search for: glnbrntt

...line(). Notice, incidentally, that > line(1:9+rnorm(9,,1e-1),1:9+rnorm(9,,1e-1)) Call: line(1:9 + rnorm(9, , 0.1), 1:9 + rnorm(9, , 0.1)) Coefficients: [1] -0.9407 1.1948 I.e., it is not likely an issue with exact integers or perfect fit. -pd > On 29 May 2017, at 07:21 , GlenB <glnbrntt at gmail.com> wrote: > >> Tukey divides the points into three groups, not the x and y values > separately. > >> I'll try to get hold of the book for a direct quote, might take a couple > of days. > > Ah well, I can't get it for a week. But the fact that i...

...line(1:9 + rnorm(9, , 0.1), 1:9 + rnorm(9, , 0.1)) >> >> Coefficients: >> [1] -0.9407 1.1948 >> >> I.e., it is not likely an issue with exact integers or perfect fit. >> >> -pd >> >> >> >>> On 29 May 2017, at 07:21 , GlenB <glnbrntt at gmail.com> wrote: >>> >>>> Tukey divides the points into three groups, not the x and y values >>> separately. >>> >>>> I'll try to get hold of the book for a direct quote, might take a couple >>> of days. >>> >>&gt...

Tukey divides the points into three groups, not the x and y values separately. I'll try to get hold of the book for a direct quote, might take a couple of days. On Mon, May 29, 2017 at 8:40 AM, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 27/05/2017 9:28 PM, GlenB wrote: > >> Bug: stats::line() does not produce correct Tukey line when n mod 6 is 2 >> or

...9,,1e-1),1:9+rnorm(9,,1e-1)) > Call: > line(1:9 + rnorm(9, , 0.1), 1:9 + rnorm(9, , 0.1)) > > Coefficients: > [1] -0.9407 1.1948 > > I.e., it is not likely an issue with exact integers or perfect fit. > > -pd > > > >> On 29 May 2017, at 07:21 , GlenB <glnbrntt at gmail.com> wrote: >> >>> Tukey divides the points into three groups, not the x and y values >> separately. >> >>> I'll try to get hold of the book for a direct quote, might take a couple >> of days. >> >> Ah well, I can't get it fo...

..., 0.1)) >>> >>> Coefficients: >>> [1] -0.9407 1.1948 >>> >>> I.e., it is not likely an issue with exact integers or perfect fit. >>> >>> -pd >>> >>> >>> >>>> On 29 May 2017, at 07:21 , GlenB <glnbrntt at gmail.com> wrote: >>>> >>>>> Tukey divides the points into three groups, not the x and y values >>>> separately. >>>> >>>>> I'll try to get hold of the book for a direct quote, might take a couple >>>> of days....

...rect quote, might take a couple of days. Ah well, I can't get it for a week. But the fact that it's often called Tukey's three group line (try a search on *tukey three group line* and you'll get plenty of hits) is pretty much a giveaway. On Mon, May 29, 2017 at 2:19 PM, GlenB <glnbrntt at gmail.com> wrote: > Tukey divides the points into three groups, not the x and y values > separately. > > I'll try to get hold of the book for a direct quote, might take a couple > of days. > > > > On Mon, May 29, 2017 at 8:40 AM, Duncan Murdoch <murdoch.dunc...

For the life of me I couldn't work out what to searc I have an m*n numeric matrix x and a numeric vector y (of length m+n-1) How do I do a calculation like this? z[i,j] = x[i,j] * y[i+j] ? Well, one can write a pair of loops, or write a single loop within which we calculate a vector at a time, but ... is there a "neat" way to do it? tiny example:

Bug: stats::line() does not produce correct Tukey line when n mod 6 is 2 or 3 Example: line(1:9,1:9) should have intercept 0 and slope 1 but it gives intercept -1 and slope 1.2 Trying line(1:i,1:i) across a range of i makes it clear there's a cycle of length 6, with four of every six correct. Bug has been present across many versions. The machine I just tried it on just now has R3.2.3:

Is the kolmogorov-smirnov test valid on both continuous and discrete data? I don't think so, and the example below helped me understand why. A suggestion on testing the discrete data would be appreciated. Thanks, a <- rnorm(1000, 10, 1);a # normal distribution a b <- rnorm(1000, 12, 1.5);b # normal distribution b c <- rnorm(1000, 8, 1);c # normal distribution c d <- rnorm(1000,

Assuming I have a matrix of data (or under some restrictions that will become obvious, possibly a data frame), I want to be able to apply a list of functions (initially producing a single number from a vector) to the data and produce a data frame (for compact output) with column 1 being the function results for the first function, column 2 being the results for the second function and so on - with

Hi everyone! #I need a loop or a function that creates a X2 variable that is X1 without the extreme values (or X1 winsorized) by industry and year. #My reproducible example: firm<-sort(rep(1:1000,10),decreasing=F) year<-rep(1998:2007,1000) industry<-rep(c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10),rep(6,10),rep(7,10),rep(8,10),rep(9,10), rep(10,10)),1000) X1<-rnorm(10000)

I found the old ace function (from acepack) valuable a couple of years ago in helping to find a transformation of the response to approximate additivity in smooth functions of the predictors. ace used alternating conditional expectations, but I'm not overly fixated on algorithms as long as it works pretty well. Is there another package that has that sort of functionality? I haven't been

I tried searching but I couldn't quite find what I was looking for. Here's a dummy data matrix (with row and column labels): > y 0 1 2 3 4 21 3 4 8 5 5 22 3 6 8 6 NA 23 4 5 11 4 3 24 4 2 1 4 6 25 6 4 4 6 6 I can get cumulative row sums as follows: > cy<-t(apply(y,1,cumsum)) > cy 0 1 2 3 4 21 3 7 15 20 25 22 3 9 17 23 NA 23 4 9 20 24 27 24 4 6 7 11 17

I have an additive model of the following form : zmdlfit <- lm(z~ns(x,df=6)+ns(y,df=6)) I can get the fitted values and plot them against z easily enough, but I also want to both obtain and plot the two additive components (the estimates of the two additive terms on the RHS) I've been looking at manuals and searching on the internet and searching the archives, but I'm apparently