similar to: Seeking help for outomating regression (over columns) and storing selected output

I'm trying to use boot.stepAIC for feature selection; I need to be able to specify the name of the dependent variable programmatically, but this appear to fail: In R-Studio with MS R Open 3.4: library(bootStepAIC) #Fake data n<-200 x1 <- runif(n, -3, 3) x2 <- runif(n, -3, 3) x3 <- runif(n, -3, 3) x4 <- runif(n, -3, 3) x5 <- runif(n, -3, 3) x6 <- runif(n, -3, 3) x7

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

Failed? What was the error message? 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 Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka

How to define bounds for a semi continous variable in lp_solve. Min 5x1 +9x2 +7.15x3 +0.1x4 subject to x1+x2+x3+x4=6.7 x1+x4 <= 6.5 And x3 can be 0 or greater than 3.6 hence x3 is a semi continous variable how to define bounds as well as semicontinous function because using set.semicont and set. bound simantaneously doesn't seem to work.Thanks in advance for the help -- View this

I have been wrestling with this function for quite a while, and am not making headway. 1) I want to apply a function to the following columns of a dataframe: myfunction. <- apply(ph5028[,c(83:107)],2,function(x) ... 2) Within each of the above columns there is a single numeric code, 1, 2 or 3 or an NA. 3) My goal is to determine the percent of time each person used a 2 code. So if a person

Until I get a fix that works, a work-around would be to rename the 'y1' column, used a fixed formula, and rename it back afterwards. Thanks for your help. SGO. -----Original Message----- From: Bert Gunter [mailto:bgunter.4567 at gmail.com] Sent: 22 August 2017 20:38 To: Stephen O'hagan <SOhagan at manchester.ac.uk> Cc: r-help at r-project.org Subject: Re: [R] boot.stepAIC

It seems that if you build the formula as a character string, and postpone the "as.formula" into the lm call, it works. instead of frm1 <- as.formula(paste(trg,"~1")) use frm1a <- paste(trg,"~1") and then strt <- lm(as.formula(frm1a),dat) regards, Heinz Stephen O'hagan wrote/hat geschrieben on/am 23.08.2017 12:07: > Until I get a fix that works, a

Hello R experts, The following problem outstrips my current programming knowledge. I have a dataframe with two fields that looks like the following: ID Contract 01 1 01 1 02 2 02 3 02 1 03 2 03 2 03 2 03 1 03 1 03 1 etc... I would like to end up with a dataframe with one row per ID where the value in the contract field would be the

Dear R Masters, I have searched high and low (the help archives and my various R reference material and help files) for a solution to what appears to me to be quite a simple problem. In the following syntax, variable n10 has three levels. I would like the symbols that appear in the graph for these three levels to be different colors. The best I have been able to do is to have the Key display

OK, here's the problem. Continuing with your example: strt1 <- lm(y1 ~1, dat) strt2 <- lm(frm1,dat) > strt1 Call: lm(formula = y1 ~ 1, data = dat) Coefficients: (Intercept) 41.73 > strt2 Call: lm(formula = frm1, data = dat) Coefficients: (Intercept) 41.73 Note that the formula objects of the lm object are different: strt2 does not evaluate the formula. So

Hello R-Masters, I have a list 's' with three elements, as shown below. I want to recode a.a, a.a2, and a.a3 to NA if the value in a.a is less than 3. I reivewed my Modern Applied Statistic Book, the online help and did some searching of R-help on the internet. I explored unlist and as.list.data.frame in an attempt to isolate the third element of the list s, but this was not helpful.

Hello, is it possible to setup a model/table schema like this: Groupable --> Membership <-- Group ^ ^ | | User UserGroup I tried the following but failed: Groupable (table with ''type'' column) has_many :memberships, :as => :groupable has_many :groups, :through => :memberships

Hello, I have a data frame that has three fields. Resp# ActCode ProdUsed 100 3 2 100 3 2 100 4 3 100 4 3 101 3 6 102 2 1 102 3 1 103 5 1 103 5 1 103

Hello R masters, This request is being made after searching through archive mailing list. I have a "many" table in a medical application where each injury a person suffers is recorded as a separate row. In my toy example below it means respondent #1 had three injuries and so on. If x2 equal 2 then I want to delete all the rows for that person from the dataframe--see Before and After

I''m trying to understand these results in irb, which implies that there''s some kind of proxy at play: >> aaron = User.find_by_login ''aaron'' => #<User:0xb7109cdc @attributes={"salt"=>"7e3041ebc2fc05a40c60028e2c4901a81035d3cd", "updated_at"=>nil, "contact_info_id"=>nil,

Hello, In the Hmisc package, functions describe and summarize can explicitly take a weight variable. My question is can a weight variable be applied when using 'summary'? For example, using...summary(var1 ~ var2) I would like to weight the data by var 3 (same length). Is this possible? Thanks a lot. Greg Blevins The Market Solutions Group, Inc.

This is a "long" way; i.e., not necessarily efficient: > qs2 [1] 2 1 1 4 4 4 1 1 1 4 2 4 3 1 4 3 3 2 4 3 > qs9 [1] 4 4 1 3 4 3 1 3 1 4 1 2 3 3 4 4 1 4 2 3 > decision <- function(a, b) { + if (a == 1 || b == 1) return(1) + if (a == 2 || b == 2) return(2) + if (a == 3 || b == 3) return(3) + if (a == 4 || b == 4) return(4) + NA + } > mapply(decision,

hey all, I have three tables like this: forum (id,title) usergroup(id,title) forum_perms(usergroup_id,forum_id,read,write,post) is there a way to deal with that kind of relation with rails such as by using has_and_belongs_to_many kind of stuff? thanx in advance Pat --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups

I the following three models which all have has_and_belongs_to_many # User <-> UserGroup <-> Permissions class UserGroup < ActiveRecord::Base has_and_belongs_to_many :users, :join_table => "user_usergroup_join" has_and_belongs_to_many :permissions, :join_table => "usergroup_permission_join", :uniq => true end I can do this: permissions =