[R] [R ] help in if else in connect the simulation in normal and gamma distribution.

I am new in R. I have to conduct simulation study on the robustness of 2 sample tests on several combination of factors (sample sizes ,variance and distribution). I have been completed write a code in normal distribution, and now i wish to add if -else in the code so that the code can

Re: [R] why data frame's logical index isnt working

> On Apr 7, 2016, at 7:44 PM, Michael Artz wrote: > > I don't get it, I thought the double index was to indicate and individual > element within a column(vector)? Character values by themselves either quoted or not are not assumed to refer to column names unless you

Re: [R] Using a function with apply Error: undefined columns selected

> On Apr 7, 2016, at 3:39 PM, John Sorkin wrote: > > I am trying to write a function that can be used to apply to process all the > columns of a data.frame. If you will run the code below, you will get the > error message undefined columns selected. I hope someone

Re: [R] why data frame's logical index isnt working

I don't get it, I thought the double index was to indicate and individual element within a column(vector)? I will stop using data.frame, thanks a lot! On Thu, Apr 7, 2016 at 9:29 PM, David Winsemius wrote: > > > On Apr 7, 2016, at 6:46 PM, Michael Artz

Re: [R] R.squared in summary.lm with weights

Perhaps I did not make clear that this is not my code - it is the code in summary.lm. I used square brackets to try to make my edited version of the text intelligible - print summary.lm and you will see where my version fits in. From: R-help

Re: [R] R.squared in summary.lm with weights

Do you mean w <- z$residuals ? Type names(z) to see the list of item in your model. I ran your code on a lm and it work fine. You don't need the brackets around mss <- Michael Long On 04/07/2016 02:21 PM, Murray Efford wrote: Following some old advice on this list, I have been reading the

Re: [R] why data frame's logical index isnt working

> On Apr 7, 2016, at 6:46 PM, Michael Artz wrote: > > data.frame.$columnToAdd["CurrentColumnName" == "ConditionMet"] <- 1 > > Can someone please explain to me why the above command gives all NAs to > columnToAdd? I thought this was possible in R to do logical

Re: [R] using apply to a data frame

> On Apr 7, 2016, at 1:25 PM, John Sorkin wrote: > > > ‪‪I would like to apply a function, fract, to the columns of a > dataframe. I tried the following > apply(data5NonEventEpochs,2,fract) > but, no surprise it did not work as apply works on matrices not data >

Re: [R] why data frame's logical index isnt working

you probably mean something like this data.frame.$columnToAdd <- (data.frame.$CurrentColumnName == data.frame.$ConditionMet) what you did is compare two character strings. They are not the same. Therefore a new column is created with the default value NA. > tmp <- data.frame(A=1:4,

Re: [R] simple question on data frames assignment

Why am I better off with true and false? On Thu, Apr 7, 2016 at 8:41 AM, Hadley Wickham wrote: > == is also vectorised, and you're better off with TRUE and FALSE > rather than 1 and 0, so I'd recommend: > > colordata$response <- colordata$color == 'blue' > > Hadley > > On

Re: [R] simple question on data frames assignment

Fyi, This statement returned the following error 'Error in "Yes" + 0 : non-numeric argument to binary operator' On Thu, Apr 7, 2016 at 8:43 AM, wrote: > Hello, > > Or even simpler, without ifelse, > > colordata$response <- colordata$color == 'blue' + 0 > > Hope this

[R] why data frame's logical index isnt working

data.frame.$columnToAdd["CurrentColumnName" == "ConditionMet"] <- 1 Can someone please explain to me why the above command gives all NAs to columnToAdd? I thought this was possible in R to do logical expression in the index of a data frame [[alternative HTML version deleted]]

Re: [R] Using a function with apply Error: undefined columns selected

Hi John, First, apply isn't guaranteed to work on data frames. There are two easy ways to do something like this, but we had better have a data frame: guppy<-data.frame(taste=rnorm(10,5), crunch=rnorm(10,5),satiety=rnorm(10,5)) If you just want to apply a function to all or a subset of columns

[R] Using a function with apply Error: undefined columns selected

I am trying to write a function that can be used to apply to process all the columns of a data.frame. If you will run the code below, you will get the error message undefined columns selected. I hope someone will be able to teach me what I am doing wrong. Thank you, John # create data frame.

Re: [R] using apply to a data frame

Use lapply or sapply. A data frame is also a list with each component representing one column; lapply/sapply will apply the function to each column. Peter On Thu, Apr 7, 2016 at 1:25 PM, John Sorkin wrote: > > ‪‪I would like to apply a function, fract, to the

[R] R.squared in summary.lm with weights

Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the observations is calculated as if the weights

Re: [R] using apply to a data frame

Inline. -- Bert Bert Gunter On Thu, Apr 7, 2016 at 1:25 PM, John Sorkin wrote: > > ‪‪I would like to apply a function, fract, to the columns of a > dataframe. I tried the following > apply(data5NonEventEpochs,2,fract) > but, no surprise it did not work as apply

Re: [R] using apply to a data frame

Dear John, Try using sapply(data5NonEventEpochs, fract) HTH Best, Luisfo Chiroque PhD Student IMDEA Networks Institute http://fourier.networks.imdea.org/people/~luis_nunez/ > El 7 abr 2016, a las 23:25, John Sorkin

[R] using apply to a data frame

‪‪I would like to apply a function, fract, to the columns of a dataframe. I tried the following apply(data5NonEventEpochs,2,fract) but, no surprise it did not work as apply works on matrices not data frames. How can I apply a fuction to the columns of a data frame? (I can't covert

Re: [R] Question about R Ver.2.7.2 compatible with JAVA 8

On 07/04/2016 11:31 AM, Rajamoorthy-CW, Thirumurugan 8361 wrote: Hi Team, We (Otsuka IT team) are planning to upgrade JAVA 8 to Biometrics department servers. On account of this, we want to make sure R Ver.2.7.2 is working fine in JAVA 8? Please let us know if R Ver.2.7.2 is compatible with

Re: [R] Storing output of loop into list()

You don't need a for loop for the first part. You can do it like this: Notice that I used unchanging data for the temps vector so that you could tell us what output you expect to see, in case what I provide is not adequate. temps <- c(13.988, 13.932, 14.039, 14.082, 13.998, 13.93, 14.028,

[R] Question about R Ver.2.7.2 compatible with JAVA 8

Hi Team, We (Otsuka IT team) are planning to upgrade JAVA 8 to Biometrics department servers. On account of this, we want to make sure R Ver.2.7.2 is working fine in JAVA 8? Please let us know if R Ver.2.7.2 is compatible with JAVA 8? Regards Thirumurugan Rajamoorthy

Re: [R] Storing output of loop into list()

Have a look at ?lapply, which will do this for you. Have you spent time with an R tutorial or two. There are many good ones on the web that discuss this sort of thig. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it."

[R] Storing output of loop into list()

Hello. I am trying to store the output from a loop into a matrix. Failed so far, thanks for the help. What I want to store into a new matrix: (just run the code once): temps<-rnorm(400,14,0.05) ttind<-NULL for(ti in 1:(length(temps)-9)) { if(temps[ti]-temps[ti+9] >= 0.1 &&

Re: [R] simple question on data frames assignment

It all makes so much sense now On Thu, Apr 7, 2016 at 10:04 AM, Jeff Newmiller wrote: > lapply(colordata2[ -1 ], f ) > > When you put the parentheses on, you are calling the function yourself > before lapply gets a chance. The error pops up because you are giving a >

Re: [R] simple question on data frames assignment

lapply(colordata2[ -1 ], f ) When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function. -- Sent from my phone.

Re: [R] dynamic reports with sweave: error when compiling the tex-file

Definitely yes I just tried a one line latex document and it bombed with \begin{ document } but is fine with \begin{document} . I think the spaces within the {} must all be removed. Latex does not like spaces John Kane Kingston ON Canada > -Original Message- > From:

Re: [R] simple question on data frames assignment

If you are not using an anonymous function and say you had written the function out The below gives me the error > 'f(colordata2$color1)' is not a function, character or symbol' But then how is the anonymous function working? f <- function(col){ ifelse(col == 'blue', 1, 0) } responses <-

Re: [R] simple question on data frames assignment

Hello, Or even simpler, without ifelse, colordata$response <- colordata$color == 'blue' + 0 Hope this helps, Rui Barradas   Citando David Barron : > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0)

Re: [R] simple question on data frames assignment

== is also vectorised, and you're better off with TRUE and FALSE rather than 1 and 0, so I'd recommend: colordata$response <- colordata$color == 'blue' Hadley On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote: > ifelse is vectorised, so just use that without the loop. > >

Re: [R] simple question on data frames assignment

Lapply is not a vectorized function. It is compact to read, but it would not be worth using for this calculation. However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a

Re: [R] simple question on data frames assignment

Hi > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael > Artz > Sent: Thursday, April 7, 2016 1:57 PM > To: David Barron > Cc: r-help@r-project.org > Subject: Re: [R] simple question on data frames assignment > > Thaks so

Re: [R] dynamic reports with sweave: error when compiling the tex-file

Did you really put spaces between the backslashes and the keywords? That would be a problem... -- Sent from my phone. Please excuse my brevity. On April 7, 2016 4:32:29 AM PDT, palad...@trustindata.de wrote: >Hello, >I took my first steps in dynamic reports with Gnu R and used sweave(). > >I

Re: [R] Analyze a file

Hi myfile <- read.table("file.txt") plot(myfile) If it does not work, you have wrong file.txt Cheers Petr > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Mohamed > Benahmed via R-help > Sent: Thursday, April 7, 2016 11:27 AM > To:

[R] Analyze a file

hi all,How to analyze a file.txt with R to obtain the corresponding graph.What is the script ? Please help me! Thanks, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see

[R] dynamic reports with sweave: error when compiling the tex-file

Hello, I took my first steps in dynamic reports with Gnu R and used sweave(). I therefore run Sweave() with an example of Friedrich Leisch, starting like this: \ documentclass [ a4paper ]{ article } \ title { Sweave Example 1} \ author { Friedrich Leisch } \ begin { document } \ maketitle

Re: [R] simple question on data frames assignment

Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at

Re: [R] simple question on data frames assignment

ifelse is vectorised, so just use that without the loop. colordata$response <- ifelse(colordata$color == 'blue', 1, 0) David On 7 April 2016 at 12:41, Michael Artz wrote: > Hi I'm not sure how to ask this, but its a very easy question to answer for > an R person. > >

[R] simple question on data frames assignment

Hi I'm not sure how to ask this, but its a very easy question to answer for an R person. What is an easy way to check for a column value and then assigne a new column a value based on that old column value? For example, Im doing colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue",

Re: [R-es] Contenido de un objeto/modelo ARIMA

Le echo un vistazo, algo aprenderé seguro. Gracias! El 7 de abril de 2016, 11:04, Isidro Hidalgo Arellano escribió: > El autor del paquete es Hyndman, tiene un libro on-line en el que puedes > echar un vistazo. La parte de modelos ARIMA en R está aquí: >

Re: [R-es] Contenido de un objeto/modelo ARIMA

Perfecto!! Gracias. :) El 7 de abril de 2016, 11:01, Carlos J. Gil Bellosta escribió: > Hola, ¿qué tal? > > Copio de la ayuda de la función: > > library(forecast) > fit <- Arima(WWWusage,order=c(3,1,0)) > > names(fit) > # [1] "coef" "sigma2""var.coef" "mask"

Re: [R-es] Contenido de un objeto/modelo ARIMA

El autor del paquete es Hyndman, tiene un libro on-line en el que puedes echar un vistazo. La parte de modelos ARIMA en R está aquí: https://www.otexts.org/fpp/8/7 Esos "TRUE" podrían significar que has utilizado coeficiente, deriva, media... no recuerdo exactamente los parámetros que utiliza la

Re: [R-es] Contenido de un objeto/modelo ARIMA

Hola, ¿qué tal? Copio de la ayuda de la función: library(forecast) fit <- Arima(WWWusage,order=c(3,1,0)) names(fit) # [1] "coef" "sigma2""var.coef" "mask" "loglik" # [6] "aic" "arma" "residuals" "call" "series" # [11] "code" "n.cond""nobs" "model"

Re: [R] identifying outliers

Hi You can check boxplot output by str > iris[1,1] <- 10 > iris[2,1] <- 8 > bbb <- boxplot(split(iris[,1], iris$Species)) > str(bbb) List of 6 $ stats: num [1:5, 1:3] 4.3 4.8 5 5.3 5.8 4.9 5.6 5.9 6.3 7 ... $ n: num [1:3] 50 50 50 $ conf : num [1:2, 1:3] 4.89 5.11 5.74 6.06 6.34 ... $

[R-es] Contenido de un objeto/modelo ARIMA

Buenos días, Os cuento: Cargo la librería "Forecast" y ejecuto su función Arima(...) sobre una serie temporal: mimodelo <- Arima(miST$miserie, ...); Ahora si ejecuto las siguientes sentencias, voy obteniendo los resultados contenidos en "mimodelo", pero algunos de ellos no sé lo que son:

Re: [R] Problem with <= (less than or equal): not giving the expected result

Hi > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of S Ellison > Sent: Wednesday, April 6, 2016 12:35 PM > To: Rainer Johannes ; r-help@r-project.org > Subject: Re: [R] Problem with <= (less than or equal): not giving the

Re: [R] Memory problem

Dear Sir, Yes I am using the plyr and in the end I am writing the output to the data.frame. Earlier I had the problem of process time and hence I made some changes in the code and now I am fetching all the required inputs needed for valuation purpose using ddply, store the results in a

[R] identifying outliers

Thanks for writing this great piece of code. x = rnorm(100) boxplot(x) # you shouldn't see any outliers here although sometimes yow will # lets add some outliers intentionally x = c(21, 20, 25, x) # now 10, 15 and 20 are outliers myboxplot <- boxplot(x) # now you should see your