search for: xpred

..."printPredicateOperand"; let ParserMatchClass = CondCodeOperand; let DecoderMethod = "DecodePredicateOperand"; } James On Mon, 14 Dec 2015 at 13:44 Sky Flyer <skylake007 at googlemail.com> wrote: > But one question! > > imagine I define *cond* as a type of Xpred (Xpred:$cond) > and in the instruction, for instance *bits<6> cond*. > How can I assign the first i32imm to the 4 MSB of cond and the second > i32imm to the 2 LSB? :-/ > > Now: > > Xpred:$cond > bits<6> cond; > Inst{5-0} = cond; > > Desired: > > Xp...

Hello James, that was also what I've planned to do but just wasn't sure. Thanks for that. On Mon, Dec 14, 2015 at 11:52 AM, James Molloy <james at jamesmolloy.co.uk> wrote: > Hi, > > You can't nest operands like that - it must be a flattened list. So: > > def *Xpred* : PredicateOperand<OtherVT, (ops *i32imm, i32imm*, i32imm), > (ops (i32 14), (i32 zero_reg))> {...} > > On Mon, 14 Dec 2015 at 10:21 Sky Flyer via llvm-dev < > llvm-dev at lists.llvm.org> wrote: > >> Hi All, >> >> In ARMInstFormats.td predicate is defin...

Dear R users, I'm working with the rpart package and want to evaluate the performance of user defined split functions. I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. If I am correct this means that for xpred.rpart, xval denots the number V of one V-fold cross validation while for rpart.control it denotes the numbe...

I just updated rpart to the latest version (3.1.0). There are a number of changes between this and previous versions, and some of the code I've been using with earlier versions (e.g. 3.0.2) no longer work. Here is a simple illustration of a problem I'm having with xpred.rpart. iris.test.rpart<-rpart(iris$Species~., data=iris[,1:4], parms=list(prior=c(0.5,0.25, 0.25))) + ) > xpred.rpart(iris.test.rpart) Error in as.double.default(fit$parms) : (list) object cannot be coerced to vector type 14 > The problem seems to be with the parms parameter in the ca...

R-users E-mail: r-help@r-project.org Hi! R-users. http://finzi.psych.upenn.edu/R/library/mvpart/html/xpred.rpart.html says: data(car.test.frame) fit <- rpart(Mileage ~ Weight, car.test.frame) xmat <- xpred.rpart(fit) xerr <- (xmat - car.test.frame$Mileage)^2 apply(xerr, 2, sum) # cross-validated error estimate # approx same result as rel. error from printcp(fit) apply(xerr, 2, sum)/var(car.t...

...predicate is defined this way: *def pred : PredicateOperand<OtherVT, (ops i32imm, i32imm),* *(ops (i32 14), (i32 zero_reg))> {...}* I use the same definition in my code. But I have another version of predicate which is exactly the same but it is a condition code plus a quantifier! (e.g. Xpred = (pred + i32imm)). I was wondering how we can define a sub sub operand, something like this: def *Xpred* : PredicateOperand<OtherVT, (ops *pred*, i32imm), (ops (i32 14), (i32 zero_reg))> {...} I don't know how clear I explained, but can someone recommend a solution? Cheers, ES ------...

....00686 ... $ lag(ret3) : num NA 0.0278 0.0152 0.0211 -0.0104 ... $ (lag(ret1)*lag(ret2)): num NA 2.13e-04 0.00 4.43e-05 1.15e-04 ... $ (lag(ret1)*lag(ret3)): num NA 4.32e-04 2.21e-04 7.03e-05 -1.75e-04 ... $ (lag(ret2)*lag(ret3)): num NA 3.82e-04 0.00 2.81e-04 -7.16e-05 ... > ar1.xpred.fitted <- regFit(ar1.xpred.model, data = regdata, use = "lm") > ar1.xpred.train.pred <- predict(ar1.xpred.fitted, regdata, se.fit = FALSE) > ar1.xpred.test.pred <- predict(ar1.xpred.fitted, regdata.test, se.fit = FALSE) Fehler: variable 'lag(ret1)' was fitted wi...

Hello! I'm still working on my problem, which also occurs with the predict.lm() function. - Providing newdata, which is a data.frame with all variables being "numeric", as str() shows, R tells me the following: ar1.xpred.test.pred <- predict(ar1.xpred.fitted, regdata.test, se.fit = FALSE) Fehler: variable 'lag(ret1)' was fitted with type "numeric" but type "nmatrix.1" was supplied Zus?tzlich: Warnmeldung: 'newdata' had 23 rows but variable(s) found have 89 rows Estimating a...

...ing that 'newdata' had 1 row but variable(s) found have 3 rows. (if I run this outside of rollapply I don't get this warning) Also, I don't see the predicted value or its se with print(fm2[[1]]). Again, if I run this outside of rollapply I am able to extract the predicted value. Xpred=c(70.67) myfn2 = function(mydata){ dd = as.data.frame(mydata) l = lm(dd[,1]~dd[,2], data=dd) c = coef(l) p = predict(l, data.frame(Xvar=Xpred),se=T) ret=c(l,c,p) } fm2 = rollapply(data.z, width=3, FUN= myfn2, by.column = FALSE, align = "right")...

Dear R users, I know cross-validation does not work in rpart with user defined split functions. As Terry Therneau suggested, one can use the xpred.rpart function and then summarize the matrix of the predicted values into a single "goodness" value. I need only a confirmation: set for example xval=10, if I correctly understood a single column of the matrix obatined by xpred.rpart gives (for a cp level), for each of the 10 groups o...

Hi, I created an rpart object and pruned the tree using 1-SE rule. I used 10-fold cross validation while creating the tree. Then, I extracted the cross-validated predictions for my data points using xpred.rpart and obtained some statistics like precision, recall, overall error rate, etc. However, these values change each time I run xpred.rpart because of the random shuffling going on before cross validation (I think so). What should I do in this case? I am inclined to treat them as random variables...

I am trying to find out what type of sampling scheme is used to select the 10 subsets in 10-fold cross-validation process used in rpart to choose the best tree. Is it simple random sampling? Is there any documentation available on this? Thanks, Penny. -- View this message in context: http://r.789695.n4.nabble.com/cross-validation-in-rpart-tp3389329p3389329.html Sent from the R help mailing list

...35 0 61 2 11.23 2 5 1 32 5.201916 0 54 2 11.35 2 6 1 33 6.22861 0 65 2 11.35 2 5 1 Here we have 33 observations and 1 event. The ?estimated rate? is 0.1225562. My questions are: (1) Is the ?estimated rate? the estimated hazard rate ratio? (2) How does rpart calculate this rate? (3) Suppose I use xpred.rpart(fit, xval=10) to perform 10-fold cross-validation using (a) the complete stagec data set and (b) only a subset of it, say, using the columns Age, EET, and G2 only. For the i-th patient, I am likely to obtain a different estimated rate. How can I meaningfully compare both rates? How can say wh...

Hi, I have done an analysis using 'rpart' to construct a Classification Tree. I am wanting to retain the output in tree form so that it is easily interpretable. However, I am wanting to compare the 'accuracy' of the tree to a Random Forest to estimate how much predictive ability is lost by using one simple tree. My understanding is that the error automatically displayed by the two

Hello list, I'm having a problem with custom functions in rpart, and before I tear my hair out trying to fix it, I want to make sure it's actually a problem. It seems that, when you write custom functions for rpart (init, split and eval) then rpart no longer cross-validates the resulting tree to return errors. A simple test is to use the usersplits.R function to get a simple, custom

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Greetings, I have two questions that I could not answer from the documentation. A - ecdf and confidence intervals : Is there a (simple) way to generate confidence intervals (95%) for a ecdf? B - cross-validation of rpart trees : a colleague is using S to generate decision tree and mentioned to me the use of cross-validation. Is this function enabled in R ? if so, how should one proceed to