search for: inx

...630649 1216399235 231414 11 11 4 9 11-11 1216637080 1216631541 5539 11 11 4 10 11-11 1216646563 1216640763 5800 11 11 4 11 11-11 1216656338 1216651635 4703 11 11 " df1 <-read.table(textConnection(text), header=TRUE) inx <- df1$commutation > 1000 comm1000 <- cumsum(inx) result <- split(df1[!inx, ], list(comm1000[!inx], df1$v_source[!inx], df1$v_destine[!inx])) result <- sapply(result, function(x) c(x$val_user[1], x$v_source[1], x$v_destine[1], nrow(x), mean(x$comm))) result <- na.exclude(t(result...

Hello, Maybe something like the following. splitDF <- function(data, col, s){ n <- nrow(data) inx <- which(data[[col]] %in% s) lapply(seq_along(inx), function(i){ k <- if(inx[i] < n) (inx[i] + 1):(inx[i + 1]) data[k, ] }) } splitDF(DF, "name", split_str) Hope this helps, Rui Barradas On 5/19/2018 12:07 PM, Christofer Bogaso wrote: > Hi, &gt...

Hi, I am struggling to split a data.frame as will below scheme : DF = data.frame(name = c('a', 'v', 'c'), val = 0); DF split_str = c('a', 'c') Now, for each element in split_str, R should find which row of DF contains that element, and return DF with all rows starting from next row of the corresponding element and ending with the preceding value of the

...e and password. ## Assume it will always contain pairs: ## username \t password \n username \t password \n ## Match the username to 'user' and then pickup the password as the next ## item. i.1 <- seq(1, length(input), by=2) usr <- input[i.1] psd <- input[i.1 + 1] inx <- usr==user if(dbg) cat("\tUser=", usr[inx], "\n") if(sum(inx) < 1) stop(paste("User, '", user, "', not present in file: ", path, sep='')) if(sum(inx) > 1) stop("Unable to locate unique USER in input") if(dbg) c...

Good Afternoon, I believe that my to the problem, the R has a more effective solution. in place the use the loop I have the following set of data, and needs to extract some sections. user pos communications source v_destine 7 1 109 22 22 7 2 100 22 22 7 3 214 22 22 7 4 322 22 22 7 5 69920 22 161 7 6

...nderstand what the usage of `_data` in transform.data.frame. Could you please help me understand what `_data` means and min doesn't work row by row? > transform.data.frame function (`_data`, ...) { e <- eval(substitute(list(...)), `_data`, parent.frame()) tags <- names(e) inx <- match(tags, names(`_data`)) matched <- !is.na(inx) if (any(matched)) { `_data`[inx[matched]] <- e[matched] `_data` <- data.frame(`_data`) } if (!all(matched)) do.call("data.frame", c(list(`_data`), e[!matched])) else `_data` } &lt...

...make it work consistently, see below: transform.data.frame <- function (`_data`, ...) { e <- eval(substitute(list(...)), `_data`, parent.frame()) tags <- names(e) ## NEW LINE ----------------------------------------------- if (is.null(tags)) tags <- character(length(e)) inx <- match(tags, names(`_data`)) matched <- !is.na(inx) if (any(matched)) { `_data`[inx[matched]] <- e[matched] `_data` <- data.frame(`_data`) } if (!all(matched)) do.call("data.frame", c(list(`_data`), e[!matched])) else `_data` } transform(data...

Hello I have a data frame like this: dput(states) structure(list(Date = c("24/07/2012", "25/07/2012", "26/07/2012", "27/07/2012", "28/07/2012", "24/07/2012", "25/07/2012", "26/07/2012", "27/07/2012", "28/07/2012"), State = c(1L, 1L, 1L, 1L, 1L, -1L, -1L, -1L, 1L, -1L)), .Names = c("Date",

...unlike 'rename', whose arguments have a reverse syntax compared to 'transform'). Here is my attempt at a rename function similar to 'transform'. Improvements are welcome. ren <- function(`_data`, ...){ # Influenced by transform.data.frame e <- unlist(list(...)) inx <- match(e, names(`_data`)) if(any(is.na(inx))) stop("Couldn't find these columns in the data: ", paste(as.vector(e)[is.na(inx)], collapse=" ") ) names(`_data`)[inx] <- names(e) return(`_data`) } d0 <- data.frame(Year = c(2009, 2009, 2009),...

...??? 22 NAHAS????????? 32 GEO??????????????? 42 ", header=TRUE)) MinMaxArray? <- data.frame(MIN = 25,MAX=35) myMatrix[myMatrix$Age<=MinMaxArray$MAX & myMatrix$Age>=MinMaxArray$MIN,] #Error in myMatrix$Age : $ operator is invalid for atomic vectors #Then, I tried Rui's code; ?inx <- MinMaxArray[[ "MIN" ]] <= myMatrix[, "Age"] & myMatrix[, "Age"]<= MinMaxArray[[ "MAX" ]] ?myMatrix[ inx , ] ???? Name Age [1,]??? 1? 27 [2,]??? 3? 30 [3,]??? 4? 32 #Previously, this was also working fine. Not sure what is happening. A.K.

Hello, I have 2 variable: one is an "id" sequence from 1:1000 and the other is variable with real values "VI" from -15.0 to 20.0 and I want to detect id values that have indicator values less than a certain threshold, for example (x=1) BUT that are in sequence equal or longer than 5. For instance, in the following column I want to recognize the sequence from "id" 4

Hi R Users, I have been trying to work out how to rename column names using grep, basically I have generated these column names using tapply: [1] "NAME" "X1.1" "X2.1" "X3.1" "X4.1" "X5.1" "X6.1" "X7.1" "X8.1" [10] "X1.2" "X2.2" "X3.2" "X4.2"

Hi all, My code looks like the following: inname = read.csv("ID_error_checker.csv", as.is=TRUE) outname = read.csv("output.csv", as.is=TRUE) #My algorithm is the following: #for line in inname #if first string up to whitespace in row in inname$name = first string up to whitespace in row + 1 in inname$name #AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the

Hello, Use grep to get the row indices and then subset with a *negative* index to remove those rows. rn <- scan(what = character(), text = " 70/556 71.1/280 72.1/556 72.1/343 73.1/390 73.1/556 ") mat <- matrix(rnorm(6*6), nrow = 6) row.names(mat) <- rn inx <- grep("73\\.", row.names(mat)) new_mat <- mat[-inx, ] new_mat Hope this helps, Rui Barradas On 5/22/2018 11:48 AM, Ahmed Serag wrote: > Dear R-experts, > > > How can I remove a certain feature or observation by a part of its name. To be clear, I have a matrix wi...

...rnals.h> #include <Rdefines.h> using namespace Puppy; //Declare Functions extern "C" { unsigned int evaluateSymbReg(std::vector<Tree>& ioPopulation, Context& ioContext, const std::vector<double>& inX, const std::vector<double>& inF); void RSymbReg(SEXP RPopSize, SEXP RNbrGen, SEXP RNbrPartTournament, SEXP RMaxDepth, SEXP RMinInitDepth, SEXP RMaxInitDepth, SEXP RInitGrowProba, SEXP RCrossoverProba, SEXP RCrossDistribProba, SEXP RMutStd...

Dear R-experts, How can I remove a certain feature or observation by a part of its name. To be clear, I have a matrix with 766 observations as a rows. The row names are like this 70/556 71.1/280 72.1/556 72.1/343 73.1/390 73.1/556 Now I would like to remove all the rows that contain the text 73.1 Any ideas or suggestion please ? Regards ********************** Ahmed Serag Analytical

Is it just me? After the mail list server upgrade, the average delivery time for messages to the users list is between 4 and 5 days. The Dev list seems fine! Doug -- Ben Franklin quote: "Those who would give up Essential Liberty to purchase a little Temporary Safety, deserve neither Liberty nor Safety."

...frame <- function (`_data`, ...) { > > e <- eval(substitute(list(...)), `_data`, parent.frame()) > > tags <- names(e) > > ## NEW LINE ----------------------------------------------- > > if (is.null(tags)) tags <- character(length(e)) > > inx <- match(tags, names(`_data`)) > > matched <- !is.na(inx) > > if (any(matched)) { > > `_data`[inx[matched]] <- e[matched] > > `_data` <- data.frame(`_data`) > > } > > if (!all(matched)) > > do.call("data.fr...

...bset with a *negative* index > to remove those rows. > > rn <- scan(what = character(), text = " > 70/556 > 71.1/280 > 72.1/556 > 72.1/343 > 73.1/390 > 73.1/556 > ") > > mat <- matrix(rnorm(6*6), nrow = 6) > row.names(mat) <- rn > > inx <- grep("73\\.", row.names(mat)) > > new_mat <- mat[-inx, ] > new_mat > > > Hope this helps, > > Rui Barradas > > On 5/22/2018 11:48 AM, Ahmed Serag wrote: >> Dear R-experts, >> >> >> How can I remove a certain feature or...

Thank you to everyone in this forum that has been helping me with the basic R skills while I learn to apply them. I would like to take the coefficient of two coordinates. One of them comes from two different columns in a table: >A x y a 1 3 b 2 2 c 3 1 the other is set and for this question I'll just call it (1,1) I've been trying to find a way to return the