similar to: sorting question

Why when I assign 0 to an element of an integer vector does the type change to numeric? Here is a particularly perplexing example: > v <- 0:10 > v [1] 0 1 2 3 4 5 6 7 8 9 10 > class(v) [1] "integer" > v[1] <- 0 > class(v) [1] "numeric" #!! > -- View this message in context:

Why when I assign 0 to an element of an integer vector does the type change to numeric? Here is a particularly perplexing example: > v <- 0:10 > v [1] 0 1 2 3 4 5 6 7 8 9 10 > class(v) [1] "integer" > v[1] <- 0 > class(v) [1] "numeric" #!! > -- View this message in context:

Hi, I have a list of numbers (classified as a list) that contains integer(0) empty vectors. How do I convert those integer(0) into NAs? Thanks [[alternative HTML version deleted]]

When operating in daemon mode, will rsync check to see which file is newer and decide which direction to transfer based on that? Example, We have 2 terminal servers, A and B. I want users to be able to work on either server and have their home directories on both stay in sync. If a user logs into server A and changes files, the server would know to transfer the file from A to B. If they log

I'd like to write some objects (eg arrays) to a log file. cat() flattens them out. I'd like them formatted as in 'print' but print only writes to stdout. Is there a simple way to achieve this result? Thanks -- View this message in context: http://www.nabble.com/print%28%29-to-file--tp24397445p24397445.html Sent from the R help mailing list archive at Nabble.com.

The situation is that I know there is a function and know approximately what the name is, and want to find the exact name. Is there a way of searching for near-matches (similar to unix apropos). For example, I know there is a function called something like allequal (or allequals or AllEquals or...). But ?allequal, etc, return nothing, only if I remember the name can I get help via ?all.equal.

I'm sure I'm missing something obvious but I'm not seeing how to simply "vectorize" a function of two or more variables. Say I have f <- function(x,y) if (x>0) y else -y Now I have vectors x and y of equal length and I'd like to apply f element-wise. I.e. conceptually z <- f(x,y) where x, y, z are vectors of the same length Some magic involving apply?

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

Suppose I have an n-diml array A and I want to extract the first "row" -- ie all elements A[1, ...] Interactively if I know 'n' I can write A[1,,,,,] with (n-1) commas. How do I do the same more generally, eg in a script? (I can think of doing this by converting A to a vector then extracting the approp elements then reshaping it to an array, but I wonder if there isn't a

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?"

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 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

Hello, I have a table (1) of the form q1 q3 q4 q8 q9 A 5 2 0 1 3 B 2 0 2 4 4 I have another table (2): q1 q2 q3 q4 q5 q6 q7 q8 q9 C 10 7 4 2 6 9 3 1 2 I would like to divide the numbers in table (1) by the number of the appropriate column in table (2): q1 q3 q4 q8 q9 A 5/10 2/4 0/2 1/1 3/2 B 2/10 0/4 2/2 4/1 4/2

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

Dear R users, I would like to create the following 3 folders (FUND1, FUND2, MARINE) within the 'parent.dir' as defined below. FUND1 <- "FD1 Q4 2009" FUND2 <- "FD2 Q4 2009" MARINE <- "MARINE Q4 2009" parent.dir <- "D:/....................." folders <- c("FUND1", "FUND2", "MARINE") for (i in

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

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