similar to: for loop help

What is the syntax for this? If you have: vector = c(1,2,3,4), how would you output this to column A of an excel spreadsheet? -- View this message in context: http://r.789695.n4.nabble.com/how-to-you-output-a-vector-to-a-column-in-excel-tp2530470p2530470.html Sent from the R help mailing list archive at Nabble.com.

If you have 5 data frames and you append them to a list, how do you access the first data frame, not the first value of the first data frame while iterating in a for loop? list = c(d1,d2,d3,d4,d5) where d1..d5 are dataframes. for(i in 1: length(list)){ print(list[1]) } -- View this message in context: http://r.789695.n4.nabble.com/extracting-objects-from-lists-tp2539412p2539412.html Sent from

arima1 = arima(data.ts[1:200], order = c(1,1,1)) arima2 = arima(data.ts[5:205], order = c(1,1,1)) arima3 = arima(data.ts[10:210], order = c(1,1,1)) arima4 = arima(data.ts[15:215], order = c(1,1,1)) arima5 = arima(data.ts[20:220], order = c(1,1,1)) arima6 = arima(data.ts[25:225], order = c(1,1,1)) arima7 = arima(data.ts[30:230], order = c(1,1,1)) arima8 = arima(data.ts[35:235], order = c(1,1,1))

I have a data file with a given time series of price data and I would like to split the time series into a test set and training set. I would then like to build an ARIMA model on the training set and apply this model on test set. Below is some code: [CODE] data= read.table("A.txt",sep=",") attach(data) training = data[1:120, 6] test = data[121:245, 6] ts1 = ts(training) ts2 =

If I have a model line = lm(y~x1) and I want to use a for loop to change the number of explanatory variables, how would I do this? So for example I want to store the model objects in a list. model1 = lm(y~x1) model2 = lm(y~x1+x2) model3 = lm(y~x1+x2+x3) model4 = lm(y~x1+x2+x3+x4) model5 = lm(y~x1+x2+x3+x4+x5)... model10. model_function = function(x){ for(i in 1:x) { } If x =1, then the list

temp = "~aparch(" temp1 = paste(temp,1, sep = "") temp2 = paste(temp1,1, sep = ",") temp3 = paste(temp2, ")",sep = "") temp 3 is a character but I want to convert to formula object. How do I do this? -- View this message in context: http://r.789695.n4.nabble.com/how-to-convert-string-to-object-tp2999281p2999281.html Sent from the R help mailing

s= "Hey" a = "Hello" table = rbind(s,a) write.table(table,paste("blah",".PROPERTIES",sep = ""),row.names = FALSE,col.names = FALSE) In my table, how do I output only the words and not the words with the quotations? -- View this message in context: http://r.789695.n4.nabble.com/How-to-unquote-string-in-R-tp3025654p3025654.html Sent from the R

Hi, I have two dataframes: The first, df1, contains some missing data: cola colb colc cold cole 1 NA 5 9 NA 17 2 NA 6 NA 14 NA 3 3 NA 11 15 19 4 4 8 12 NA 20 The second, df2, contains the following: cola colb colc cold cole 1 1.4 0.8 0.02 1.6 0.6 I'm wanting all missing data in df1$cola to be replaced by the value of df2$cola.

Hi all, I sent a request round last week asking for help with using a "for" loop to read and separate a large dataset. The response I got worked great, but now I have another problem with using my loop. Basically I have a number of different files containing columned data. There are 132 datasets, named such that I have something in the form... precip_colxxx.txt ...where xxx is a

Dear List: It appears that simulating data where all dataframes are stored as a list will only work for relatively small analyses. Instead, it appears that creating N individual dataframes, saving them, and loading them when needed is the best way to save memory and make this a feasible task. As such, I now have a new(er) question with respect to dealing with individual files within a loop. To

Thanks Don, for (i in 1:10){ nm <- paste0("V", i) d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) } is exaclty what I needed. Best regards, Luca 2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>: > Your code doesn't make sense to me in a couple of ways. > > Inside the loop, the first line assigns a value to an

Hello, instead of ifelse, the following is exactly the same and much more efficient. d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0) Hope this helps, Rui Barradas On 4/28/2018 8:45 PM, Luca Meyer wrote: > Thanks Don, > > for (i in 1:10){ > nm <- paste0("V", i) > d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) > }

Your code doesn't make sense to me in a couple of ways. Inside the loop, the first line assigns a value to an object named "t". Then, the second line does the same thing, assigns a value to an object named "t". The value of the object named "t" after the second line will be the output of the ifelse() expression, whatever that is. This has the effect of making

I forgot to explain why my suggestion. The logical condition returns FALSE/TRUE that in R are coded as 0/1. So all you have to do is coerce to integer. This works because the ifelse will return a 1 or a 0 depending on the condition. Meaning exactly the same values. And is more efficient since ifelse creates both vectors, the true part and the false part, and then indexes those vectors in

Hi Rui Thank you for your suggestion, I have tested the code suggested by you against that supplied by Don in terms of timing and results are very much aligned: to populate a 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the one with ifelse employed 80 secs, hence unfortunately not really any significant time saved there. Nevertheless thank you for your contribution.

If I have a list: list = c(~garch(1,1), ~arma(1,1)) and I run typeof(list[1]), the output is a list object. But I want each element in the list to be a language object. How do I transform these list objects to language objects? -- View this message in context: http://r.789695.n4.nabble.com/how-to-convert-list-to-language-object-tp2966813p2966813.html Sent from the R help mailing list archive at

Hi all, I was wondering if you can help me with the following situation: I have a data frame that includes weather station data for 30 years in the form: YEAR, MONTH, DAY, TEMP 1970, 01, 01, -15 ... 1999, 12, 31, -21 I would like to add another variable "JULIAN" that assigns the integers 1 to 365 (and 1 to 366 for leap years) for each day of a year over multiple years. Here is what

Hello How to create plots dynamically with results of several analysis ? I got many outputs from lm fuction like: mp1.lm mp2.lm mp3.lm mp4.lm mp5.lm ... I'd like to make experimental versus predicted response plots of all analysis in a 'for loop': for( i in 1:10){ x11() plot( mp*i*$experimental_response, fitted( mp*i* ) ); abline(0,1) } I tried: paste( 'mp', i,

Thank you for both replies Don & Rui, The very issue here is that there is a search that needs to be done within a text field and I agree with Rui later comment that regexpr might indeed be the time consuming piece of code. I might try to optimise this piece of code later on, but for the time being I am working on the following part of building a neural network to try indeed classifying some

Hi, I am trying to debug the following code: for (i in 1:10){ t <- paste("d0$V",i,sep="") t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0) } and I would like to see what code is actually processing R, how can I do that? More to the point, I am trying to update my variables d0$V1 to d0$V10 according to the presence or absence of some text (contained in the file d1)