similar to: On matrix calculation

Hello again, Let say I have an non-negative integer vector (which may be random): Vec <- c(0, 13, 10, 4) And I have a date: > Date <- as.Date(Sys.time()) > Date [1] "2013-03-09" Using these 2 information, I want to get following date-vector: New_Vec <- c("2013-03-01", "2014-04-01", "2014-01-01", "2013-07-01") Basically the

Dear all, let say I have following data: dat <- structure(list(V1 = structure(c(1L, 4L, 5L, 3L, 3L, 5L, 6L, 6L, 4L, 3L, 5L, 6L, 5L, 5L, 4L, 4L, 6L, 2L, 3L, 4L, 3L, 3L, 2L, 5L, 3L, 6L, 3L, 3L, 6L, 3L, 6L, 1L, 6L, 5L, 2L, 2L), .Label = c("C", "G", "I", "O", "R", "T"), class = "factor"), V2 = c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L,

Hello again, Let say I have following vector: set.seed(1) Vec <- sample(LETTERS[1:5], 10, replace = TRUE) Vec Now with each repeated letter, I like to add suffix programatically. Therefore I want to get following vector: c("B", "B1", "C", "E", "B2", "E1", "E2", "D", "D1", "A") Can somebody

Hello again, Let say I have following string: Vec <- c("sada", "asdsa", "sa") Now I want to make each element of this vector with equal length. Basically I want following vector: c("sada ", "asdsa", "sa ") Therefore we can get: > nchar(c("sada ", "asdsa", "sa ")) [1] 5 5 5 Is there any

Hello again, Let say I have following vector: Vec <- c("0.0365780769", "(1.09738648244378)", "(0.812507787221523)", "0.5778069963", "(0.452456601362355)", "-1.8900812605", "-1.8716093762", "0.0055217041", "-0.4769192333", "-2.4133018880") Now I want to convert this vector to numeric vector. I

Dear all, I would like to ask one question related to statistics, for specifically on defining dummy variables. As of now, I have come across 3 different kind of dummy variables (assuming I am working with Seasonal dummy, and number of season is 4): > dummy1 <- diag(4) > for(i in 1:3) dummy1 <- rbind(dummy1, diag(4)) > dummy1 <- dummy1[,-4] > > dummy2 <- dummy1 >

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

Hello again, Ley say I have 1 matrix and 1 data frame: > mat <- matrix(1:15, 5) > match_df <- data.frame(Seq = 1:5, criteria = sample(letters[1:5], 5, replace = T)) > mat [,1] [,2] [,3] [1,] 1 6 11 [2,] 2 7 12 [3,] 3 8 13 [4,] 4 9 14 [5,] 5 10 15 > match_df Seq criteria 1 1 c 2 2 e 3 3 c 4 4 c 5

Hello again, Let say I have 1 matrix: Mat <- matrix(1:12, 4, 3) rownames(Mat) <- letters[1:4] Now I want to subscript of Mat in following way: Subscript_Vec <- c("a", "e", "b", "c") However when I want to use this vector, I am geting following error: Mat[Subscript_Vec, ] Error: subscript out of bounds Basically I want to get my final matrix

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

Dear all, Does R use double precision for calculation as default? If not, how to enforce double precision calculation in R for my current calculation session? I Use R-2.14.0 with windows XP. Thanks,

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =

Dear all, can somebody please help me how to calculate "McFadden R-square" for a LOGIT model? Corresponding definition can be found here: http://publib.boulder.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_plum_statistics_rsq_mcfadden.htm Here is my data: Data <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1,