search for: eure

Hi R users, I have a dataframe in the below format xyz 01/03/2007 15.25 USD xyz 01/04/2007 15.32 USD xyz 01/02/2008 23.22 USD abc 01/03/2007 45.2 EUR abc 01/04/2007 45.00 EUR

*Hola!!! resulta que tengo unos datos de divisas ordenados por fechas (días) los que he convertido a formato tipo YYYY-MM-DD donde DD siempre es 01:* * * * EUR.resto$date<-as.Date(EUR.resto$date) EUR.resto$mo <- substr(EUR.resto$date,6,7) EUR.resto$yr <- substr(EUR.resto$date, 1,4)

Hi, i export data from an csv file like this :  Data <- read.csv2("c:/Art.csv",sep=",") # import data into R > Data <- Data [1:5,1:5]# extracting the first 5 rows and columns > Data   Policy.Number AXA.Entity Country LoB ccy.data 1         6e+13        BNL     BNL   P      EUR 2         6e+13        USA     BNL   P      EUR 3         6e+13         UK     BNL  

Dear R users: I want to convert some character vectors into numeric vectors. > head(price) [1] "15450 EUR" "7900 EUR" "13800 EUR" "3990 EUR" "4500 EUR" [6] "4250 EUR" >head(mileage) [1] "21000?km" "119000?km" "36600?km" "92000?km" "140200?km" [6] "90000?km" in

I have a vector of character strings such as mainnames<-c("CAD","AUD") and another vector say checknames<-c("CAD.l1","AUD.l1","JPY.l1","EUR.l1","CAD.l2","AUD.l2","JPY .l2","EUR.l2") I want a new vector of character strings that is just

Does anyone know whether there is an R version of the S-Plus software that can be downloaded from the website of the book Elements of Statistical Learning by Hastie, Tibshirani and Friedman? Rob Potharst -- ********************************************************** Dr. Rob Potharst Lecturer in Computer Science Erasmus University email: potharst at few.eur.nl P.O. Box 1738

  Hi, i export data from an csv file like this :  Data <- read.csv2("c:/Art.csv",sep=",") # import data into R > Data <- Data [1:5,1:5]# extracting the first 5 rows and columns > Data   Policy.Number AXA.Entity Country LoB ccy.data 1         6e+13        BNL     BNL   P      EUR 2         6e+13        USA     BNL   P      EUR 3         6e+13         UK     BNL  

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Folks, I have three dataframes storing some information about two currency pairs, as follows: R> a EUR-USD NOK-SEK 1.23 1.33 1.22 1.43 1.26 1.42 1.24 1.50 1.21 1.36 1.26 1.60 1.29 1.44 1.25 1.36 1.27 1.39 1.23 1.48 1.22 1.26 1.24 1.29 1.27 1.57 1.21 1.55 1.23 1.35 1.25 1.41 1.25 1.30 1.23 1.11 1.28 1.37 1.27 1.23 R> b EUR-USD NOK-SEK 1.23 1.22 1.21 1.36 1.28 1.61 1.23 1.34 1.21 1.22

This is a question about how to calculate similarities/distances among items that are classified by hierarchical attributes for the purpose of visualizing the relations among items by means of clustering, MDS, self-organizing maps, and so forth. I have a set of ~260 items that have been classified using two sets of hierarchically-organized codes on the basis of form and content. The data looks

Hi I am trying to plot two simple graphs with a grid in background. The axis and grid appears in correct position, but the actual data are not there.... Can somebody provide me a hint what is missing? The code is: pln <- read.table(file="PLN.txt", header=TRUE, dec=",") par(mfrow=c(1,2)) plot(pln[,1], type="l", lwd="2", ylab="EUR/PLN",

Hi guys I have a reasonably basic question with zoo usage, but I havent been able to find a satisfactory workaround yet. Heres a simple example of what I'm talking about (the following data frame contains numeric columns that contains NAs): > head(ebs) time src tstamp code bid ask 1 2009-03-03 13:03:29.536 perf.Tib_listener 14980321164 EBS.REC.EURJPY=EBS.NaE 123.48 NA 2 2009-03-03

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Hello; I am trying following but getting a warning message : Warning in rbind.zoo(...) : column names differ, no matter whatever I do. Also I do not want to specify column names manually, since I am just writing a wrapper function around getSymbols to get chunks of data from various sources - oanda, dividends etc. I tried giving col.names = T/F, header = T/F and skip = 1 but no help. I think

Hi, i. After estimating some coefficients using robust regression with rlm() or lqs(), I wonder if there exist some measures of the goodness-of-fit as those for standard linear model(R2)... or evenly if it's a statistics non-sense to look for since I do not find any mention of that in differents chapters on robust and resistant regression or in severals R documentation (Fox, Ripley and

Hi All, I can get Grandstream 100 SIP phones for EUR 75. I'm not sure about the pricing for these in Europe, so I'd like to hear from people here whether that is a reasonable price for them? TIA & BRgds -- Francesco Peeters ---- GPG Key = AA69 E7C6 1D8A F148 160C D5C4 9943 6E38 D5E3 7704 If your program doesn't recognize my signature, please visit

Surprisingly, I played around with some test code and below actually creates equations that look correct. tempmat<-matrix(10,nrow=6,ncol=6) restrictmat<-diag(6) colnames(tempmat)<-c("AUD.l1","CHF.l1","CAD.l1","GBP.l1","EUR.l1","JPY.l 1")