plyr

I have a data frame with 3 columns: custId, saleDate, DelivDateTime. > head(events22) custId saleDate DelivDate 1 280356593 2012-11-14 14:04:59 11/14/12 17:29 2 280367076 2012-11-14 17:04:44 11/14/12 20:48 3 280380097 2012-11-14 17:38:34 11/14/12 20:45 4 280380095 2012-11-14 20:45:44 11/14/12 23:59 5 280380095 2012-11-14 20:31:39 11/14/12 23:49 6 280380095 2012-11-14 19:58:32 11/15/12 00:10 Here's the dput: > dput(events22) structure(list(custId = c(280356593L, 280367076L, 280380097L, 280380095L, 280380095L, 280380095L, 280364279L, 280364279L, 280398506L, 280336395L, 280364376L, 280368458L,

This question already has an answer here: Calculate group mean (or other summary stats) and assign to original data 4 answers Is there a faster way to do this? I guess this is unnecessary slow and that a task like this can be accomplished with base functions. df <- ddply(df, "id", function(x) cbind(x, perc.total = sum(x$cand.perc))) I'm quite new to R. I have looked at by() , aggregate() and tapply() , but didn't get them to work at all or in the way I wanted. Rather than returning a shorter vector, I want to attach the sum to the original dataframe. What is the best way to do this? Edit: Here

I have a dataset like this: df = data.frame(group = c(rep('A',4), rep('B',3)), subgroup = c('a', 'b', 'c', 'd', 'a', 'b', 'c'), value = c(1,4,2,1,1,2,3)) group | subgroup | value ------------------------ A | a | 1 A | b | 4 A | c | 2 A | d | 1 B | a | 1 B | b | 2 B | c | 3 What I want is to get the percentage of the values of each subgroup within each group, i.e. the output should be: group | subgroup | percent ------------------------ A | a | 0.125 A | b | 0.500 A | c | 0.250 A | d | 0.125 B | a | 0.167 B | b | 0.333 B | c | 0.500 Example for group A, subgroup A: the value was 1, the sum of

The following code produces bar plots with standard error bars using Hmisc, ddply and ggplot: means_se <- ddply(mtcars,.(cyl), function(df) smean.sdl(df$qsec,mult=sqrt(length(df$qsec))^-1)) colnames(means_se) <- c("cyl","mean","lower","upper") ggplot(means_se,aes(cyl,mean,ymax=upper,ymin=lower,group=1)) + geom_bar(stat="identity") + geom_errorbar() However, implementing the above using helper functions such as mean_sdl seems much better. For example the following code produces a plot with 95% CI error bars: ggplot(mtcars, aes(cyl, qsec)) + stat_summary(fun.y = mean, geom = "bar") + stat

I want to sum rows that have the same value in one column: > df <- data.frame("1"=c("a","b","a","c","c"), "2"=c(1,5,3,6,2), "3"=c(3,3,4,5,2)) > df X1 X2 X3 1 a 1 3 2 b 5 3 3 a 3 4 4 c 6 5 5 c 2 2 For one column (X2), the data can be aggregated to get the sums of all rows that have the same X1 value: > ddply(df, .(X1), summarise, X2=sum(X2)) X1 X2 1 a 4 2 b 5 3 c 8 How do I do the same for X3 and an arbitrary number of other columns except X1? This is the result I want: X1 X2 X3 1 a 4 7 2 b 5 3 3 c 8 7 ddply(df, "X1", numcolwise(sum)) see ?numcolwise for details and examples. aggregate can

Suppose you have a data frame with the following structure: df <- data.frame(a=c(1,2,3,4), b=c("job1;job2", "job1a", "job4;job5;job6", "job9;job10;job11")) where the column b is a semicolon-delimited list (unbalanced by row). The ideal data.frame would be: id,job,jobNum 1,job1,1 1,job2,2 ... 3,job6,3 4,job9,1 4,job10,2 4,job11,3 I have a partial solution that takes almost 2 hours (170K rows): # Split the column by the semicolon. Results in a list. df$allJobs <- strsplit(df$b, ";", fixed=TRUE) # Function to reshape column that is a list as a data.frame simpleStack <- function(data){ start <- as

I have a dataframe consisting of an ID , that is the same for each element in a group, two datetimes and the time interval between these two. One of the datetime objects is my relevant time marker. Now I like to get a subset of the dataframe that consists of the earliest entry for each group. The entries (especially the time interval) need to stay untouched. My first approach was to sort the frame according to 1. ID and 2. relevant datetime. However, I wasn't able to return the first entry for each new group. I then have been looking at the aggregate() as well as ddply() function but I could