similar to: more flexible "ave"

Hi, I have been trying to calculate summary error and coding statistics on a by subject basis and seem to be writing a lot of code to do a simple thing. I won't go into my messy version. What I am asking is if anyone knows of a single command that could take the following data and get error rates over a given vector. Given data from ss with,,, subj a rt code 1 1 200 good 1 1 321 good

Hello, I am doing a confirmatory factor analysis with package Lavaan. My model consists of 5 factors and I would like to calculate the factor-reliability and the average variance extracted (AVE) for every factor. What function can I use to receive this measures? The Lavaan Package doesn't have this measures. Best regards, Ina [[alternative HTML version deleted]]

Hello, I have many subsets of x. I want to get the standard deviation for each subset with the same factor levels. For "FUN=mean" and "FUN=median" I am using "ave()". Can anybody tell me the match of "ave()" for using "FUN=SD"? At the beginning I used aggregate(), also for mean and median. But aggregegate make ""arithmetical

Dear R developers, In the help text for ave() you can read in Arguments section: "... Grouping variables, typically factors, all of the same length as x" ...and in Value: "If ... is g1,g2,". I think a description in the Arguments section of how two or more grouping variables is specified together with an Example would be very helpful. In absence of a help text I just

Consider the following simple data set and a call to ave: > tdata <- data.frame(f1=c(1,1,1,1,2,2,2,2), f2=c(1,2,1,2,1,1,1,1), y=1:8) > with(tdata, table(f1, f2)) f2 f1 1 2 1 2 2 2 4 0 > with(tdata, ave(y, f1, f2, FUN=max)) [1] 3 4 3 4 8 8 8 8 Warning message: In FUN(X[[i]], ...) : no non-missing arguments to max; returning -Inf There are no missing values in the

Hi: How do I find average across three columns such as weight1, weight2, weight3 (20 people, each person has three different measures)? They have NA's also. Thanks Tom, Univ of Pittsburgh, ttrut at yahoo.com

[R 2.0.0 on Linux] I tried: > df <- data.frame( grp1=factor( c('A' ,'A' ,'A' ,'D', 'D' ) ) , grp2=factor( c('a1','a2','a2','d1','d1') ) ) > df grp1 grp2 val 1 A a1 1 2 A a2 2 3 A a2 4 4 D d1 8 5 D d1 16 I got: > with( df, ave( val, grp1, grp2, FUN=sum ) )

Hi Patrick Hausmann <c18g at uni-bremen.de> napsal dne 15.11.2007 18:59:06: > Hello Petr, > > one question solved, the next is standing in front of me... If you > have a minute to look .... great! > > > x > V1 V2 F1 > 1 A 2 0.1552277 > 2 A 3 0.1552277 > 3 A 4 0.1552277 > 4 B 3 0.8447723 > 5 B 2 0.8447723 > 6 C 6 0.2500000

Dear list, I'm still trying to calculate the sd for V2 for each group in V1 if V3 is '0': > x V1 V2 V3 1 A01 2.40 0 2 A01 3.40 1 3 A01 2.80 0 4 A02 3.20 0 5 A02 4.20 0 6 A03 2.98 1 7 A03 2.31 0 8 A04 4.20 0 # Work x$vmean <- ave(x$V2, x$V1, x$V3 == 0, FUN = mean) # Work x$vsd2 <- ave(x$V2, x$V1, FUN = sd) # Doesn't work x$vsd <- ave(x$V2, x$V1, x$V3

Dear list, # I have a DF like this: sleep$b <- c(rep(8,10), rep(9,10)) sleep$me <- with(sleep, ave(extra, group, FUN = mean)) sleep # I would like to create a new variable # holding the b-th value of group 1 and 2. # This is not what I want, it takes always the '8' from group '1' # and not the '9' sleep$gr <- with(sleep, ave(extra, group, FUN = function(x)

I've been using the ave function to compute some statistics on a data frame. After a while I noticed that, for some reason, it was returning numerical statistics as strings instead of numbers. I delved into the code of the functions and traced the problem to the following fact: ave uses split<- to do its work. Specifically, it does "split(x, g) <- lapply(split(x, g),

Hello again. Following a great idea from Jim and Dennis, I began to use the function ave() in my big data. I proved to use this command in my data with FUN=max(x,na.rm=T), and I received a Warning that translated looks like Warning in max (x, na.rm = T): no arguments to max; returning -Inf Is this happening 'cause I´m using a big table : 254417 rows 9 columns? [[alternative HTML version

Dear R users, It may be very simple but it is being difficult for me. I'd like to calculate the difference in percent between to measures. My data looks like this: set.seed(123) df1 <- data.frame(measure = rep(c("A1", "A2", "A3"), each=3), water = sample(c(100:200), 9), tide = sample(c(-10:+10), 9)) df1 # What I want to calculate is:

Dear R experts--- Has anyone written parallel versions of "by" (i.e., mcby) and "ave" (i.e. mcave) ? I did ask a question like this a year ago, and then the answer was no. for those who are googling the group for the answer to this question, in the meantime, the poor man's version of "by" is mclapply( split( ds, factor ), FUN ) I don't know the poor

library(plyr) n<-100000 grp1<-sample(1:750, n, replace=T) grp2<-sample(1:750, n, replace=T) d<-data.frame(x=rnorm(n), y=rnorm(n), grp1=grp1, grp2=grp2) system.time({ d$avx1 <- ave(d$x, list(d$grp1, d$grp2)) d$avy1 <- ave(d$y, list(d$grp1, d$grp2)) }) # user system elapsed # 39.300 0.279 40.809 system.time({ d$avx2 <- ave(d$x, interaction(d$grp1, d$grp2, drop =

Hi there, I checked google for aov. usually one uses summary to see whether the p-value is small. but I want to put aov in my script. how can I get the p-value, (0.1115, 0.6665, 0.6665 in the following example)? thanks YU > datafilename="http://personality-project.org/r/datasets/R.appendix2.data" > data.example2=read.table(datafilename,header=T) > aov.ex2 =

> Dear list, > I have a data frame like: > > > log2.ratios[1:3,1:4] > ID a1 a2 a3 > 1 GS1-232B23 -0.0207500 0.17553833 0.21939333 > 2 RP11-82D16 -0.1896667 0.02645167 -0.03112333 > 3 RP11-62M23 -0.1761700 0.08214500 -0.04877000 > 4 RP11-62M23 0.2761700 -0.15214500 -0.05877000 > the 3rd and

I know this is a little OT for the list and blatent advertising but inkeeping with the thread topic, I've just finished a Windows playlist generator for internet/radio stations. Suitable for highly automated situations. Features anti-song repeat & anti-artist repeat options. No song maintenance required. Works on a rotation or "clock" basis. Produces a personalised HTML

Full_Name: Brendan Barnwell Version: 2.9.0 OS: Windows XP Pro Submission from: (NULL) (71.102.131.29) The ave() function returns an incorrect datatype. Specifically, ave(x, g, f) always returns a vector with the same mode as x, rather than using the mode of the vector returned by f. Observe: > x [1] "A" "B" "C" "A" "B" "C"

Hi, This is probably really obvious, by I can't seem to find anything on it. Is there a fast version of ave for when the data is already sorted in terms of the factor, or if the breaks are already known? Basically, I have: X = 0.1, 0.2, 0.32, 0.32, 0.4, 0.56, 0.56, 0.7... Y = 223, 434, 343, 544, 231.... etc of the same, admittedly large length. Now note that some of the values of X are