similar to: special recursive filter

I have 2 vectors A and B. For each element of B I'd like to find the index of the next higher or equal element in A. And I'd like to store it effectiv. E.g.: A <- c(1,3,7,8,10) B <- c(5,8) result: 3, 4 I have a possibility but for long vectors it works not very effectiv: ans <- sapply(B, function(x) which.max(A[A < x]) ) as.integer(ans + 1) Is there anyone how has a better

Hello, I've written a small function that's supposed to save me some time, and it's ending up killing it- the intention is to iteratively subset a dataset fram on framevec, fit a model (either lm or nls depending on type) and return the r2 or AIC from the model, respectively. Although as far as I can tell in my code the plots are dependent on the fit of the model to the data and the

Hi! I am trying to build a package which calls an c function. It compiles it and installs it. But if i run R CMD check test_ the last output are: * checking examples ... ERROR Running examples failed. x<-1:10 and in the test-Ex.out file i find: > test(x) Error in .C("mytest", as.double(x), as.integer(lengthx)) : C/Fortran function name not in load table Execution halte

I have a list 'ans' from the following code: tt <- rnorm(50) rr <- rnorm(50) ans <- lm(rr~tt) ans[1] is "$coefficients", ans[2] is "$residuals", ans[3] is "$effects", ... and so on up to ans[12]. Is there an easy way to display just these names and not the data they contain? I thought I saw my advisor type "ans$" and they were displayed,

Dear all, I had a R segmentation fault, and then invoked debug mode and ran step by step. When I reached "terms(Y~X1*X2*...*X16)", I would then have "segmentation" fault. However, if I just ran this under regular "R interactive" mode, it would be fine though taking long time. My questions are: 1. Is there a known limit of terms for a formula? 2. Why does the

Dear all, I had a R segmentation fault, and then invoked debug mode and ran step by step. When I reached "terms(Y~X1*X2*...*X16)", I would then have "segmentation" fault. However, if I just ran this under regular "R interactive" mode, it would be fine though taking long time. My questions are: 1. Is there a known limit of terms for a formula? 2. Why does the

Hi, I am trying to get different combinations of a vector of different size and find their sum. For example, if I have a vector (i,j) where i and j can be anything from 0 to 5, i get these combinations (0,0), (0,1), (1,0), (1,1), (2,0), ...... (5,5) and find sum of these as 0, 1, 1, 2, ..... , 10. I used outer functions to get this and it worked. What if I have a vector (i,j,k) where all i, j ,

Hello! I have a data set: set.seed(123) y<-data.frame(week=seq(as.Date("2010-01-03"), as.Date("2011-01-31"),by="week")) y$var1<-c(1,2,3,round(rnorm(54),1)) y$var2<-c(10,20,30,round(rnorm(54),1)) # All I need is to create lagged variables for var1 and var2. I looked around a bit and found several ways of doing it. They all seem quite complicated - while in

Hi, I just started with writing functions in R and so some questions popped up. I provide some values as argument to my function such as: function(a,b,c){} Now i want that the function first checks if the arguments are valid for the function. E.g argument "a" has to be a number in the range 0-1. How can that easily done? So far I have: a <- as.numeric(a) if(0 <= a &&

Hi all, How to store recursive resutls from a function for each step without using global operators <<-? Thanks ahead. Xiaohui Chen Dept. of Statistics UBC, Canada _________________________________________________________________ Don?t waste time standing in line?try shopping online. Visit Sympatico / MSN

Dear all,   could you please give me some pointers on how I could make R screen for a value if it falls within a certain range? I looked at the subset function, but is not doing it, perhaps because I only have 1 value to screen?   aptreciate the input   ex: a <-16.5 I would like to screen to see if a is within the range of 15 to 20, (which it is:-)), and I would like the code to return a value

Just to inform: I posted that before in R-sig-ecology but as it might be interesting also for other useRs, I post it also to the general r-user list: Hello Alexandre, thank you very much. I also found another way to extract summarizing information from lm results over e.g. 1000 repeated model runs: results2 <- t(as.data.frame(results)) summary(results2) Although some questions popped up in

Hi, I'm presently testing to upgrade from 3.0.22 to 3.0.25 and have an ads problem. net -d 10 -s /etc/samba/smb.conf ads join 3.0.25: ======= [2007/07/18 12:03:11, 4] libsmb/namequery_dc.c:ads_dc_name(131) ads_dc_name: using server='ADS-2K3.ADS2K3.Q-LEAP.DE' IP=192.168.53.212 [...] [2007/07/18 12:03:11, 3] libads/ldap.c:ads_connect(394) Connected to LDAP server 192.168.53.212

Hi all, we got a new test system here and I just also tested btrfs raid6 on that. Write performance is slightly lower than hw-raid (LSI megasas) and md-raid6, but it probably would be much better than any of these two, if it wouldn''t read all the during the writes. Is this a known issue? This is with linux-3.9.2. Thanks, Bernd -- To unsubscribe from this list: send the line

There is an issue with the new Kerberos 1.5. It does not currently support building static libraries. I'm willing to leave the port at 1.4.3 until MIT fixes the static library build. OTOH, if folks want 1.5, without static library support, the 1.5 port is ready to commit. I may update the port to build 1.5 if static libraries are not wanted and build 1.4.3 if they are wanted. Static

List serv subscribers, I am wondering if there is an efficient means of doing recursive matrix multiplication in R. My data resides in a 4 X 2541 matrix from which I need to extract 2541 2X2 matrices based on each row. If I attempt something like this: function(AO) {A<-AO {for(t in 1:4)

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966

I have the following problem: I am using igraph and entering a graph from an external (.txt) file using read.graph("a.txt","ncol") In a.xtx the name of the nodes contains special characters (such Polish names, etc). In the R console, the visualization of the names is not correct and I am loosing some of the special characters. Any help will be greatly appreciate. Renato

Hi, I'm now to R and hope (actually, I'm quite sure) you can help me. I made an experiment and measured two values. As I know the errors of these values I want to use them with the linear regression, eg Value A 1.1+-0.02 1.9+-0.05 3.05+-0.03 4.0+-0.01 5.1+-0.06 Value B 4.2+-0.14 5.3+-0.05 6.8+-0.11 7.9+-0.01 8.5+-0.02 lm(B~A) does the linear regression but how can I

Hi! I'm looking for a streaming client for the icecast2-server. It should just feed mp3-files from a playlist to the icecast2-server. The old shout doesn't work: hout -P a -e 8000 -x -p /big/m3u/list.m3u => [3:21] Size: 3231748 Bitrate: 128000 (41432 bytes/dot) [Server error: [HTTP/1.0 401 Authentication Required ] WWW-Authenticate: Basic realm="Icecast2 Server"