similar to: What is 'freeny.x' object?

Dear all, I have a system of simultaneous equations with 2 unknowns as follows: x*y + (1-x) = 0.05 x*(y - .5)^2 + (1-x)*0.6 = 0.56^2 Ofcourse I can do it manually however wondering whether there is any direct way in R available to get the solution of this system? Thanks and regards, [[alternative HTML version deleted]]

> On 23 Aug 2017, at 20:51 , Ista Zahn <istazahn at gmail.com> wrote: > > On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: >> ummm, Ista, it's 2^n. > > ummm yes ughhhh. > You didn't really say otherwise: sum(choose(n,0:n)) == 2^n by the binomial expansion of (1+1)^n (but you knew that) This points to a different

ummm, Ista, it's 2^n. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Wed, Aug 23, 2017 at 8:52 AM, Ista Zahn <istazahn at gmail.com> wrote: > On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso >

On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ummm, Ista, it's 2^n. ummm yes ughhhh. My point is, if the number of groups is large, check it before hand. If you can check it without embarrassing yourself in public like I did that's even better. Best, Ista > > Cheers, > Bert > > > Bert Gunter > > "The trouble

Dear all, I was encountering with a typical Matching problem and was wondering whether R can help me to solve it directly. Let say, I have 2 vectors of equal length: vector1 <- LETTERS[1:6] vector2 <- letters[1:6] Now I need to match these 2 vectors with all possible ways like: (A,B,C,D,E) & (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) & (b,a,c,d,e), however there

Full thread here: https://github.com/tidyverse/broom/issues/287 Reproducible example: is.object(freeny$y) # [1] TRUE attr(freeny$y, 'class') # [1] "ts" class(freeny$y) # [1] "ts" # ts attribute wiped by model.frame class(model.frame(y ~ ., data = freeny)$y) # [1] "numeric" attr(model.frame(y ~ ., data = freeny)$y, 'class') # NULL # but still:

Dear all, let say, I have following list object: listObj <- vector("list", length = 3) listObj[[1]] <- rnorm(3) listObj[[2]] <- rnorm(4) listObj[[3]] <- rnorm(5) Now I want to convert above list into a Matrix. Ofcourse I can do it using "Reduce("rbind", listObj)". However as you notice that as elements of that list are arbitrary length vectors,

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 ==

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

Hi Say I want to add manually an intercept in the function lm. Even if almost all results will be identical, few stats are different as DF counting will be different as intercept will not be included in "automatic" case, while it will be in "manual" case. See: ###usual lm on freeny fr<-lm(freeny.y~freeny.x) ###manual lm on freeny man<-cbind(1,freeny.x)

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,

@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 data frame: > data.frame(x=rnorm(18), y=rep(c("a", "b", "c"), each=6)) x y 1 -1.072152537 a 2 0.382985265 a 3 0.058877377 a 4 -0.006911939 a 5 -2.355269051 a 6 -0.303095553 a 7 0.484038422 b 8 0.733928931 b 9 -1.136014346 b 10 0.503552090 b 11 1.708609658 b 12 -0.294599403 b 13

Hi My goal is to do a (multiple) regression, just knowing that my Y variables will be the say k first variables of a matrix/data frame. I thought I should do it with eval(parse)) but encounter a strange problem. See: lm(y~.-y, data=freeny) #that's what I want to do in the one equation case #Problem is I don't know name of the variable... only that it is the first one... #so idea is to

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! I would love to be able to include an external variable to a lm call, I mean something: if(TRUE) a<-freeny.x[,4] else a<-NULL lm(freeny.y~freeny.x[,-4] +a) but it does not work with a<-NULL, whereas lm(freeny.y~freeny.x[,-4] +NULL) I don't understand why and did not find an answer in the manuals... do you see it? Any idea? Thanks!!

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