similar to: Does Chi Square test for R differ from S-Plus?

Displaying 20 results from an estimated 300 matches similar to: "Does Chi Square test for R differ from S-Plus?"

2002 Aug 05
2
options(digits) (PR#1879)
[this message needed manual improvement by the mailing list administrator since it was `HTMLified' .. ``please do not''] Apologies for bothering you about a fairly trivial matter. I have been getting some inconsistencies with the display digits in R V1.5. I have been using the hypergeometric distribution function, and have found that when printing out the results from this
2005 Mar 22
1
error with polr()
Dear Sir, I get an error message when I use polr() in MASS package. My data is "ord.dat". I made "y" a factor. y y1 y2 x lx 1 0 0 0 3.2e-02 -1.49485 2 0 0 0 3.2e-02 -1.49485 3 0 0 0 1.0e-01 -1.00000 4 0 0 0 1.0e-01 -1.00000 5 0 0 0 3.2e-01 -0.49485 6 0 0 0 3.2e-01 -0.49485 7 1 1 0 1.0e+00 0.00000 8 0 0 0 1.0e+00 0.00000 9 1 1 0
2007 Oct 11
1
constraining correlations
Hello, I've searched for an answer to no avail. I am wondering if anyone knows how to constrain certain correlations to be equal. I have family data with 2 twins per family plus up to 2 siblings. I would like to somehow constrain all the sibling correlations (twin-sib and sib-sib) to be the same while allowing the twin-twin correlation to be different. Here is some simulated code:
2009 Nov 28
1
R function that duplicates Octave's poly function?
By any chance is anyone aware of an R function that duplicates Octave's poly function? Here is a description of Octave's poly function: Function File: poly (A) If A is a square N-by-N matrix, `poly (A)' is the row vector of the coefficients of `det (z * eye (N) - a)', the characteristic polynomial of A. As an example we can use this to find the eigenvalues
2011 Apr 29
1
question of VECM restricted regression
Dear Colleague I am trying to figure out how to use R to do OLS restricted VECM regression. However, there are some notation I cannot understand. Please tell me what is 'ect', 'sd' and 'LRM.dl1 in the following practice: #OLS retricted VECM regression data(denmark) sjd <- denmark[, c("LRM", "LRY", "IBO", "IDE")] sjd.vecm<-
2020 May 12
4
S3 method dispatch for methods in local environments
Dear All, In R 3.6.3 (and earlier), method dispatch used to work for methods stored in local environments that are attached to the search path. For example: myfun <- function(y) { out <- list(y=y) class(out) <- "myclass" return(out) } print.myclass <- function(x, ...) print(formatC(x$y, format="f", digits=5)) myfun(1:4) # prints: [1]
2011 Aug 06
1
help with predict for cr model using rms package
Dear list, I'm currently trying to use the rms package to get predicted ordinal responses from a conditional ratio model. As you will see below, my model seems to fit well to the data, however, I'm having trouble getting predicted mean (or fitted) ordinal response values using the predict function. I have a feeling I'm missing something simple, however I haven't been able to
2010 May 11
1
how to extract the variables used in decision tree
HI, Dear R community, How to extract the variables actually used in tree construction? I want to extract these variables and combine other variable as my features in next step model building. > printcp(fit.dimer) Classification tree: rpart(formula = outcome ~ ., data = p_df, method = "class") Variables actually used in tree construction: [1] CT DP DY FC NE NW QT SK TA WC WD WG WW
2008 Oct 01
3
"tapply versus by" in function with more than 1 arguments
Hi. I searched the list and didn't found nothing similar to this. I simplified my example like below: #I need calculate correlation (for example) between 2 columns classified by a third one at a data.frame, like below: #number of rows nr = 10 #the third column is to enforce that I need correlation on two variables only dataf =
2017 Jul 28
3
problem with "unique" function
I have the joint distribution of three discrete random variables z1, z2 and z3 which is captured by "z" and "prob" as described below. For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. Also, the probability adds up to 1. > head(z) z1 z2 z3 [1,] -0.46667 -1.0000 -1.0000 [2,] -0.33333 -0.9333 -0.9333 [3,] -0.20000 -0.8667 -0.8667
2018 May 15
2
source line number for instruction
Hi, I want to retrieve a function from a source line-number, is this something possible to achieve? I tried the following code snippet with LLVM-7 but it doesn't seem to workhttp://llvm.org/docs/SourceLevelDebugging.html#ccxx_frontend <http://llvm.org/docs/SourceLevelDebugging.html#ccxx_frontend:> Is there specific documentation I can refer to implement this? Thanks -- *Rtr. PP
2005 Oct 14
1
Predicting classification error from rpart
Hi, I think I'm missing something very obvious, but I am missing it, so I would be very grateful for help. I'm using rpart to analyse data on skull base morphology, essentially predicting sex from one or several skull base measurements. The sex of the people whose skulls are being studied is known, and lives as a factor (M,F) in the data. I want to get back predictions of gender, and
2018 May 15
0
source line number for instruction
Not sure what you tried/how it didn't work - could you explain more? You could scan through all the llvm::Functions in an llvm::Module and look at the debug info associated with them (getSubprogram) then check the location of that debug info. - Dave On Tue, May 15, 2018 at 12:46 AM Ridwan Shariffdeen via llvm-dev < llvm-dev at lists.llvm.org> wrote: > Hi, > > I want to
2011 Dec 23
1
Help creating a symmetric matrix?
Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal -
2017 Jul 28
0
problem with "unique" function
Most likely, previous computations have ended up giving slightly different values of say 0.13333. A pragmatic way out is to round to, say, 5 digits before applying unique. In this particular case, it seems like all numbers are multiples of 1/30, so another idea could be to multiply by 30, round, and divide by 30. -pd > On 28 Jul 2017, at 17:17 , li li <hannah.hlx at gmail.com> wrote:
2018 May 24
1
Getting variable names from LLVM Pass
Hi Michael, Thanks for the help, it seems like you said its not going to be easy. But I will have a try at this, thank you for the link to LibTooling. Best On Mon, May 21, 2018 at 6:26 PM Dean Michael Berris <dean.berris at gmail.com> wrote: > > > > On 21 May 2018, at 18:38, Ridwan Shariffdeen via llvm-dev < > llvm-dev at lists.llvm.org> wrote: > > > >
2009 Apr 05
4
extract the p value of F statistics from the lm class
Dear R users I have run an regression and want to extract the p value of the F statistics, but I can find a way to do that. x<-summary(lm(log(RV2)~log(IV.m),data=b)) Call: lm(formula = log(RV2) ~ log(IV.m), data = b[[11]]) Residuals: Min 1Q Median 3Q Max -0.26511 -0.09718 -0.01326 0.11095 0.29777 Coefficients: Estimate Std. Error t value Pr(>|t|)
2018 May 21
2
Getting variable names from LLVM Pass
Hi, I want to retrieve the variable names used in a statement, I tried the following snippet, but it only gives me the variable named in llvm bitcode. I need the variable name in source code. for (auto op = I.op_begin(); op != I.op_end(); op++) { Value* v = op->get(); StringRef name = v->getName(); } Is there specific documentation I can
2010 Jun 16
2
Accessing the elements of summary(prcomp(USArrests))
Hello again, I was hoping one of you could help me with this problem. Consider the sample data from R: > summary(prcomp(USArrests)) Importance of components: PC1 PC2 PC3 PC4 Standard deviation 83.732 14.2124 6.4894 2.48279 Proportion of Variance 0.966 0.0278 0.0058 0.00085 Cumulative Proportion 0.966 0.9933 0.9991 1.00000 How do I access the
2008 Feb 07
1
Don't understand removing constant on 1-way ANOVA
I am playing with the a 1-way anova with and without the "-1" option. I have a simple cooked up example below but it behaves the same on a more complex real example. From what I can tell: 1) the estimated means of the different levels are correctly estimated either way (although reported as means with the -1 and as contrasts without the -1 as expected) 2) the residuals are