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> -----Original Message----- > From: NICOLAS DEIG > > hello, > > i am encoutering problems with a function of R concerning the > classification trees. > > In the library {tree}, when I use the function "tree" to grow a ^^^^^^^ That should be "package"... > classification tree it should give me an object of class >

Dear friends, After running the lm() model, we can get summary resluts like the following: Coefficients: Estimate Std. Error t value Pr(>|t|) x1 0.11562 0.10994 1.052 0.2957 x2 -0.13879 0.09674 -1.435 0.1548 x3 0.01051 0.09862 0.107 0.9153 x4 0.14183 0.08471 1.674 0.0975 . x5 0.18995 0.10482 1.812 0.0732 . x6 0.24832 0.10059 2.469 0.0154 * x7

hi, I am encoutering a very little problem that seemed to be so easy to solve.... I need to divide the array > A<-c(1:200) into two subsets at random. Therefore I use the function "sample" in R: > S<-sample(A,100) for a random sample of size 100. Then I need the values in A that are not selected in S to be put in another array, there is my problem! Is there anyway to do

Dear R-list, I'm not sure what I've found about a function in DAAG package is a bug. When I was using cv.lm(DAAG) , I found there might be something wrong with it. The problem is that we can't use it to deal with a linear model with more than one predictor variable. But the usage documentation hasn't informed us about this. The code illustrates my discovery: > library(DAAG)

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I

hello r-helpers: there is a .txt file: x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 y1 17 5 77 18 19 24 7 24 24 72 52 100 2 6 72 18 17 15 4 12 18 35 42 97.2 17 2 58 10 5 3 4 3 3 40 28 98 17 2 69 14 13 12 4 6 6 50 37 93 2 3 75 20 38 18 6 12 18 73 67 99 14 4 59 16 18 9 4 3 15 47 40 99.95 17 4 87 18 17 12 4 15 12 69 46 100 14 3 74 15 9 12 1 15 12 44 35 98 17 6 76 15 33 21 15 9 18 46 41 100 17 5 76 17 22 18 1

Dear R users, I’m trying to understand how correlated predictors impact the Relative Importance measure in Stochastic Boosting Trees (J. Friedman). As Friedman described “ …with single decision trees (referring to Brieman’s CART algorithm), the relative importance measure is augmented by a strategy involving surrogate splits intended to uncover the masking of influential variables by others

Hello, I have the following data frame (DF): V5 V5.1 V5.2 V5.3 V5.4 V5.5 2 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183 3 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183 4 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183 5 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677

Dette er en melding med flere deler i MIME-format. --=_alternative 004C4E4A00257091_= Content-Type: text/plain; charset="US-ASCII" Yes. so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? and there is a difference in (x1*x2*x3*x4*x5*x6*x7*x8)^2 and (x1*x2*x3*x4*x5*x6*x7*x8) althoug the resulting formulas are the same, or? This fikses my problem, but R still crashes for the

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > And some more informastion I forgot. > R does not crash if I write out the formula: > > set.seed(123) > x1 <- runif(1000) > x2 <- runif(1000) > x3 <- runif(1000) > x4 <- runif(1000) > x5 <- runif(1000) > x6 <- runif(1000) > x7 <- runif(1000) > x8 <- runif(1000) > y <-

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > Yes. > so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? Yes in the sense that the simplified formula given by terms() is the same. > and there is a difference in > (x1*x2*x3*x4*x5*x6*x7*x8)^2 > and > (x1*x2*x3*x4*x5*x6*x7*x8) > althoug the resulting formulas are the same, or? The first is reduced to the

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Hey all, I learned that using the equals sign "=" to assign objects is generally OK, but will not work in some cases. As I always use "<-" for assignments, I have not encoutered any problems. Could somebody provide an example or explanation, why getting used to "=" is not a good idea? Or is it? Thanks ahead, Berry ------------------------------------- Berry

On 29/03/2020 14:31, Patrick DERWAEL wrote: > Hi Phil > > Here it is: > > > [root at totorbex ~]# lspci -nn > .00:00.0 Host bridge [0600]: Intel Corporation Atom/Celeron/Pentium > Processor x5-E8000/J3xxx/N3xxx Series SoC Transaction Register [8086:2280] > (rev 21) > 00:02.0 VGA compatible controller [0300]: Intel Corporation > Atom/Celeron/Pentium Processor

Hi List, Being new to R, I am trying to apply boot.stepAIC() for Model selection by bootstrapping the stepAIC() procedure. I had gone through the discussion in various thread on the variable selection methods. Understood the pros and cons of various method, also going through the regression modelling strategies in rms. I want to read Final model or Formula or list of variables automatically

> From: Peter Dalgaard > > Hallgeir.Grinde at elkem.no writes: > > > Dette er en melding med flere deler i MIME-format. > > --=_alternative 004613C000257091_= > > Content-Type: text/plain; charset="US-ASCII" > > > > And some more informastion I forgot. > > R does not crash if I write out the formula: > > > > set.seed(123)

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

Hi Phil Here it is: [root at totorbex ~]# lspci -nn .00:00.0 Host bridge [0600]: Intel Corporation Atom/Celeron/Pentium Processor x5-E8000/J3xxx/N3xxx Series SoC Transaction Register [8086:2280] (rev 21) 00:02.0 VGA compatible controller [0300]: Intel Corporation Atom/Celeron/Pentium Processor x5-E8000/J3xxx/N3xxx Integrated Graphics Controller [8086:22b1] (rev 21) 00:10.0 SD Host controller

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