similar to: Determine longest run of number

Classification: UNCLASSIFIED Caveats: NONE Thanks for your quick response. The program you mentioned below available from R is based on number of runs (up or down) not based on a longest length of runs of same events. To be more specific, for example, from a series, HHTHTTTTHHH, the number of runs are 5, and the longest length of runs of the same events is 4. I'll check for the website you

Hi Everyone, I'm new to programming R and have accomplished my goal, but feel that there is probably a more efficient way of coding this. I'd appreciate any guidance that a more advanced programmer can provide. My goal -- I would like to find the length of the longest word in a string containing many words separated by spaces. How I did it -- I was able to find the length of the

I need to determine the length of the longest series of consecutive numbers (1's to be specific) and the start time of that series. For example, in the following sample, the first column is "time" and the second column indicates the presence of the target behavior. I would like a function that would return "21" as the start time and "17" as the length. sample

...if that is possible My task is to find the longest streak of continuous days a user participated in a game. Instead of writing an sql function, I chose to use the R's rle function, to get the longest streaks and then update my db table with the results. The (attached) dataframe is something like this: day user_id 2008/11/01 2001 2008/11/01 2002 2008/11/01 2003

I have a matrix with NAs and want to extract the longest column. > is.matrix(foo) [1] TRUE > dim(foo) [1] 2000 75 > GetLength <- function(x) {length(na.omit(x))} > junk <- apply(foo, 2, GetLength) > junk [1] 1004 512 432 523 691 396 607 838 [9] 730 389 388 445 609 333 637 1024 [17] 1163 823 718 466 799 459 701 833 [25] 456 549 376 728 539 384

Classification: UNCLASSIFIED Caveats: NONE Hello, R users, Has anybody written a function for random test based on the length of longest run of same events. I really appreciate your help. Kyong Park Classification: UNCLASSIFIED Caveats: NONE [[alternative HTML version deleted]]

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let

why? > rle Run Length Encoding lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ... values : chr [1:1650061] "4bbf9e94cbceb70c BG bg" "4fbbf2c67e0fb867 SK sk" ... > as.data.frame(rle) Error in as.data.frame.default(vertices.rle) : cannot coerce class '"rle"' into a data.frame it seems that rle.df <-

Does anyone know off hand what the longest known serious security issue (i.e. remote compromise) has been with FreeBSD that went unpatched ? e.g. security hole is reported to security-officer@FreeBSD.org. X days later, fix and advisory committed. What has been the largest X ? My jaw dropped when I saw http://www.eeye.com/html/Research/Upcoming/index.html ---Mike

Hi, I have a small problem with the function split() and would appreciate your help. I have a table called ?XXX? with 2 columns and 49 rows. The 49 rows belong to 8 different levels (electrode1, ...,electrode8). I want to split the table always at the row where ?electrode1? starts again so that I can export 7 individual dataframes (numbered ?dataframe1? to ?dataframe7?) which contain always

Hello everyone, I have two vectors having only one element different: vector1 vector2 vector1 TWC TWC TWC VFC TWX NA VIA/B VFC VFC

Hello, I already download rm160.sit to a iMac9.1 computer. But I can not run R-icon for installation of R. The computer told me --the application "R" could not be opend because "Carbonlib" could not be found. I donot know why I could not install R to the computer. Thank you. Jinbo __________________________________________________ Faith Hill - Exclusive Performances,

Hello R community: I'm encountering unexpected behaviour as I download string data from MySQL using the RODBC package. The unexpected behaviour is that the last character of the string is truncated. As far as I can see, this happens when some strings are of length 501 or more. Here's how to reproduce this behaviour. ## In MySQL, pick a database and run this: CREATE TABLE test ( v1

Dear R-Help i have a newbie question, how edit size fonts and colors in the plots (scatter.acm or dudi.acm function) of multiple correspondence analysis in the ADE4 Package? thanks? Rafael Gutierrez Estad??stico Unidad de Tecnolog??a Cerro Matoso S.A. Tel. 4-7723350 Fax. 4-7723236 ************************************************************************************************* Este correo

On Dec 18, 2012, at 2:07 PM, Chandler Carruth <chandlerc at google.com> wrote: > On Tue, Dec 18, 2012 at 1:13 PM, Eli Bendersky <eliben at google.com> wrote: > On Tue, Dec 18, 2012 at 1:09 PM, Craig Topper <craig.topper at gmail.com> wrote: > > But its pretty easy to change the tabstop within the editor to make it > > readable. > > > > True, in

Hi. The description of na.contiguous says: "Find the longest consecutive stretch of non-missing values in a time series object. (In the event of a tie, the first such stretch.)" But this seems not to be the case if one of the tied longest stretches is at the start of the sequence/series. In the following example, there are three stretches of length 3, so I expect the result

Hi, According to its man page substring() "expands (its) arguments cyclically to the length of the longest _provided_ none are of zero length". So, as expected, I get an error here: > substring("abcd", first=2L, last=integer(0)) Error in substring("abcd", first = 2L, last = integer(0)) : invalid substring argument(s) But I don't get one here:

> > Maybe it's naive, but I would expect it to be easy for each backend to > > expose a constant N which is the length of the longest mnemonic, and then > > for the printer to pad to N+1 or N+2…. > > That would probably work for X86, but other targets (ARM in particular) > often have operands which are printed/parsed as suffices on the mnemonic > itself. Because

Hi, In R < 3.0.0, we used to get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) A B A B A "bcdef" "2345" "bcd" "23" "b" But in R >= 3.0.0, we get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) [1] "bcdef"

Hello R-users, I need to create a vector inserting an 1 after each value of another vector. For example: vec1<-c(2,3,4) I need to create a vector with the values 2,1,3,1,4 Does anyone know how create this vector without loops (vec1 could have 1000 elements) Thank you, Juan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read