search for: xlev

I am trying once again. By just changing f <- match(xlevs[f], nlevs) to f <- match(xlevs, nlevs)[f] , function 'factor' in R devel could be made more consistent and back-compatible. Why not picking it? -------------------------------------------- On Sat, 25/11/17, Suharto Anggono Suharto Anggono <suharto_anggono at yahoo.com> wrote: S...

...ade ago there was a problem with model.frame when the variable names were long: https://stat.ethz.ch/pipermail/r-help/2002-August/024492.html I have similar symptoms with R 2.15.3 on Windows 7: Browse[2]> x <- model.matrix(formula(myform), p$data) Error in model.frame.default(object, data, xlev = xlev) (from mice.R#601) : object is not a matrix My attempt at a work-around did not help: Browse[2]> x <- model.matrix(formula(myform), as.matrix(p$data)) Error in model.frame.default(object, data, xlev = xlev) (from mice.R#601) : 'data' must be a data.frame, not a matrix or...

...devel, I saw attempts to speed up subsetting and 'match', and to cache results of conversion of small nonnegative integers to character string. That's good. I am sorry for pushing, still. Is the partial new behavior of function 'factor' with respect to NA really worthy? match(xlevs, nlevs)[f] looks nice, too. - Using f <- match(xlevs, nlevs)[f] instead of f <- match(xlevs[f], nlevs) for remapping - Remapping only if length(nlevs) differs from length(xlevs) Applying changes similar to above to function 'levels<-.factor' will not change 'levels<-.fact...

...ata.frame( y=runif( 3), disc=factor( c( 'cat', 'dog', 'cat'), levels=c( 'cat', 'dog', 'earwig'))) test> lm.predbug.2_ lm( y~disc, data=scrunge.data.2) test> predict(lm.predbug.2, newdata=scrunge.data.2) Error in model.frame.default(object, data, xlev = xlev) : factor disc has new level(s) earwig A cure for this seems to be to add the commented line below towards the end of `model.frame.default': <<...>> if (length(xlev) > 0) { for (nm in names(xlev)) if (!is.null(xl <- xlev[[nm]])) {...

...e in the global environment and variables that are in the data frame given with the formula, but not calculation results from the intermediate calling environment. Results from traceback(): 9: nrow(aus) 8: eval(expr, envir, enclos) 7: eval(predvars, data, env) 6: model.frame.default(object, data, xlev = xlev) 5: model.frame(object, data, xlev = xlev) 4: model.matrix.default(frml, data, ...) 3: model.matrix.formula(1:nrow(aus) ~ ., data = aus) 2: model.matrix(1:nrow(aus) ~ ., data = aus) 1: fac.design(nlevels = c(2, 6, 2)) If I reset model.matrix.formula to model.matrix.default, the problem disa...

Hi, I'm trying to get sparse.model.matrix to retain unused levels. I can't seem to get this working through the most obvious routes such as specifying drop.unused.levels = FALSE in the model.frame or trying to pass all levels in xlev,which is an argument to sparse.model.matrix (see code below). Any help would be gratefully received. Cheers, Jarrod fac<-factor(rep(1:10,10), levels=1:100) mf<-model.frame(~fac, drop.unused.levels = TRUE) mf2<-model.frame(~fac, drop.unused.levels = FALSE) dim(model.matrix(~fac, mf)...

I'm working on predict.survreg and am confused about xlevels. The model.frame method has the argument, but none of the standard methods (model.frame.lm, model.frame.glm) appear to make use of it. The documentation for model.matrix states: xlev: to be used as argument of model.frame if data has no "terms" attribute. But the terms attribute ha...

...lt;- "model.frame" mf <- eval(lmcall, parent.frame()) xvars <- as.character(attr(Terms, "variables"))[-1] if ((yvar <- attr(Terms, "response")) > 0) xvars <- xvars[-yvar] if (length(xvars) > 0) { xlev <- lapply(mf[xvars], levels) xlev <- xlev[!sapply(xlev, is.null)] } else xlev <- NULL resp <- model.response(mf) qtx <- model.matrix(Terms, mf, contrasts) cons <- attr(qtx, "contrasts") dnx <- colnames(qtx) a...

...incorrect fitted values in version 1.5.1 and an error in 1.6.0 Error in "contrasts<-"(*tmp*, value = "contr.treatment") : contrasts apply only to factors In addition: Warning message: variable ihalf is not a factor in: model.frame.default(object, data, xlev = xlev) The problem in 1.5.1 seems to be model.matrix return inconsistently ordered interaction terms. There are abviously several work arounds. ### Start of example dfr <- expand.grid(i=0:154,j=0:154) dfr <- dfr[with(dfr,i+j<=154),] dfr$logj <- with(dfr,log(j+0.5)) dfr$jmin6 <-...

My idea (like in https://bugs.r-project.org/bugzilla/attachment.cgi?id=1540 ): - For remapping, use f <- match(xlevs, nlevs)[f] instead of f <- match(xlevs[f], nlevs) (I have mentioned it). - Remap only if length(nlevs) differs from length(xlevs) . On use of 'order' in function 'factor' in R devel, factor.Rd still says 'sort.list' in "Details" section. My comments on the p...

...gt; mf <- eval(lmcall, parent.frame()) > xvars <- as.character(attr(Terms, "variables"))[-1] > if ((yvar <- attr(Terms, "response")) > 0) > xvars <- xvars[-yvar] > if (length(xvars) > 0) { > xlev <- lapply(mf[xvars], levels) > xlev <- xlev[!sapply(xlev, is.null)] > } else xlev <- NULL > resp <- model.response(mf) > qtx <- model.matrix(Terms, mf, contrasts) > cons <- attr(qtx, "contrasts") > dnx...

...ot;),levels=c("blue","green","yellow")))) y x1 x2 1 0.90 1 blue 2 2.05 2 blue 3 3.02 3 green 4 NA 4 green 5 5.20 NA green > fit<-lm(y~x1+x2,data=data,na.action=na.exclude) > predict(fit,data) Error in model.frame.default(object, data, xlev = xlev) : factor x2 has new level(s) yellow Interpretation: Since level "yellow" was not used (is in some sense missing) in the data, predict.lm blocks. This should not happen. Maybe a warning should be given. But predict.lm should not quit with error. Here is problem 3: > print(...

...> fit.lm <- lm(y ~ x + z, data=xyz, subset=(z != "C")) > fit.glm <- glm(y ~ x + z, family=gaussian, data=xyz, subset=(z != "C")) > zz <- z[1:10] > xx <- rnorm(10) > predict(fit.lm,data.frame(x=xx,z=zz)) Error in model.frame.default(formula, data, xlev = xlev) : factor z has new level(s) C > predict(fit.glm,data.frame(x=xx,z=zz)) 1 2 3 4 5 6 2.570970 -18.007372 18.108771 12.498562 9.566029 10.518460 7 8 9 10 -9.132206 11.440242 -16.054621 -3.0...

When I alter the levels of a factor, why does it alter the names too? f <- factor(c(A="one",B="two",C="one",D="one",E="three"), levels=c("one","two","three")) names(f) -- gives [1] "A" "B" "C" "D" "E" levels(f) <-

...nd then I would like to predict the response variable (Throughput) for unseen factor levels. When I try to predict I get the following error: > throughput.pred <- predict(throughput.fit,experiments,type="response") Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : factor 'No_databases' has new level(s) 200, 400, 600, 800, 1000 Of course these are new factor levels, it is exactly what I am trying to achieve i.e. extrapolate the values of Throughput. Can anyone please advice? Below I include all details. Thanks in advance, Bes...

...Send mail to mime@docserver.cac.washington.edu for more info. --1133331701-635408861-993124011=:10449 Content-Type: TEXT/PLAIN; charset=US-ASCII I'm not sure if the warning messages Warning messages: 1: variable 67 is not a factor in: model.frame.default(Terms, newdata, na.action = act, xlev = attr(object, 2: variable 70 is not a factor in: model.frame.default(Terms, newdata, na.action = act, xlev = attr(object, 3: variable 71 is not a factor in: model.frame.default(Terms, newdata, na.action = act, xlev = attr(object, reported by model.frame.default (thru predict.rpart) are corr...

....data) # terms will be re-ordered test> predict( lm.bug, scrunge.data) Error in "contrasts<-"(*tmp*, value = "contr.treatment") : contrasts apply only to factors In addition: Warning message: variable discrete is not a factor in: model.frame.default(object, data, xlev = xlev) This actually turns out to be a bug in `model.frame.default', to do with an inconsistency between `predvars' and `vars' when `model.frame.default' is called inside `predict'. AFAICS it can be fixed by including the commented line below in `model.frame.default':...

.... 1. I'm using a regression tree to predict the selling prices of 10 new records (homes). The following code is resulting in an error message: pred <- predict(model, newdata = outOfSample[, -6]) The error message is: Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = attr(object, : factor Sq. Feet has new levels 1375, 1421, 1547, 1621, 1868, 2211, 2265, 2530, 2672, 3365 Does anybody know what is causing this? I've pasted a snippet of my original dataset (Crankshaw) and my out-of-sample dataset below. Below it appears all code which I entered leading...

...e(formula = y ~ x2 + x3 + x1, data = df, drop.unused.levels = TRUE) 11: eval(expr, envir, enclos) 10: eval(mf, parent.frame()) 9: lm(formula = y ~ x2 + x3 + x1, data = df, method = "model.frame") 8: eval(expr, envir, enclos) 7: eval(fcall, env, parent.frame()) 6: model.frame.lm(fob, xlev = object$xlevels) 5: model.frame(fob, xlev = object$xlevels) 4: stats:::add1.lm(object, scope = scope, scale = scale) 3: addterm.lm(fit, scope$add, scale = scale, trace = max(0, trace - 1), k = k, ...) 2: addterm(fit, scope$add, scale = scale, trace = max(0, trace - 1), k = k, ...) 1: stepAIC(...

...and therefore they have days==NA. I want to predict how many days these sites will take to complete (I've simplified the above discussion to remove many of the other factors involved.) nconst<-subset(const,is.na(const$days)) x<-predict(g,nconst) Error in model.frame.default(object, data, xlev = xlev) : factor city has new level(s) ALBANY This is because we haven't yet completed a site in Albany. If I just had one to worry about I could easily fix it (choose a nearby market with similar characteristic) but I am dealing with a several hundred cities. Instead, for the cities n...