similar to: if-else that returns vector

?ifelse 'ifelse' returns a value with the same shape as 'test' which is filled with elements selected from either 'yes' or 'no' depending on whether the element of 'test' is 'TRUE' or 'FALSE'. This is actually rather startling, because elsewhere in the S (R) language, operands are normally replicated to the length of the longer.

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 ==

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,

@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

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),

Hi, As an example, I want to create below kind of custom Function which either be mclapply pr lapply Lapply_me = function(X = X, FUN = FUN, ..., Apply_MC = FALSE) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { list(...) = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, ...)) } } However when Apply_MC =

?s 12:13 de 13/07/2024, Christofer Bogaso escreveu: > Hi, > > I ran below code > > Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv') > head(Dat) > Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) > head(predict(Model,

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 =

Hi Christofer, Before you made the change that I suggested, your program was stopping at the statement: list(...) = list(..) .etc This means that it never tried to execute the statement: return(lapply(X,FUN,...)) Now that you have made the change, it gets past the first statement and tries to execute the statement: return(lapply(X,FUN,...)). That attempt is generating the error message because

Hi, I ran below code Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv') head(Dat) Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) head(predict(Model, type="response")) My_Predict = 1/(1+exp(-1 * (as.vector(coef(Model))[1] * as.vector(coef(Model))[2] *

Hi Christofer, You cannot assign to list(...). You can do the following myList <- list(...)[!names(list(...)) %in% 'mc.cores'] HTH, Eric On Sun, Mar 4, 2018 at 6:38 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > Hi, > > As an example, I want to create below kind of custom Function which > either be mclapply pr lapply > > Lapply_me =

... but do note: glm(lot1 ~ log(u), data = clotting, family = gaussian) is a plain old *linear model*, which is of course a specific type of glm, but not one that requires the machinery of glm() to fit. That is, the above is exactly the same as: lm(lot1 ~ log(u), data = clotting) and gives exactly the same sigma() ! (and I would therefore hazard the guess that the poster may misunderstand

Hello again, Let say I have following data: Dat <- structure(list(AA = structure(c(3L, 1L, 2L, 1L, 2L, 3L, 3L, 2L, 3L, 1L, 1L, 3L, 3L, 2L, 2L, 3L, 2L, 1L, 1L, 1L), .Label = c("A", "B", "C"), class = "factor"), BB = structure(c(2L, 3L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 1L, 1L, 2L, 3L, 1L, 3L, 2L, 1L, 2L, 2L, 3L ), .Label = c("a", "b",

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

Hi again, I am looking for some way to alternately use 2 related functions, based on some ifelse() condition. For example, I have 2 functions mclapply() and lapply() However, mclapply() function has one extra parameter 'mc.cores' which lapply doesnt not have. I know when mc.cores = 1, these 2 functions are essentially same, however I am looking for more general way to control them

Hello again, Let say I have following user defined function: fn <- function(x, y) { Vec1 <- letters[1:6] Vec2 <- 1:5 return(list(ifelse(x > 0, Vec1, NA), ifelse(y > 0, Vec2, NA))) } Now I have following calculation: > fn(-3, -3) [[1]] [1] NA [[2]] [1] NA > fn(3, -3) [[1]] [1] "a" [[2]] [1] NA Here I can not understand why in the second case, I get

Hello all, Let say I have following (numeric) vector: > x [1] 11.00 11.25 11.35 12.01 11.14 13.00 13.25 13.35 14.01 13.14 14.50 14.75 14.85 15.51 14.64 Now, I want to create a 'Date' variable (i.e. I should be able to do all calculations pertaining to date/time and also time-series plotting etc.) like 2012-12-31 11:00:00 AM, 2012-12-31 11:25:00 AM, 2012-12-31 11:35:00 AM,

On 04/03/2018 10:39 AM, Christofer Bogaso wrote: > Hi again, > > I am looking for some way to alternately use 2 related functions, > based on some ifelse() condition. > > For example, I have 2 functions mclapply() and lapply() > > However, mclapply() function has one extra parameter 'mc.cores' which > lapply doesnt not have. > > I know when mc.cores =

Dear all, Let say I have following data-frame: Dat <- structure(list(dat = c(-0.387795842956327, -0.23270882099043, -0.89528973290562, 0.95857175595512, 1.61680582493783, -1.17738110289352, 0.210601060411423, -0.827369747447338, -0.36896112964414, 0.440288648776096, 1.28018410608809, -0.897113649961341, 0.342216546981718, -1.17288066266219, -1.57994101992621, -0.913655547602414,

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