similar to: distance function (analogue)

Hello, I am trying to get a linear model of y ~ log(x). *> lm (y~log(x))* However, I always get an error report: /Error in model.frame.default(formula = y ~ log(x), drop.unused.levels = TRUE) : variable lengths differ (found for 'log(x)')/ *Here was my y:* > y [1] 0.4500000 0.0500000 0.5000000 0.4000000 0.0000000 0.5000000 0.4000000 [8] 0.0500000

Dear Mailing-List, I think this is a newbie question. However, i would like to integrate a loop in the function below. So that the script calculates for each variable within the dataframe df1 the connecting data in df2. Actually it takes only the first row. I hope that's clear. My goal is to apply the function for each data in df1. Many thanks in advance. An example is as follows: df1

Hi, xtabs() creates a table of counts. I want a table of proportions -- that is, I want to divide every vector (along a particular dimension) by its sum. The tiny example below does that. The call to xtabs() creates a matrix "A" with dimensions ("x1","x2","y"). I transform "A" using aperm() and aaply() to get the matrix "B". The

Dear list, I just made a very simple mistake, but it was hard to spot. And I think that I should warn other people, because it is probably so simple to make... === R code === # Let us create a matrix: (a <- cbind(c(0,1,1), rep(1,3))) # [,1] [,2] # [1,] 0 1 # [2,] 1 1 # [3,] 1 1 # That is a MISTAKE: a/colSums(a) # [,1] [,2] # [1,] 0.0000000 0.3333333

Hi, I sometimes play around with extreme parameters for distributions and found that qbeta is not always monotone as the following example shows. I don't know whether this is serious enough to submit a bug report (as this example is near to the limitations of floating point arithmetic). Josef > x <- qbeta((0:100)/100,0.01,5) > x [1] 0.000000e+00 1.253990e-201 1.589622e-171

See ?rpart.control. I get: > golf.rp = rpart(Outlook ~ ., golf, control=rpart.control(minsplit=1)) > golf.rp n= 14 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 14 9 rain (0.2857143 0.3571429 0.3571429) 2) Temperature< 71.5 6 2 rain (0.1666667 0.6666667 0.1666667) 4) Temperature< 64.5 1 0 overcast (1.0000000 0.0000000 0.0000000) * 5)

Dear all, I have two data frames- x1 and y1 with same row names and column names(actually the names of the patients). x1 a b c d e a 1.0000000 0.4730679 0.6226994 0.6036036 0.6433333 b 0.4730679 1.0000000 0.6227273 0.6303855 0.5730858 c 0.6226994 0.6227273 1.0000000 0.7290503 0.6900585 d 0.6036036 0.6303855 0.7290503 1.0000000

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

Hello again, Let say I have a matrix: Mat <- matrix(1:12, 4, 3) And a vector: Vec <- 5:8 Now I want to do following: Each element of row-i in 'Mat' will be divided by i-th element of Vec Is there any direct way to doing that? Thanks for your help

Hello! In the example below the Reduce() function by default assigns "a" to be the last accumulated value and "b" to be the current value in "x". I could not find this documented anywhere as the default settings for the Reduce() function. Does any sort of documentation for this behavior exist? x <- c(0,0,0,0,0,1,0,0,0,5,0,0,0,7,0,0,0,8,5,10)

Hi, My objective is to calculate "Relative (Cumulative) Frequency of Event Occurrence" - something as follows: Sample.Number 1st.Fly 2nd.Fly Did.E.occur? Relative.Cum.Frequency.of.E 1 G B No 0.000 2 B B Yes 0.500 3 B G No 0.333 4 G B No 0.250 5 G G Yes 0.400 6 G B No 0.333 7 B B Yes 0.429 8 G G Yes 0.500 9 G B No 0.444 10 B B Yes 0.500 Please refer to the code below:

Hello, might be rather easy for R pros, but I've been searching to the dead end to ... twsource.area <- table(twsource, area, useNA="ifany") gives me a nice cross tabulation of frequencies of two factors, but now I want to convert to pecentages of those absolute values. In addition I'd like an extra column and an extra row with absolute sums. I know, Excel or the

Hello, (Power Book G4, Mac OS X, R 2.5.0) I would like to repeat the function range for 85 Vectors (V1-V85). I tried with this code: i<-0 > repeat { + i<-i+1 + if (i<85) next + range (Vi, na.rm = TRUE) + if (i==85) break + } I presume that the Vi is wrong, because in this syntax i is not known as a variable. But I don´t know how to say that it is a variable here. Would be nice if

Hello and sorry that I still haven?t found a solution for my problem. I need to extract the lower and upper triangle from a square matrix including the diagonal. This diagonal is not zero in that special case. I tried with as.dist w<-as.dist(w, diag = TRUE) > w 1 2 3 4 5 1 0 2 2 0 3 3 8 0 4 4 9 14 0 5 5 10 15 20 0 but found no way to keep the diagonal that is in the

Dear all, (Mac OS X 10.4.11, R 2.6.0) I have a quantitative dataset with a lot of Na?s in it. So many, that it is not possible to delete all rows with NA?s and also not possible, to delete all variables with NA?s. Is there a function for a principal component analysis, that can deal with so many NA?s. Thanks in advance Birgit Birgit Lemcke Institut f?r Systematische Botanik

Dear R users, I am looking for more efficient way to compute the followings -------------------------------------------------------------------------- a <- matrix(c(1,1,1,1,2,2,2,2),4,2) b <- matrix(c(1,2,3,4),4,1) Eventually, I want to get this matrix, `c`. c <- matrix(c(1/1,1/2,1/3,1/4,2/1,2/2,2/3,2/4),4,2) --------------------------------------------------------------------------

Hello everyone I'm trying to use Scheffe's complete cubic model (mixture design). In the bibliographies, they indicate that the term is of the type: A * B * (A-B). But I see that trying to adjust the three cubic terms results in singularities. I know this implies not having the inverse matrix: solve (t (X)% *% X) does not exist. The bibliographies show all three cubic terms. So my

Hi, Assuming that you provided the sample data from the file. temp <- readLines(textConnection("#Hogd/met, Temp, 005[M], Value #Hogd/met, Difftemp, 051[M], Value BA0+ 1 MTEMP005 1 [deg.C] 2 MDTMP051 1 [deg.C] EOH 891231, 2400, -1.5, -0.21, 900101, 0100, -1.4, -0.25, 900101, 0200, -1.6, -0.28, 900101, 0300, -1.7, -0.25, 900101, 0400, -2.1, -0.0999999, 900101, 0500, -2.3, -0.0899999,

Dear all, Please, I have a doubt regarding symbol plotting with data originating from a table. Please, see below: I have a tab delimited file called table1.txt with 4 columns: ypos animal var1 var2 5 cat gina <= lady gina \u2264 lady 7 dog bill >= tony bill \u2265 tony 9 fish dude <= bro dude \u2264 bro #I then load in the data to R: table1<-read.table("table1.txt",

Hello! I am R beginner and I have a question obout a simple matching. I have to datasets that i read in with: MalVar29_37<-read.table("MalVar29_37.csv", sep = ";") FemVar29_37<-read.table("FemVar29_37.csv", sep = ";") They look like this and show binary variables: V1 V2 V3 V4 V5 V6 V7 V8 V9 1 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1