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Hi, there, I have a simple data manipulation question for you. Thank you for your help! Suppose that I have this data about people appearing in a class Mary Mary Mary Sam Sam John John John John Then I want to find out what exact time(s) the student appears at the moment such as Mary 1 Mary 2 Mary 3 Sam 1 Sam 2 John 1 John 2 John 3 John 4 the fifth row shows tha Sam show the second times

Dear R Helpers, I am missing something very elementary here, and I don't seem to get it from the help pages of the ave, seq and seq_along functions, so I wonder if you could offer a quick help. To use an example from an earlier post on this list, I have a dataframe of this kind: dat = data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4))) dat$freq =

Hi, R function aggregate can only take summary stats functions, can I aggregate text columns? For example, for the dataframe below, > a <- rbind(data.frame(id=1, name='Tom', hobby='fishing'),data.frame(id=1, name='Tom', hobby='reading'),data.frame(id=2, name='Mary', hobby='reading'),data.frame(id=3, name='John',

Hello! I have a matrix a with 2 variables (see below) that contain character strings. I need to create a 3rd variable that contains True if the value in column x is equal to the value in column y. The code below does it. a<-data.frame(x=c("john", "mary", "mary", "john"),y=c("mary","mary","john","john"))

Hi, I have a data frame with 3 columns: ID, year and score. How can I select for each unique ID, the year that has the max score? For example, for data frame ID, year, score tom, 1995, 88 rick, 1994, 90 mary, 2000, 97 tom, 1998, 60 mary, 1998,100 I shall have ID, year, score tom, 1995, 88 rick, 1994, 90 mary, 1998,100 Thanks, Richard [[alternative HTML version deleted]]

Hello, I am having a problem with the lmtp delivery of mail addressed to multiple recipients. ?The first recipient receives the mail correctly but the next recipients do not. ?ltmp apparently tries to create a link from the subsequent recipients' mail store to the first recipient mail store and fails. This returns 451 smtp codes to the sender which then resends at some later time. Please see

Hallo, I have got an existing data frame and want to add a new column. The existing data frame was created like this: > df <- rbind( c("Fred", "Mary", 4), c("Fred", "Mary", 7), + c("Fred", "Mary", 9), c("Barney", "Liz", 3), + c("Barney", "Liz", 5) ) > df

Hello, I have an unbalanced panel data set that looks like: ID,YEAR,HEIGHT Tom,2007,65 Tom,2008,66 Mary,2007,45 Mary,2008,50 Harry,2007,62 Harry,2008,62 James,2007,68 Jack,2007,70 Jordan,2008,72 That is, James, Jack, and Jordan are missing a YEAR. Is there any command that will "fill in" the missing YEAR such that the end result will be balanced and look like: ID,YEAR,HEIGHT

I will be out of the office starting Wed June 9th and returning Wed June 16th. Please contact Mary at mary at accessgate.net cell 407-267-1463 or Jonathan at jsnyder at accessgate.net cell 407-267-0056 or call our main number 888-227-9337.

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...0 0 1 1 1 1 Josh 1 0 0 0 0 0 Probably, this has been addressed before - I just don't know how to search for the answer with the right search terms. Any help is appreciated, even just a link to a page where this is addressed! Thank you! Annemarie -- Annemarie Verkerk, MA Evolutionary Processes in Language and Culture (PhD student) Max Planck Institute for Psycholinguistics P.O. Box 310, 6500AH Nijmegen, The Netherlands +31 (0)24 3521 185 http://www.mpi.nl/research/research-projects/evolutionary-processes

Hi I am having trouble using read.ssd. Can someone help? The code that I have written is *sashome<-"C:/Mary/Datasets"* *read.ssd(read.ssd(file.path(sashome, "core", "sashelp"), "surv_1v",* *sascmd = file.path(sashome, "sas.exe"))* Here the path that I have given is correct, where the dataset surv_1v.sas7bdat is kept. The message that R gives

I have two users on my network, Mary and Bob, who work together in a shared share. They both belong to the group Accounting. Bob is a savvy Linux user who accesses the share via NFS4. Mary toils away using Windows accessing the share via the Samba server. Mary will create a directory on the share and dump a number of files in which Bob and Mary will split the load. Bob, being a LInux user, will

Hi I have a character vector with thousands of names which looks like this: > V=c("Fred", "Mary", "SAM") > V [1] "Fred" "Mary" "SAM" > class(V) [1] "character" I would like to change it to a list: > L=as.list(V) > L [[1]] [1] "Fred" [[2]] [1] "Mary" [[3]] [1] "SAM" but I need to

Dear R experts, I'm reading data from an online database via API and it gets delivered in this messy comma separated structure, > RAW.API <-

Hi UserRs, I know that there has to be an easy way to do this in R (probably easy enough that once someone clues me in I'll smack myself on the forehead for not figuring it out myself), but my searches on my own have not yielded any hints. I have many fields in my dataset that participants entered as "free lists" - i.e., the field constitutes a varying number of names each

Hi all, let's say I have matrix People Desc Value Mary Height 50 Mary Weight 100 Fanny Height 60 Fanny Height 200 Is there a quick way to form the following matrix? People Height Weight Mary 50 100 Fanny 60 200 (Assuming I don't know the length of people/desc and let's say these are characters matrix.. I tried

Wow ! so many thanks Arun and Rui works like a charm problem solved 2013/4/13 arun <smartpink111@yahoo.com> > Hi, > Try this; > library(reshape2) > res<-dcast(Input,people~place,value.var="time") > res[is.na(res)]<-0 > res > # people beach home school sport > #1 Joe 5 3 0 1 > #2 Marc 0 4 2 0 > #3 Mary

Hello I have a dataset of people spending time in places. But most people don't hang out in all the places. it looks like: > Input<-data.frame(people=c("Marc","Marc","Joe","Joe","Joe","Mary"), + place=c("school","home","home","sport","beach","school"), +

Hello, I am having trouble getting the output from the tapply function formatted so that it can be made into a nice table. Below is my question written in R code. Does anyone have any suggestions? Thank you. Geoff #Input the data; name <- c('Tom', 'Tom', 'Jane', 'Jane', 'Enzo', 'Enzo', 'Mary', 'Mary'); year <- c(2008, 2009,