ha borrado un adjunto en formato HTML...

I want to find the DOY (Day of Year) of some dates. I think to substract the date 1. January from the data to achive this. Something like: > d <- as.Date("2006-03-13") - as.Date("2006-01-01") +1 > d Time difference of 72 days So far so good. But d is a 'difftime' object. How do I get an Integer value from that? I tried severel things, incuding the

Buenas tardes, Al tratar de hacer en rstudio documentos pdf, word o html con un archivo rmd me sale un mensaje tal que así: "Rendering R Markdown documents requires updated versions of the following packages: knitr, rmarkdown. Do you want to install these packages now?" He probado a reinstalar esos paquetes (que ya tenía instalados) pero cuando vuelvo a aplicar knit to pdf,

Estimados Usuarios-R: Estoy convirtiendo un programa en Matlab a R. El original lo saqu de: Lai, M., & Poltera, Y. (2009). Lecture with computer exercises: Modelling and simulating social systems with matlab. Tech. rep., Swiss Federal Institute of Technology (December 2009). 27. Ahora estoy convirtiendo la siguiente funcin: function sizes = gridsizes(N,varargin) % gridsizes(N) calculates

Hi! Is not there an standard R function to retrieve the day of the year (since 1st Jan of the same year)? I know I can make my own using julian, but find it weird that having days(), months() etc doy() does not exist as an standard function. Also, is the following not a bit inconsistent? > a <- chron("20100506",format="ymd") > a [1] 100506 > years(a) [1] 2010

Hola, saludos desde México. Me estoy iniciando en el uso de R, que me parece un muy buen programa. Yo doy la clase de bioestadística en la facultad de Medicina de Mérida, Yucatán en México. Me gustaría hacer contacto con personas que estén usando R en el área biomédica, especialmente en docencia, para compartir experiencias, dudas, ejemplos, etc. Saludos José Arturo [[alternative HTML version

Hola a todos, Gracias por las respuestas. Los nombres ciéntificos son los nombres de las columnas de un csv. Usando vuestros comentarios y sugerencias, estoy tratando de "automatizarlo", ya que tengo unas 40 especies. De momento sin mucho éxito por que no consigo dar con la forma de que me abrevie las especies correctamente y mezclo campos La aproximación de Jorge me valdría (y he