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Good Afternoon, I have the following code, but it seems that something must be doing wrong, because it is giving the results I want. The idea is to create segments while the value of Commutation is less than 1000. for example, from the small set of data below text=" val_user pos v v_star v_end commutation v_source v_destine 1 1 96-96 1173438391 1173438391 0

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

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

I don't like to have my password exposed by typing at all. I also don't like to enter it each time that I wish to open a database (or when I run scripts automatically across a Linux cluster). My solution has been to keep a file in my HOME directory containing the username and password for the databases. This file has read and write permission set so only I (and root) can read it; this is

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

Hi All, I meant to take the min row by row. But the result is apparently not what I want. Changing min to pmin solve the problem. > df=data.frame(X=1:10, Y=1:10) > transform(df, Z=min(X,10-Y)) X Y Z 1 1 1 0 2 2 2 0 3 3 3 0 4 4 4 0 5 5 5 0 6 6 6 0 7 7 7 0 8 8 8 0 9 9 9 0 10 10 10 0 I try to look at the source code to understand what transform() does. I know

Dear r-devel, See below: transform(data.frame(a = 1), 2, 3) #> a #> 1 1 transform(data.frame(a = 1), b=2, 3) #> a b X3 #> 1 1 2 3 We need a small modification to make it work consistently, see below: transform.data.frame <- function (`_data`, ...) { e <- eval(substitute(list(...)), `_data`, parent.frame()) tags <- names(e) ## NEW LINE

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",

A question that has come up a few times on r-help is how to rename columns of a data.frame. There are several ways to do this by hand (see the list archives). There is also a 'rename' function in the reshape package. I often use the 'transform' function shortly after reading in a data file and wanted to have a renaming function that has a syntax consistent with the

HI, The code was working perfectly fine yesterday and today, until half an hour ago.? Couldn't find any problems in the code. Still, I am getting error message. myMatrix <- data.matrix(read.table(text=" Name??????????? Age ANTONY??????? 27 IMRAN????????? 30 RAJ????????????????? 22 NAHAS????????? 32 GEO??????????????? 42 ", header=TRUE)) MinMaxArray? <- data.frame(MIN =

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

Hello R Community, I've been working on a project that uses pre-made C++ libraries (using STL) in R and build a package. However, I've been getting an unusual segfault that I'm unable to trace its origin. After many attempts of debugging using gdb, commenting out parts (or all of my code) and valgrind, I'm unable to make heads or tails about what I'm doing wrong. In fact, when

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."

Note that ?transform.data.frame says arguments need to be named, so you are testing unspecified behaviour. I guess this falls in a similar category as the note If some of the values are not vectors of the appropriate length, you deserve whatever you get! Experiments for a related Problem Report (<https://bugs.r-project.org/show_bug.cgi?id=17890>) showed that packages

Hello, Please always cc the list. As for the question, yes, it does. If you want to remove just the ones with exactly 73.1 use the pattern grep("^73\\.1$", etc) Explanation: Beginning of string: ^ End of string: $ Escape special characters: \\ (needed because the period is a special character.) Hope this helps, Rui Barradas On 5/22/2018 12:50 PM, Ahmed Serag wrote: > Thank

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