similar to: ave and grouping

Hello all, I have a df like this: w <- c(1.20, 1.34, 2.34, 3.12, 2.89, 4.67, 2.43, 2.89, 1.99, 3.45, 2.01, 2.23, 1.45, 1.59) g <- rep(c("a", "b"), each=7) df <- data.frame(g, w) df # 1. Mean for each group tapply(df$w, df$g, function(x) mean(x)) # 2. Range for each group - fix value 0.15 tapply(df$w, df$g, function(x) x[(x > mean(x) -

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, Can someone give me hint to change a data.frame. I want to split a column in more columns depending on the value of a other column. Thanks for the reaction, Andre Example: > dat x1 x2 1 1 a 2 1 b 3 1 c 4 2 d 5 2 e 6 2 f 7 3 g 8 3 h 9 3 i in > dur d1 d2 d3 1 a d g 2 b e h 3 c f i [[alternative HTML version deleted]]

Hi all, I would like to calculate the percent of the total per group for this data.frame: df <- data.frame(site = c("a", "a", "a", "b", "b", "b"), gr = c("total", "x1", "x2", "x1", "total","x2"), value1 = c(212, 56, 87, 33, 456, 213)) df

Dear R users, It may be very simple but it is being difficult for me. I'd like to calculate the difference in percent between to measures. My data looks like this: set.seed(123) df1 <- data.frame(measure = rep(c("A1", "A2", "A3"), each=3), water = sample(c(100:200), 9), tide = sample(c(-10:+10), 9)) df1 # What I want to calculate is:

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) ?

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

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 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

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

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 <-

Dear list, I want to apply the "table" function to every pair of variables in df and the return should be a list. setwd(123) asd <- data.frame(a1=sample(1:4, 20, replace=TRUE), a2=sample(1:4, 20, replace=TRUE), a3=sample(1:4, 20, replace=TRUE), a4=sample(1:4, 20, replace=TRUE)) with(asd, table(a1, a2)) with(asd, table(a1,

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)

Dear list, as a result of a logical operation I want to assign a new variable to a DF with NA-values. z <- data.frame( x = c(5,6,5,NA,7,5,4,NA), y = c(1,2,2,2,2,2,2,2) ) p <- (z$x <= 5) & (z$y == 1) p z[p, "p1"] <-5 z # ok, this works fine z <- z[,-3] p <- (z$x <= 5) & (z$y == 2) p z[p, "p2"] <-5 z # this failed... - how

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Dear group, Here is my data frame: futures <- structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10", "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Dear R Users, I have the following data frame: v1 <- c(rep(10,3),rep(11,2)) v2 <- sample(5:10, 5, replace = T) v3 <- c(0,1,2,0,2) df <- data.frame(v1,v2,v3) > df v1 v2 v3 1 10 9 0 2 10 5 1 3 10 6 2 4 11 7 0 5 11 5 2 I want to add a new column v4 such that its values are equal to the value of v2 conditional on v3=0 for each subgroup of v1. In the above example, the

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