search for: ungroup

...shing in a journal that is very picky about their graphics formats. It is more efficient for me to get the final picky details of each graph done by our graphics department, who use Windows CorelDraw. The problem is that I have not yet been able to give them a vector graphics format that can be ungrouped in CorelDraw for editing. Bitmaps cannot be ungrouped <www.eatbywest.com/pub/vectorbitmap/>. The standard postscript that I am producing in R cannot be ungrouped either, and comes into CorelDraw looking fragmented. While I may be doing something odd in the production of my postscript fi...

...ay: y A B C 0 1 1 2 0 1 2 1 1 1 1 2 0 1 1 2 1 1 1 2 1 1 2 1 0 1 2 2 . . . And so on. How can I make my data look like the following: y A B C 2 1 1 2 1 1 2 1 0 1 2 2 . . . In other words how can I change my ungrouped cases into grouped cases? Thanks! Chris [[alternative HTML version deleted]]

...me (Galton's data on heights of parents and children), in grouped form, parent child frequency 73.5 72.2 1 73.5 73.2 3 72.5 68.2 1 72.5 69.2 2 72.5 70.2 1 72.5 71.2 2 72.5 72.2 7 72.5 73.2 2 72.5 74.2 4 71.5 65.2 1 71.5 66.2 3 71.5 67.2 4 71.5 68.2 3 71.5 69.2 5 71.5 70.2 10 ... and need the ungrouped equivalent, where each input observation appears the number of times represented by its frequency. In SAS this would be data galton2; set galton; do i=1 to frequency; output; end; I can replicate this in R, but surely there must be an easier way. (In APL2 it would be frequen...

...n so something like this XX <- data.frame(x=1:10,y=rbinom(10,3,0.5)) sapply(XX,rep,XX[,2]) HTH (if this do what you really want) Thomas -----Urspr?ngliche Nachricht----- Von: Michael Friendly [mailto:friendly at yorku.ca] Gesendet: Do 25.09.2003 19:14 An: r-help Cc: Betreff: [R] ungrouping grouped data I'm sure this is probably simple, but I can't find an answer... I have a data frame (Galton's data on heights of parents and children), in grouped form, parent child frequency 73.5 72.2 1 73.5 73.2 3 72.5 68.2 1 72.5 69.2 2 72.5 70.2 1 72.5 71.2 2 72.5...

I'm hoping the community knowledge can help me out here. I have found great value in R, especially using it to generate charts, but I am still scaling the learning curve in a number of ways. I am looking plot one grouped line and one ungrouped line in a lattice plot. I can plot one grouped line like this (the line's color in each panel becomes dependent on the newpool value): xyp<-xyplot(cpucap~date|zone,data=df,type="l",groups=newpool,auto.key=list(points=F,lines=T), main=paste(df$server[1]," CPU C...

Thanks for helping me out with the thing of quotation marks. I finally could save time running programs at the university and home. One additional question for all of you. I'm starting to be familiar with multinomial regression models. Actually I'm doing some exercise with the pneumo data that appears in Faraway library. The database has 24 observations and I want to study the probability

Dear Masters, first I'd like to wish u all a great 2011 and happy holydays by now, second (here it come the boring stuff) I have a question to which I hope u would answer: I run a logistic regression by glm(), on the following data type (y1=1,x1=x1); (y2=0,x2=x2);......(yn=0,xn=xn), where the response (y) is abinary outcome on 0,1 amd x is any explanatory variable (continuous or not)

...9;s data, with both regression lines and data ellipses, and I must be doing something wrong, because the lines do not intersect at \bar{x}, \bar{y}. The problem is likely in the line for x.y, but I don't know how to specify that correctly. The data is read in grouped form( galton), and then ungrouped (galton2): galton <- read.table("~/sasuser/data/galton.txt", header=TRUE) # ungroup galton2<-galton[rep(1:nrow(galton), galton$frequency), 1:2] attach(galton) sunflowerplot(child, parent, number=frequency, xlim=c(61,75), ylim=c(61,75), xlab="Child height", ylab=&q...

...me gustaría cambiar el nombre de los topics a la hora de visualizarlos y en vez de que se llamen 1,2,3...darles un nombre. ¿Es posible en este código? lda22 <- LDA(reviews_dtm, k = 22, control = list(seed = 1234)) lda22 %>% tidy() %>% group_by(topic) %>% top_n(10, beta) %>% ungroup() %>% mutate(term = reorder_within(term, beta, topic)) %>% ggplot(aes(term, beta, fill = factor(topic))) + geom_col(show.legend = FALSE) + facet_wrap(~ topic, scales = "free_y") + coord_flip() + scale_x_reordered() Muchas gracias

I've noticed that the implementation of Kendall's Tau in R is O(N^2). The following reference describes how it can be done in O(N log N): A Computer Method for Calculating Kendall's Tau with Ungrouped Data William R. Knight Journal of the American Statistical Association, Vol. 61, No. 314, Part 1 (Jun., 1966), pp. 436-439 http://www.jstor.org/pss/2282833 I'm interested in contributing an implementation of this algorithm to R. However, I have a few questions: 1. Would it likely be...

...e me to submit graphs that are "editable in PowerPoint". I would be grateful for suggestions as to how I should do this. The best route seems to be to copy-and-paste the figures from the windows() device as a metafile. However, on converting the graphs to an editable format and "ungrouping" them, it appears that every line, tickmark, dot and dash is treated as a separate entity by PowerPoint (for example even the horizontal and vertical parts of "+" are separated). Alternatively, reading in the files from a saved metafile is even more problematic: having used par(...

Hi Folks, I have just started using dplyr and could use some help getting unstuck. It could well be that dplyr is not the package to be using, but let me just pose the question and seek your advice. Here is my basic data frame. head(h) subject ageGrp ear hearingGrp sex freq L2 Ldp Phidp NF SNR 1 HALAF032 A L A F 2 0 -23.54459 55.56005 -43.08282

...= lazyeval::interp(~round(((a - mean(a[IS_GG == "YES"], na.rm = TRUE))/mean(a[IS_GG == "YES"], na.rm = TRUE)) * 100, 1), a = as.name(col)) dataset %>% group_by(TEST_SET_NAME) %>% mutate_(.dots = setNames(list(mutate_call), col_name)) %>% ungroup() } # run function results_demo<- demo_fn(dat = dem) # expected results structure(list(id = c("L1", "L2", "L3", "M1", "M2", "M3"), TEST_SET_NAME = c("A", "A", "A", "B", "B", &quot...

I have used plor() to model a rather large 3-category dataset (~1500 data points, ~15 independent variables); from the resulting model (with a deviance slightly below the residual degrees of freedom), the training data are placed in only the two extreme categories. Though the result appears to indicate that there's only a relative 'narrow' bin for the medium group, [and when the

...S_GG == > > "YES"], na.rm = TRUE))/mean(a[IS_GG == "YES"], na.rm = TRUE)) * > > 100, 1), a = as.name(col)) > > dataset %>% group_by(TEST_SET_NAME) %>% mutate_(.dots = > setNames(list(mutate_call), > > col_name)) %>% ungroup() > > } > > > # run function > results_demo<- demo_fn(dat = dem) > > # expected results > > structure(list(id = c("L1", "L2", "L3", "M1", "M2", "M3"), TEST_SET_NAME = > c("A", > > "...

...onstructed two types of data sets. On Rfile-01.Rdata (attached here) are data frames, data.1 and data.2. The information is identical but the data are arranged differently between these two data frames. Data frame, data.1, groups frequencies by subject, day and treatment. Data frame, data.2, is ungrouped. Consistency of these data frames is substantiated by computing these tables: ftable(xtabs(response ~ expt + subject + day, data = data.1)) ftable(xtabs(as.numeric(response) - 1 ~ expt + subject + day, data = data.2)) If I ignore the repeated measurement aspect of th...

I am trying to test goodness of fit for my legalistic regression using several options as shown below.  Hosmer-Lemeshow test (whose function I borrowed from a previous post), Hosmer–le Cessie omnibus lack of fit test (also borrowed from a previous post), Pearson chi-square test, and deviance test.  All the tests, except the deviance tests, produced p-values well above 0.05.  Would anyone please

It has always been my understanding that deviance for GLMs is defined by; D = -2(loglikelihood(model) - loglikelihood(saturated model)) and this can be calculated by (or at least usually is); D = -2(loglikelihood(model)) As is done so in the code for 'polr' by Brian Ripley (in the package 'MASS') where the -loglikehood is minimised using optim; res <-

...This is the code: dat2 <- dat1 %>% group_by(country) %>% group_by(grp = cumsum(is.na(lag(Y))), add = TRUE) %>% mutate(first_year_at_1 = match(1, Y) * any(is.na(Y)) * any(tail(Y, 3) == 1L), X = {x <- integer(length(Y)) ; x[first_year_at_1] <- 1L ; x}) %>% ungroup() It doesn?t really generate what I described above. Any help here would be much appreciated. Below you can see my sample data with the desired outcome ?X? dummy in it. Thank you! > dput(data) structure(list(year = c(1991L, 1992L, 1993L, 1994L, 1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 20...

...represented by a filled rectangle (with the color given by bg=...). This is to be expected. However, I find no options to disable this and produce a graph with transparent background. Does such an option exist? If not, then this could be a useful addition, because it is tricky to edit the metafile, ungroup de elements, select the background rectangle, delete it, regroup all the other elements and save the graph in a graphic pachage just to get transparent background on the graph... This could be simply bg=False. Thanks, Philippe Grosjean P.S.: I really appreciate the changes made to exported graphs...