similar to: Lists of data.frame

I have a xts object with logical data . > tail(q1) ..1 2010-02-19 TRUE 2010-02-22 FALSE 2010-02-23 FALSE 2010-02-24 FALSE 2010-02-25 FALSE 2010-02-26 FALSE > I want to build a xts that records the dates that there is a change. If TRUE -> FALSE it is "down" if FALSE -> TRUE it is "up" I can do it like this... q2 <- matrix(ncol=2, nrow =

I would like to know how many years (including partial years) are between two dates. So difftime("20/11/1962", "20/5/1964", units="years") would be about 1.5 But units of years are not available. cheers Worik [[alternative HTML version deleted]]

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

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

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, 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 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

Hello, I have a question on how to tabulate the time series data. I use RStudio, but if can be done in any other R editor, it should work in RStudio as well. > a1<-11:22 > a1ts<-ts(a1, frequency=4, start=c(1978,1)) > a1ts Qtr1 Qtr2 Qtr3 Qtr4 1978 11 12 13 14 1979 15 16 17 18 1980 19 20 21 22 If I click the variable "a1ts" on the

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 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 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

----- 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. >

I have a data set with quarterly time series for several variables. The time index is recorded in column 1 of the dataframe as a character vector "Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want to produce line plots with ggplot2, but it seems I need to convert the time index from character to date class. Is that right? If so, how

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 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

It seems obvious to me that the empty string "" is length 0. cheers Worik [[alternative HTML version deleted]]

I have trouble with this: as.Date("Sep-1981", format="%b-%Y") Returns "NA" >From documentation for strftime '%b' Abbreviated month name in the current locale. (Also matches full name on input.) '%Y' Year with century. What am I doing wrong? cheers Worik [[alternative HTML version deleted]]