search for: 14.5

I've got a list of 5 matrices that are each 5 x 6. I'd like to end up with a 5 x 6 matrix that contains the mean value of the 5 original matrices. I can do this by brute force, but there must be a better way than making each matrix into a vector and then remaking a matrix thanks very much for any help david freedman ll=list(structure(c(9.7, 17.6, 20.8, 24.1, 33.8, 14.5, 25.7, 29.8,

Dear all, I am quite new to R; facing certain problems: Say, I have a text file( named as "try"): Year C1 C2 C3 C4 C5 C6 Y1 3.5 13.8 9.5 6.8 0.4 24.2 Y2 3.8 13.9 9.9 7.6 0.7 12.8 Y3 4.5 14.5 14.2 9.2 0.6 14.5 Y4 5.9 16.2 24.6 12.7 0.2 24.3 Y5 7.2 20.4 40.6 18.2 0.8 28.2 Y6 5.9 18.6 37.4 14.5 0.3 36.9 Y7 8.0 16.1 88.6 24.1 0.1 34.6 Y8 13.6 21.1 56.3 19.0 0.7 33.3 I wish to import the

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

Dear R users, A have a dataframe (matrix) with two collumns (plot, and diameter (d)). I want all diameters values for different combination of plots. For example I want all d values for all posible combination, 100C2 (all d values for plot 1 with all d values in the plot 2.......with all d values from plot 1 with all d values from plot 100, ...... with all d values from plot 99 with all d values

Dear all, I think I must be going crazy.... If I do > mean(rpois(1000000, 14)) I get: [1] 13.50420 and again: > mean(rpois(1000000, 14)) [1] 13.49896 > mean(rpois(1000000, 14)) [1] 13.50161 > mean(rpois(1000000, 15)) [1] 14.49250 > mean(rpois(1000000, 15)) [1] 14.49897 > mean(rpois(1000000, 14.5)) [1] 13.99689 > mean(rpois(1000000, 14.5)) [1] 13.9963 This was on a platform

Hi, I have a data frame with two columns of data, one an indexing column and the other a data column. My issue is, this data frame is incomplete and there are missing lines. I want to know how I can find and add data into these missing lines. See example below ## Example data data <- data.frame(index=c(1:4,6:10), data=c (1.5,4.3,5.6,6.7,7.1,12.5,14.5,16.8,3.4)) index data 1 1

Hi, Suppose i have generated an object using the following : fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) And when i print fit, i get the following : n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 17 absent (0.7901235 0.2098765) 2) Start>=8.5 62 6 absent (0.9032258 0.0967742) 4) Start>=14.5 29 0 absent (1.0000000

R-helpers: I am experiencing some odd behavior with the 'matrix' function that I have not experienced before and was wondering if there is something that I was missing in my code. --------------------------------- > sessionInfo() R version 2.6.1 (2007-11-26) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United

Dear R gurus: Here is the following from Montgomery's Design and Analysis of Experiments, 5th edition. > str(rout1.df) 'data.frame': 16 obs. of 3 variables: $ resp: num 18.2 18.9 12.9 14.4 27.2 24 22.4 22.5 15.9 14.5 ... $ A : Factor w/ 2 levels "-1","1": 1 1 1 1 2 2 2 2 1 1 ... $ B : Factor w/ 2 levels "-1","1": 1 1 1 1 1 1 1 1 2

Thanks for all the helpful responses. I include the data file and the syntax file for reference. Again, if I use the fields function, as is, I get the message: Error in svd(tempM) : error 159 in dsvdc using traceback, I get: > traceback() 4: stop(paste("error ", z$info, " in dsvdc")) 3: svd(tempM) 2: Krig(x, Y, cov.function = rad.cov, m = m, decomp = decomp,

is there a way to use the actual index value for plotting zoo objects this is the way that the index is set up and a sample range of what I would like to plot 01/01/06 00:00:00 - 01/01/06 23:45:00 { library(zoo) # chron library(chron) fmt.chron <- function(x) { chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) }} x <- structure(c(15.57, 15.5,

Date Fo Co 6/27/2007 57.1 13.9 6/28/2007 57.7 14.3 6/29/2007 57.8 14.3 6/30/2007 57 13.9 7/1/2007 57.1 13.9 7/2/2007 57.2 14.0 7/3/2007 57.3 14.1 7/4/2007 57.6 14.2 7/5/2007 58 14.4 7/6/2007 58.1 14.5 7/7/2007 58.2 14.6 7/8/2007 58.4 14.7 7/9/2007 58.7 14.8 Hello all: I am a newbie to R, and I was wondering how can I

I have a set of 52 weekly values, one is the desired high lake stage for a week, and the other is the desired low lake stage for each week. It looks like this: week High Low 1 16 14.5 2 16 14.5 3 15.95 14.45 4 15.84 14.34 5 15.73 14.23 6 15.61 14.11 7 15.5 14 8 15.38 13.88 9 15.25 13.75 10 15.13 13.63 .... 52 15 13.5 I can plot out two nice curves

Greetings, I would like to plot three lines on the same figure, and I am lost. There is an answer to a similar thread… but I tried matplot and it is beyond me. An example of the data follows: Year EM IM BM 1983 9.1 16.8 -7.7 1984 12.0 18.0 -6.0 1985 13.6 19.1 -5.5 1986 12.4 17.3 -4.9 1987 14.6 20.3 -5.7 1988 20.6 23.3 -2.6 1989 25.0 27.2 -2.2 1990 28.4 30.2 -1.8 1991 33.3 31.2 2.1 1992 40.6

Hi: I ran a live test broadcast today and noticed a couple of things that concerned me. Now, I'll admit I was pushing my P2/266 pretty hard, but it kept things going so I'm guessing that it's not unreasonable for me to expect this to work. I was sending two streams encoded from the sound card, one at 22khz mono quality -1 and the other at 44.1 stereo quality 0. I was also saving

Hi all, I have investigated the compile-time overhead of "Polly Scop Detection" pass based on LNT testing results. This mail is to share some results I have found. (1) Analysis of "SCOP Detection Pass" for PolyBench (Attached file PolyBench_SCoPs.log) Experimental results show that the "SCOP Detection pass" does not lead to significant extra compile-time

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees of freedom,

Dear all, I'd like to subset a df within a function, and use select for choosing the variable. Something like (simplified example): mydf <- data.frame(a= 0:9, b= 10:19) ttt <- function(vv) { tmpdf <- subset(mydf, select= vv) mean(tmpdf$vv) } ttt(mydf$b) But this is not the correct way. Any help? Thanks in advance Juli

Some day I may figure out how ggplot2 works. I am trying to plot 5 columns of data on a graph (similar to a simple matplot) =========================================================================== library(ggplot2) bmi <- structure(list(pct = 2:21, P10 = c(14.6, 14.5, 14.2, 13.9, 13.7, 13.7, 13.9, 14.2, 14.5, 14.8, 15.3, 15.9, 16.6, 17.2, 17.8, 18.1, 18.3, 18.4, 18.5, 18.6), P25 =

Dear all I am intrested in making iterative estimation (thro' loop statements) of, say, linear regression model. For this purpose, I have written the following programme and that I have made use of a sample data (viz., exp.txt): ? Programme: ? # Linear regression modelling with sample data (try5.txt) # Repeated estimation through loop statement