similar to: Expanding a data structure

I cannot find out how to build data structures as lists of data structures. I want to do... r-help@r-project.org d1 <- data.frame(x=1, y=2) d2 <- data.frame(x=1, y=3) d3 <- data.frame(x=21, y=3) q1 <- data.frame(q=d1, n="a") q2 <- data.frame(q=d2, n="a") q3 <- data.frame(q=d3, n="a") v <- vector() or list() or whatever v[1] <- q1 v[2]

Example: Will ts objects be obsolete or modified? > a [,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035> lag(a,4) [,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

# Dear List, # I want to characterize a time series according to its Quarter components. # My data ("a.ts": http://docs.google.com/View?id=dfvvwzr2_478cr9k4cdb)? look like: #???????????????? Qtr1????????? Qtr2????????? Qtr3????????? Qtr4 #?? 1948 -0.0714961837? 0.0101747827? 0.0654816569 -0.0227830729 #?? 1949 -0.1175517556? 0.1151378692? 0.1015777858 -0.1971535900 #?? 1950?

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"

Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. > x <- zoo(11:21) > z <- zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) > x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 > lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 > z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4

useRs ? I?m working with the attached data that contains one year?s worth of sub-daily observations of flow (?Q?) and specific conductance (?SC?, a surrogate for concentration) at a point in a stream. The R code posted below shows the extent of data processing thus far. My goal is to create a covariance matrix that takes on the following form: Q1 Q2 ? Q365 SC1 SC2 ? SC365 Q1 Q2 ? Q365

I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101

Hi, I have a question about how to obtain the union of several data frame rows. I'm trying to create a common key for several tests composed of different items. Here is a small scale version of the problem. These are keys for 4 different tests, not all mutually exclusive: id q1 q2 q3 q4 q5 q6 1 A C 2 B D 3 A D B 4 C D B D I would like

I am generating an ftable (by running ftable on the results of a xtabs command) and I am getting the following. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?Var1? Var2 date ? ? ? ? ? ? ? ? group? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2007-01-01? ? ? ? ? q1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?1? ? 9 ? ? ? ? ? ? ? ? ? ? ?q2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101

hi, for example, I have an string "q1 q2 q3". how can I encode it in rails to produce the result like "q1%20q2%20q3". Thanks! Best Regards, Zhenjian -------------- next part -------------- An HTML attachment was scrubbed... URL: http://wrath.rubyonrails.org/pipermail/rails/attachments/20060717/3824f992/attachment.html

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

----- Original Message ----- From: "Silvano" <silvano at uel.br> To: <r-help at r-project.org> Sent: Thursday, June 30, 2011 9:07 AM Subject: Tables and merge > Hi, > > I have 21 files which is common variable CODE. > Each file refers to a question. > > I would like to join the 21 files into one, to construct > tables for each question by CODE. >

Hi R users, I have following data frame df<-data.frame(q1=c(0,0,33.33,"check"),q2=c(0,33.33,"check",9.156), q3=c("check","check",25,100),q4=c(7.123,35,100,"check")) and i would like to replace every element that is less then 10 with . (dot) in order to obtain this: q1 q2 q3 q4 1 . . check . 2 . 33.33 check 35

Dear R-helpers, I know of two ways to reading data within an R program, using textConnection and stdin (demo program below). I've Googled about and looked in several books for comparisons of the two approaches but haven't found anything. Are there any particular advantages or disadvantages to these two approaches? If you were teaching R beginners, which would you present? Thanks, Bob

Hi All, There are a variety of functions that can be applied to a variable (column) in a data frame: mean, min, max, sd, range, IQR, etc. I am aware of only two that work on the rows, using q1-q3 as example variables: rowMeans(cbind(q1,q2,q3),na.rm=T) #mean of multiple variables rowSums (cbind(q1,q2,q3),na.rm=T) #sum of multiple variables Can the standard column functions (listed in the

I got the code for how to do regression without an intercept out of the back of my book and the next part of the homework asks me to do it with an intercept. The problem is, Q1 disappears whenever I try. Here is my code: Without the intercept: load("tsa3.rda") > > Q=factor(rep(1:4,21)) > reg=lm(log(jj)~0+trend+Q,na.action=NULL) > model.matrix(reg) trend Q1 Q2 Q3 Q4 1

Dear R users, Something is strange in summary and IQR. Suppose, I have a data set and I would like to find the Q1, Q2, Q3 and IQR. x<-c(2,4,11,12,13,15,31,31,37,47) > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 2.00 11.25 14.00 20.30 31.00 47.00 > IQR(x) [1] 19.75 However, I test the same data set in SAS "proc univariate", and SAS shows that