similar to: Read a Time Serie CSV

Hello, I have a data of Patents from OECD in delimited text format with IPC being one column, I want to filter the data by selecting only certain IPC in that column and delete other rows which do not have my required IPCs. Please, can anybody guide me doing it, also the IPC codes are string variables. The data is somewhat like below, but its a huge dataset containing more than 11 million rows

Hello, this is not really an R-related question, but since the posting guide does not forbid asking non-R questions (even encourages it to some degree), I though I'd give it a try. I am currently doing some secondary analyses of the PISA (http://pisa.oecd.org) student data. I would like to treat missing values properly, that is using multiple imputation (with the mix package). But I am not

Dear R Users - R is a wonderful software package. CRAN provides a variety of tools to work on your data. But R is not apt to utilize all the public databases in an efficient manner. I observed the most tedious part with R is searching and downloading the data from public databases and putting it into the right format. I could not find a package on CRAN which offers exactly this fundamental

Dear list, # I have a DF like this: sleep$b <- c(rep(8,10), rep(9,10)) sleep$me <- with(sleep, ave(extra, group, FUN = mean)) sleep # I would like to create a new variable # holding the b-th value of group 1 and 2. # This is not what I want, it takes always the '8' from group '1' # and not the '9' sleep$gr <- with(sleep, ave(extra, group, FUN = function(x)

The need comes from the PISA data. (http://www.pisa.oecd.org) In the data there are many cases and each of them carries a numeric variable that signifies it's weight. In SPSS the command would be "WEIGHT BY" In simpler words here is an R sample ( What is get VS what i want to get ) > data.recieved <- data.frame( + kindergarten_attendance = factor(c(2,1,1,1), labels =

I've been using R for several months now, in fact it was my intro to "real" computer programming. It used to frustrate me no end, but lately I've noticed my reaction has changed. Nowadays its much more like "wow, is that cool, or what?". So I guess I've gotten over the hump of the learning curve. Comparing what I now do with R to my previous hodge-podge of

Hi there, I have a data.frame (pwt6) which I would like to transform: country year gdp MEX 1950 2 MEX 1951 5 BOL 1950 4 BOL 1951 12 ITA 1950 45 ITA 1951 2 This should be the result: year MEX.gdp BOL.gdp ITA.gdp 1950 2 4 45 1951 5 12 2 Right now I have this code (better - no code): country.label<-names(table(pwt6$country)) result<-data.frame(year=NULL) for(i in country.label) ?

Dear R-users, again two question... # Question 1 Adding two lines of text to a plot, I am using: # ------------------------------- plot(k[,1], k[,2], pch=16, ylim=range((min(k[,2])-0.2):(max(k[,2])+1))) a <- paste("Cor.:" ,cor(k[,2],k[,1])) b <- paste(nrow(k), "Countries") text(90, max((k[,2])-0.51), a) text(90, max((k[,2])-0.83), b) #

Hello, I have a csv-file which looks like: #### pwt6_r.csv #### code;year;rgdpch AGO;1998;1234 ALB;1998;3576 ARG;1998;#NA SVN;1996;13439 SWE;1996;21492 AGO;1960;#NA ALB;1960;2345 ARG;1960;4634 #### pwt6_r.csv #### To import this file i call: pwt6<-read.csv("d:/pwt6_r.csv",header=T,na.strings="#NA",sep=";") Now I want to generate a new data.frame which include

Dear list, this works fine: x <- split(iris, iris$Species) x1 <- lapply(x, function(L) transform(L, g = L[,1:4] * 3)) but I would like to multiply each Species with another factor: setosa by 2, versicolor by 3 and virginica by 4. I've tried mapply but without success. Any thoughts? Thanks for any idea! Patrick

Hello, I'm running Wine 0.9.10 on a Fedora Core 5 at work. I'd like to use the Socks5 proxy we've got with application run under Wine. The application I'm using in Wine works fine using direct connection to the internet but this application has no settings I can change to inform it of the socks proxy. So I'm using Sockscap32 2.38 from www.permeo.com . It works fine under

Sorry, but I'm just an ignorant Linux user! I have a silly question. Situation: Now that R 2.1.0 is out (Thank you, to all concerned!) I'm about to go off and download it and all the packages. What this means in practice is trotting off to my "local" (and brand new) Internet Cafe with some CDs, where I can get the lot over a fast link and burn it onto a CD or two. However,

Try the code below: df <- read_delim("C:/Users/lruan1/Desktop/1112.csv", "|", escape_double = FALSE, trim_ws = TRUE) df_new <- subset(df,df$IPC == 'H04M001/02'| df$IPC == 'C07K016/26' ) You can add more condition with "|" in the subset function. Good luck! On Wed, Jan 3, 2018 at 2:53 PM, Saptorshee Kanto Chakraborty < chkstr at

Hello, I am having a problem with the get.hist.quote command (tseries library) in the Windows version. This problem is not happening is the Linux version (Mandrake 8.2). Attached is the error message, for an example included in the help file. Also the R.Version() details is attached. Please, do you know if there is a workaround ? Thanks, Carlos. ++++++++++++++++++++++++ ERROR MESSAGE

Dear r-help, I'm having this DF df <- data.frame(id = 1:6, xout = c(1234, 2134, 234, 456, 324, 345), xin= c(NA, 34,67,87,34, NA)) and would like to calculate the fraction (xin_t / xout_t-1) The result should be: # NA, 2.76, 3.14, 37.18, 7.46, NA I am sure there is a solution using zoo... but I don't know how... Thanks for any help! Patrick

Hello, I am sure this is a trivial question, but I don't get it... Having a dataframe (b.70) like that: > b.70[1:4,2:3] ARG AUS 1960 19041 25949 1961 19675 25451 1962 19302 26463 1963 18121 27644 Now I want to calculate the log for each year and save the result in a new dataframe called 'x'. x<- data.frame(w=NULL) for(i in 1:3)

Hello, I would like to switch this dataframe: > k country 1960 1961 1962 1963 99 ARG 7493 7733 7581 7108 246 AUS 10484 10342 10809 11357 295 AUT 7438 7808 7938 8212 393 BDI 587 502 555 608 442 BEL 8223 8638 9021 9311 in this structure: year ARG AUS AUT BDI BEL 1960 7493 10484 7438 587 8223 1961 7733 10342 7808 502 8638 1962 7581 10809 7938 555

On Mon, 16 Jun 2003 09:56:07 +0200 Patrick Hausmann <p.hausmann at mr-regionalberatung.de> wrote: > Dear Frank, > > I can reproduce the problem from Philippe Glaziou (see attachment). > Maybe this can help. > Best, > Patrick > > Am Wollelager 11 > 27749 Delmenhorst > Tel. 04221 96373-0 > Fax 04221 96373-29 > > http://www.mr-regionalberatung.de >

Dear list, I am trying to split a string using regexp: x <- "2 Value 34 a-c 45 t" strsplit(x, "[0-9]") [[1]] [1] "" " Value " "" " a-c " "" " t" But I don't want to lose the digits (pattern), the result should be: [[1]] [1] "2" " Value " "34" " a-c "

Dear list, I'm trying to replace NA-values with the preceding values in that column. This code works, but I am sure there is a more elegant way... df <- data.frame(id = c("A1", NA, NA, NA, "B1", NA, NA, "C1", NA, NA, NA, NA), value = c(1:12)) rn <- c(rownames(df[!is.na(df$id),]), nrow(df)+1) rn <-