search for: cat2

Hello, I can get the median for each factor, but I'd like another column to go with each factor. The nm column is a long name for the lvls column. So unique work except for the order can get messed up. Example: x = data.frame(val=1:10,lvls=c('cat2',rep("cat1",4),rep("cat2",4),'cat1'),nm=c('longname2',rep("longname1",4),rep("longname2",4),'longname1')) x val lvls nm 1 1 cat2 longname2 2 2 cat1 longname1 3 3 cat1 longname1 4 4 cat1 longname1 5 5 cat1 lon...

...o interact. An example of such a model follows at the bottom of this message, but here is a section of its summary table: Estimate Std. Error z value Pr(>|z|) (Intercept) 1.220186 0.539475 2.262 0.0237 * var1 0.028182 0.050850 0.554 0.5794 cat2 -0.112454 0.781137 -0.144 0.8855 cat3 0.339589 0.672828 0.505 0.6138 var1:cat2 0.007091 0.068072 0.104 0.9170 var1:cat3 -0.027248 0.064468 -0.423 0.6725 I am having trouble interpreting this output. I think I understand that...

...6 categories, and the other some other of the 6 categories (and not necessarily exclusive). So let's built some results after i used the table function. m1 <- matrix(c(32, 35, 36, 12, 15, 16), nrow = 2, ncol = 3, byrow = TRUE, dimnames = list(c("land", "water"), c("cat2", "cat5", "cat6"))) > m1 cat2 cat5 cat6 land 32 35 36 water 12 15 16 m2 <- matrix(c(45, 46, 47, 48, 21, 22, 23, 24), nrow = 2, ncol = 4, byrow = TRUE, dimnames = list(c("land", "water"), c("cat1", "cat2", "c...

...redicted 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 determine what that is. Thanks in advance for your help. Adam dd <- datadist(all.data2.stand) options(datadist='dd') bp.cat2 <- all.data2.stand$bp.cat2 u <- cr.setup(bp.cat2) u b.mean <-rep(all.data2.stand$b.mean, u$reps) r.mean <-rep(all.data2.stand$r.mean, u$reps) mean.ova.energy <- rep(all.data2.stand$mean.ova.energy, u$reps) y <- (u$y) # constructed binary response cohort <- u$cohort attach(al...

Hello, I'm looking for a solution for the following problem: 1) I have a folder with several csv files; each contains a set of measurement values 2) The measurements of each file belong to a position in a two dimensional matrix (lets say "B02.csv" belongs to position 2,2 3) The size of the matrix is fix 4) I cannot assure to have a csv file for each position 5) Each position

...00, 500)) cat<-(runif(2000)<=.5)+0 for(i in 1:length(cat)){ if(cat[i] == 0){ pop[i,"cat1"] = "this" } else{ pop[i,"cat1"] = "that" } } cat<-(runif(2000)<=.5)+0 for(i in 1:length(cat)){ if(cat[i] == 0){ pop[i,"cat2"] = "here" } else{ pop[i,"cat2"] = "there" } } #Extract sample sam <- pop[sample(1:2000,50),] #Combine data sets x <- make.groups(pop, sam) #Create trellis display densityplot(~var1 + var2 | cat1 * cat2, data=x, groups=which, plot.points=FA...

...a brute force "for" loop approach which seems okay as it is a small dataset. It looks a bit inelegant and given all the warnings in the Intro to R, etc, about using loops I wondered if anyone could suggest something a bit simpler or more efficent? Example: cat1 <- c( 1,1,6,1,1,5) cat2 <- c( 1,2,3,4,5,6) cat3 <- c( 5,4,6,7,8,8) cat4 <- c( 1,2,1,2,1,2) years <- c( 'year1', 'year2', 'year3', 'year3', 'year1', 'year1') id <- c('a','a','b','c','c','a') df1 <- data.frame...

...r categories (or bins), and I need to find the median category. Trivial to do by hand, but I was wondering if there is a means to do it in R in an elegant way. The obvious medioan(vector) returns the median frequency for the binns, and that is not what I want. i.e,: freq cat1 1 cat2 10 cat3 100 cat4 1000 cat5 10000 I want it to return cat5, instead of cat3. Thanks a lot Martin

Hi, I have the following dataset and I would like to create a dotplot with confidence limits: CAT1 CAT2 MEAN Lower Upper 1 1 1 0.619 0.392 0.845 2 1 10 1.774 1.030 2.518 3 1 100 7.700...

...one between B and C) not be there. What I've come up with is this (instead using geom_segment): df <- data.frame(cat=LETTERS[1:4],num=rnorm(4)) dfdf <-cbind(df[seq(from=1,by=2,along.with=df),],df[seq(from=2,by=2,along.with=df),]) names(dfdf) <- c("cat","num","cat2","num2") ggplot(df, aes(x=cat, y=num)) + geom_point() + geom_segment(aes(x=cat,xend=cat2,y=num,yend=num2),data=dfdf) Hoewever, in a more extended example (where df is more complex and also contains factors used for grouping and faceting) this becomes a bit cumbersome. Is there a b...

Hello R help I have a dataframe, with 71 samples (rows) and 30 variables. I got linear models for some of the variables,? and I want to join fitted and residuals of these models to the data frame. Sometimes, these vectors have the same length of the dependant variable, but in a few cases, NA values can be found on my data, and therefore, both fitted and residuals have a few rows less than the

Hello all, I have been meaning to learn R for a while and have just subscribed to this list. I am planning to give R a shot at one of my live projects. I am looking to explore graphical features of R on my data below. Sample Data: Cat1 - Cat2 - Cat3 - Cat4 - NumPeople - Salary H - L - H - L - 100 - 50000 L - L - L - L - 40 - 30000 - H - H - - 100 - 45000 Cat1 through Cat4 are categorical variables containing High, Medium, Low or Blank values and the last two variables are continuous. 1. Is there a good way of graphically represen...

I have a categorical variable in a dataframe similar to the following... cat 1 1 3 2 4 I need to convert it to 4 dichotemous variables for each observations like... cat1 cat2 cat3 cat4 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 Thanks in advance! Shane

i am making a category tree and i iterate over the category using my counter_cache however it would show a different number than what was actually being represetned in the tree. Here is an example cat.children = [cat2, cat3] puts cat.children.size 2 puts cat.children_count 2 cat4.parent_id = cat.id cat4.save puts cat.children.size 2 puts cat.children_count 2 however cat.children.count produces the correct...

I have a controller method as such: def delete @category = Category.find_by_name(params[:name]) @element_id = @category.name Category.delete_all(["name = ?", @category.name]) end (For the purpose of this exercise, category names are unique) I have the corresponding delete.rjs file: if @element_id page.remove :id => @element_id page.visual_effect :highlight,

...;s-news at wubios.wustl.edu> Sent: Tuesday, December 16, 2003 2:04 PM Subject: [S] plot stacked bar chart in R > Hi, > I am trying to plot a stacked bar chart in R but am not able to find the > documentation. > > The input data format is like the following > > > CAT1 CAT2 Count > A a 234 > A b 758 > A c 156 > B a 753 > B b 568 > B c 684 > C a 356 > C b 564 > C c 256 > D a 123 > D b 165 > D c 754 > > please advice. > > > thanks, > > > Yun-Fang > > > --------------------------------...

Hi all, I have some data as follows. Cat1 Cat2 Cat3 COG Counts A B C COG1 10 B D COG2 20 C COG3 30 D COG4 40 I would like to sum all the counts for each category: A B C D 10 30 40 60 >CAT2COG<- list(A="COG1",B=c("COG1","COG2"),C=c("CO...

...lumn represents a category and the individual data points in each category are binary responses (in this case they are actually 1's and 0's). What I want to extract are the counts for each category and put them in a vector. To do that I used the following: cats<-c("cat1","cat2", "cat3", ...) c()->counts for(j in cats){ append(counts, sum(data$j)) -> counts } However, the 'counts' object only contains 0's after the script runs: counts [1] 0, 0, 0, .... After replacing various elements in the sc...

Hi, I have a dataset with 6 categorical variables. I have used this following code to make the variables u1-u6 ordered factors and this works well. cat1cat2 cat3 cat4 cat5 cat6 ? 0 ? ?? 1 ? ? 1????? 0 ??? 0? ?? 1 ? 1 ? ?? 1 ? ? 0 ? ?? 0 ? ? 0 ? ? 0 ....... .... ############ data<-read,table("example.txt") data <- as.data.frame(lapply(data, ordered)) ############ Now, suppose I have bigger dataset where some variables are categori...

Dear Rhelpers, Is there a faster way than below to set a vector based on values from another vector? I'd like to call a pre-existing function for this, but one which can also handle an arbitrarily large number of categories. Any ideas? Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable