similar to: Problem with ties in rank()

I am grateful to Prof. Ripley for his explanation. Indeed, rounding explains it all. I take the difference between two vectors and call it "z" method.a <- c(6.3, 6.3, 3.5, 5.1, 5.5, 7.7, 6.3, 2.8, 3.4, 5.7, 5.6, 6.2, 6.6, 7.7, 7.4, 5.6, 6.3, 8.4, 5.6, 4.8, 4.3, 4.2, 3.3, 3.8, 5.7, 4.1) method.b <- c(5.2, 6.6, 2.3, 4.4, 4.1, 6.4, 5.4, 2.3, 3.2, 5.2, 4.9,

I have a simple vector, called tmp that I want to subset based on another vector called vec. Everything works as expected except for below where the subsetting returns something other than the original data. Any ideas? > vec <- c(1,2,3,4,5,59,60,27,32,21) > tmp [1] 1.0 1.1 2.0 2.1 2.2 3.0 3.1 4.0 5.0 5.1 6.0 7.0 8.0 8.1 9.0 [16] 9.1 9.2 10.0 10.1 11.0 12.0 13.0 14.0

Dear all, I have the following data frame: > dat V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 AAAACACCCACCCCCCCCCCCCCCCCCCCCCCCC 9.0 18 12.00 18.0 15.0 12.0 6.0 2 1 ACGATACGGCGACCACCGAGATCTACACTCTTCC 18.0 8 12.00 18.0 15.0 12.0 18.0 3 1 ACTACTGCTCCCCCCCCACTCCCCCCCCCCCCCC 15.0 8 12.00 12.0 18.0 12.0 12.0 4 1 ACTTATACGGCGACCACCGAGATCTACACTCTTT 15.0

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

Hi, I am very new to R and I've been trying to work through the R book to gain a better idea of the code (which is also completely new to me). Initially I imputed my data from a text file and that seemed to work ok, but I'm trying to examine linear relationships between gdist and gair, gdist and gsub, m6dist and m6air, etc. This didn't work and I think it might have something to do

Hi all! I tried to perform Shapiro-Wilk test for my sample of 243 values. > Us [1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3 [13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8 [25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.3 5.8 -10.8 [37] 21.9 8.2 -3.2 -3.9 -2.3 12.6 -4.7 -8.0 11.8 27.4 -9.5 -20.8 [49]

I have some doubts about the validity of my procedure to estimeate linear contrasts ina a factorial design. For sake of semplicity, let's imagine a one way ANOVA with three levels. I am interested to test the significance of the difference between the first and third level (called here contrast C1) and between the first and the seconda level (called here contrast C2). I used the following

Hi , R Version 0.90.1 on Solaris2.5 and Suse Linux 6.[1,3] crashes some time after a matrix row or column has been addressed via an incorrect row/colname: R : Copyright 1999, The R Development Core Team Version 0.90.1 (December 15, 1999) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type "?license" or

I have a data frame (gom) or a matrix of trace metal data and some other observations from water column samples taken at sea (e.g., 19 samples (rows), 19 variables) I can calc. the rank individually from each column of the attached object. How can I create a matrix that contains the ranked data for each variable (either 1-19, ties=avg)? For example: >gom<-read.csv ("gomdata.csv")

Dear R users, Suppose I have the following data.frame: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 c Double 22 33 44 d Double 4 6 8 and I want to have new records: myID myType myNum1 myNum2 myNum3 e Single 12.5 18

Hola!! tengo un data.frame donde cada fila corresponde a un año y cada columna a un mes (De enero a diciembre) > head(valT) V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 1941 18.0 16.3 15.2 10.1 8.1 8.3 8.8 9.2 7.9 12.2 11.9 14.6 1942 17.2 15.9 13.6 11.6 8.7 6.2 6.4 7.2 9.7 12.0 14.1 16.7 1943 17.6 17.3 13.5 12.5 10.5 7.0 8.2 7.9 -999.9 -999.9

Hi, I'm searching for a function to subistute Values in a Matrix to new Values. For example: old value new value 1.1 6 1.2 7 . . . . . . 1.9 14 2.0 15 and 2.1 15.5 2.2 16 . . . . 2.9 19.5 3.0 20 There is a difference

Hello zfs-discuss, One of a disk started to behave strangely. Apr 11 16:07:42 thumper-9.srv sata: [ID 801593 kern.notice] NOTICE: /pci at 1,0/pci1022,7458 at 3/pci11ab,11ab at 1: Apr 11 16:07:42 thumper-9.srv port 6: device reset Apr 11 16:07:42 thumper-9.srv scsi: [ID 107833 kern.warning] WARNING: /pci at 1,0/pci1022,7458 at 3/pci11ab,11ab at 1/disk at 6,0 (sd27): Apr 11 16:07:42 thumper-9.srv

Further to the discussion between Murray Jorgensen and Brian Ripley, it seems to me better to choose tabulations that will not come and bite you. Suppose your data are sligtly irregular, e.g. (for the sake of the argument): data( warpbreaks ) warpbreaks$variant <- rep( 1:5, len=54 ) attach( warpbreaks ) tb <- table( wool, tension, variant ) tb # in this case you would like to see: tp

Further to the discussion between Murray Jorgensen and Brian Ripley, it seems to me better to choose tabulations that will not come and bite you. Suppose your data are sligtly irregular, e.g. (for the sake of the argument): data( warpbreaks ) warpbreaks$variant <- rep( 1:5, len=54 ) attach( warpbreaks ) tb <- table( wool, tension, variant ) tb # in this case you would like to see: tp

Hello all, turns out i'm having a bad R week. I am at my wits end with a function that I am trying to write. When I run the lines of code outside of a function, I get the desired output. When I wrap the lines of code into a function it doesn't work as expected. Not sure what is going on here. I suspected that the syntax of the if statement with the for loop was the culprit, but when I

hello, I'd like know how to sort a data frame in R for example how I should do to sort by Catholic with swiss data frame like below thanks Fertility Agriculture Examination Education Catholic Infant.Mortality Courtelary 80.2 17.0 15 12 9.96 22.2 Delemont 83.1 45.1 6 9 84.84 22.2

Another request for help implementing the 'apply' functions to avoid a loop structure... I am working with a data set that includes lab measurements taken at different dates for the subjects, with some subjects having more results than others. I would like to average lab results for each subject that were taken on the same day. I can do this using a for loop, but would like to know how

I'm using a very long irregular time-series of air temperature and relative humidity of this kind (this is an extract only) its.format("% Y%d%m %X) > base T H 20020601 12.00.00 27.1 47 20020601 15.00.00 29.1 39 20020601 18.00.00 27.4 39 20020601 21.00.00 24.0 40 20020602 0.00.00 22.0 73 20020602 3.00.00 19.2 49 20020602 6.00.00 19.5 74 20020602

Dear R-User, Appreciate any helps. It looks simple, but I don't have a clue. Given that I have a dataframe of tree population with three variables: sp=species , d0=initial_size grow=growth increment from initial size per year How can I calculate the future growth increment of each tree for the next 3 years. The following Rscript was written, #---------- a0 <-