similar to: finding row and column indices of date in multiple columns of a data frame

Hello, I am getting output from anova() and summary(aov()) that depends on the order of the factors in the fitted model object, and this has me baffled. I see this dependency with the data.frame below but not with an example (table 6.4) from Montgomery's DOE book. This is with R 1.3.0 on Debian GNU-Linux. Where have I gone wrong? > centerpts run sample CH50mg 1 day1 dev126 0.56 2

Hi, If I have a data frame A with the following format: Day1 Day2 Day3 Day4 1 1979-11-02 1979-11-03 1979-11-04 <NA> 2 1979-12-06 <NA> <NA> <NA> 3 1979-12-13 1979-12-14 1979-12-15 1979-12-16 4 1979-12-20 <NA> <NA> <NA> And a date "1979-12-14", for

Hi, I am an R novice and I am trying to do something that it seems should be fairly simple, but I can't quite figure it out and I must not be using the right words when I search for answers. I have a dataset with a number of individuals and observations for each day (7 possible codes plus missing data) So it looks something like this Individual A, B, C, D Day1 1,1,1,1 Day 2 1,3,4,2 Day3

Can someone help me with the proper usage of the reshape function? Let's say I have a dataset with columns like this (wide format): Id Sex Group Test Day1 Day 2 Day 3... And I want to transpose this into something like this (thin): Id Sex Group Test Time Where the new column labeled time contains all the time variables (Day 1, Day 2, Day 3...) that were in the wide

I would like to include this in a package. The S3 methods on R CMD check says * checking S3 generic/method consistency ... WARNING window: function(x, ...) window.chron: function(data, day1, hour1, day2, hour2, ...) See section 'Generic functions and methods' of the 'Writing R Extensions' manual. I have looked and can not figure it out. This function is for convience. What

HI all, raw_urine = read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\mz.spot.sam.dat.new", header = TRUE ) pvalue = read.table("Z:\\bruce.9.3.09.sample.stability.analysis\\urine\\all.urine.features.t.test.result", header = TRUE ) library(compositions) p = function(a,b){ y = pvalue[,a] if(y<0.01){ index = which(y, arr.ind=TRUE) day1 = raw_urine[index,3:7] day2 =

Dear group, Here is my environment : > ls() [1] "l" "PLglobal" "Pos100415" "Pos100416" "Pos100419" "Pos100420" "position" "select" "Trad100415" "Trad100416" "Trad100419" "Trad100420" "trade" "y" With objects : > l [1]

I have the following function defined as below match.trace <- function(dfobj, distance, day1, day2) { day1 <- substitute(dfobj$day1); day1 day2 <- substitute(dfobj$day2) distance <- substitute(dfobj$distance) xx <- NULL for (i in 0:10) xx[i+1] <- with(dfobj, cor(Lag((day1-day1[1]),i), (day2-day2[1]), use='pair')) i <- match(max(xx), xx) with(dfobj, {

Dear helpeRs- I'm using a for loop to create a series of models. I'm trying to assign a name to each model created, using the loop index. The loop gets stuck at the name of the model, giving the error "target of assignment expands to non-language object". The linear model runs without error; only the name is problematic. Here is the current loop syntax. The use of dat

Hi, i did a mistake with my first post. I have to reshape data from this matrix: id x1 x2 x3 x4 day1 day2 day3 day4 day5 day6 day7 day8 day9 1 0.129 0.797 0.231 0.615 4 4 1 1 1 1 3 3 3 2 0.420 0.376 0.501 0.282 4 4 4 4 5 4 2 5 5 3 0.377 0.486

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I have this problem with records that have to be alligned on end- and startdate. I came up with the following. but am not convinced that this is the best way to tackle this problem. Anyone care for a braindump? def head_to_tail # Remove days that have startdate >= self.startdate AND enddate <= self.enddate # # before after # # |=====| |=====|

Dear R users,   I need to draw a barplot with 2 Y axis. I have 3 days each of wich having 2 groups (and error bar for each of them). The height of the 3rd day is too tall compared to others. That's why I have to use a second Y axis for that. I am using  "barplot2" function of "gplots" library (to be able to add error bars as well). Data and  codes currently I am using is

Dear all, I have the following data in excel: day1 y 1 2 3 2 3 x1 0.2 0.3 0.4 0.3 0.2 x2 7 3.4 2 8 6 day2 y 2 4 3 2 2 x1 0.4 0.5 0.3 0.3 0.2 x2 7 8 9.1 6 5 I have the following problems: first of all, when I ask R to read the file (with the package xlsReadWrite and the command read.xls) it has a problem with the fact that the left most corner is labelled the same way, so in order for it

This patch adds support for symbol versions. It is based on what libvirt does. The generated syms file looks like: LIBNBD_1.0 { global: nbd_...; nbd_...; local: *; }; In a future stable 1.2 release, new symbols would go into a new section which would look like this: LIBNBD_1.2 { global: nbd_new_symbol; nbd_another_new_symbol; local: *; } LIBNBD_1.0; In my testing the

Hi all, I have been bumbling around with r for years now and still havent come up with a solution for plotting reliable graphs of relationships from a linear regression. Here is an example illustrating my problem 1.I do a linear regression as follows summary(lm(n.day13~n.day1+ffemale.yell+fmale.yell+fmale.chroma,data=surv)) which gives some nice sig. results Coefficients:

thanks to your guys help I am closer to solving my problem but I have some small problem. So let's say I start with >data number day hour 1 17 10 2 17 11 3 17 6 4 18 4 5 18 10 6 19 8 7 19 8 I want to split to odd days, which I am able to do, I call this object frames, which looks like: > frames $`1` c1 day1 hour1 1 1 17 10 2 2 17 11 3 3 17 6 $`2` c1 day1

Hello, I just started learning R and have a very basic question: When I try ar model and extract residuals it returns Null. Could someone explain what's going on here? Does that mean the model is null? But it seems residual has a length=# observation of the original series according to the model summary. I am learning it by myself and have no one to ask here so please allow me to ask this

Hi All I want to import several .dat files every day of the week from the same folder. Let say, on day 1, I have about 100 files in the folder. By using this code, everything works perfectly (maybe there is a more efficient way to do it): filenames <-list.files(path="pathtofile", full.names=TRUE) library(plyr) import.list <- llply(filenames, read.table, header=TRUE,

Here''s an intersting one: I have a client who wants to store hours of operation of a business for their RoR web app. I came up with a solution, but I really don''t like it. I have a MySQL table that has these fields monday_start, monday_end, tuesday_start, tuesday_end, etc., all as time fields. That''s all find and dandy. The kicker is, the client wants the hours to