similar to: Same color key for multiple lattice contour plots

Greetings, Sorry, the last message was sent by mistake! Here it is again: I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix

Hi, I have been able to generate association rules for Market Basket Analysis using the following codes: **************************************************************************** ******************************************* library("arules") rules <- read.csv("write1.csv",na.strings=c(".", "NA", "", "?"),header=TRUE)

Hi, My data.frame "A" has FID like this FID a a b b b c c d d d d Now my second data.frame "B" has age value for a, b, c, d like FID Age a      5 b      7 c      9 d      3 How can search for the Age column in "B" and replace the values in "A" so that my new "A" looks like this FID Age a      5 a      5 b      7 b      7 b      7

Hi All, I have the following problem (read the commented bit below): a<-matrix(1:9,nrow=3) a [,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 div<-1:3 apply(a,2,function(x)x/div) ##want to divide each column by div- instead each row is divided## [,1] [,2] [,3] [1,] 1 4.0 7 [2,] 1 2.5 4 [3,] 1 2.0 3

Dear R users, has anybody made an attempt to create the soil texture triangle graph in R? For an example see here: http://www.teachingkate.org/images/soiltria.gif I would like to get the lines in black and texture labels in gray to allow for plotting my texture results on top. Any examples or suggestions are very welcome! Thanks in advance, Sander. --

I have a graph with edge and vertex weights, stored in two data frames: --8<---------------cut here---------------start------------->8--- vertices <- data.frame(vertex=c("a","b","c","d"),weight=c(1,2,1,3)) edges <-

Dear all, When calling rlm with the following data, I get an error. (R v.1.8.1, WinXP Pro 2002 with service pack 1.) > d <- na.omit(data.frame(CPRATIO, HEIGHTZ, FAMILYID)) > c <- tapply(d$CPRATIO, d$FAMILYID, mean) > h <- tapply(d$HEIGHTZ, d$FAMILYID, mean) > c 1 2 3 6 7 9 10 11 6.000000 2.500000 3.250000

*Dear **Madam**/**Sir,* *What would happen if we irrigate the degraded land using the desalinated saline water in Minqin Oasis, Gansu Province, China? Can we prevent the disappearing of Minqin Qasis?* * * *Background:* *1. **The Minqin Oasis is surrounded by two deserts, Badain Jaran Desert and Tengger Desert. * *2. **Water is the root of ecological

Dear R users, It is a pleasure for me to announce the availability of the new package kwb.hantush (0.2.1)? on CRAN. Its objective is the calculation of groundwater mounding beneath an (stormwater) infiltration basin by solving the Hantush (1967) equation. For checking the correct implementation of the algorithm the R modelling results were cross-checked against alternative models assessed in

Dear R users, It is a pleasure for me to announce the availability of the new package kwb.hantush (0.2.1)? on CRAN. Its objective is the calculation of groundwater mounding beneath an (stormwater) infiltration basin by solving the Hantush (1967) equation. For checking the correct implementation of the algorithm the R modelling results were cross-checked against alternative models assessed in

Greetings,   I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix $A_{i,j}$, $1\leq i,j\leq 5$, numerically and check against the known

I've got this little program I wrote to test something, and it keeps giving the wrong result. I'm not inexperienced in C, but I can't believe strtof (et al) are broken, so I must be doing something wrong. However, I've spent hours looking at this and comparing it to the man pages and don't see what I'm doing wrong. strtod() and strtold() also give equally wrong results.

I have three samba 4.1.6 servers, all three are domain controllers. A, B , C, where A is has the FSMO role. I am getting the following error: [2015/02/02 11:49:42.359246, 0] ../source4/dsdb/samdb/ldb_modules/repl_meta_data.c:3664(replmd_op_possible_c onflict_callback) ../source4/dsdb/samdb/ldb_modules/repl_meta_data.c:3664: Unable to find replPropertyMetaData for conflicting record

Hi- I need to subset the following data by the column 'dal' for values that equal the regular interval seq(0, 150, by=0.5) exactly.... ....excluding rows with irregular 'dal' values such as c(2.888958, 2.891620), etc. data<-data.frame(id=id, dal=dal, date=date, mu.x=mu.x) $dal [1] 0.000000 0.500000 1.000000 1.500000 2.000000 2.500000 2.888958 2.891620 3.000000 3.245405

Hi, forgive me if someone has already posted about this but I have had a look and cannot find the answer, also I am very new to R and been getting the grips with this. I have been trying to use Adonis to find out if there are significant difference between groups on data that I have analyses with NMDS, and have been struggling with getting this to work and understanding what is going on. I am

I'm using R on Windows v2.0.1 with the nlme package (v3.1-53) and am finding some unexpected discrepancies in the output of intervals.lme and coef.lme. I've included a toy dataset at the end, but briefly, the data are longitudinal data from couples in marital therapy. Each spouse's relationship satisfaction is measured 4 times; I've fit both linear and quadratic models to the

Dear R family, I am a relative newbie and have been dabbling with R for a little while. Simple things really, but my employers are beginning to see the benefits of using R instead of excel. We have a remote monitoring station measuring groundwater levels. We download the date as a .csv file and up until now, we have been using excel to analyse the data. It?s been a hassle trying to wrestle

R Users: My question is probably more about elementary statistics than the mechanics of using R, but I've been dabbling in R (version 2.2.0) and used it recently to test some data . I have a relatively small set of observations (n = 12) of arsenic concentrations in background groundwater and wanted to test my assumption of normality. I used the Shapiro-Wilk test (by calling shapiro.test()

I am trying to make predictions with se's using a nlme (kew11.nlme below). I get an error indicating levels for a factor are not allowed. I have searched and read Rnews, MEMSS, MASS, R-Help, and other lists in Spanish where I found questions similar to mine but not solution. I do not really care about the method used. Any suggestions to obtain predictions with se's from an nlme

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