similar to: Regular Expression

Hi, First of all, I would like to introduce myself as I will probably have many questions over the next few weeks and want to thank you guys in advance for your help. I'm a cancer researcher and I need to learn R to complete a few projects. I have an introductory background in Python. My questions at the moment are based on the following sample input file: *Sample_Input_File*

Hi, My project is set up the following way: root directory contains the following folders: folders: "Breast_Cancer" AND "Colorectal_Cancer" AND "Lung_Cancer" AND "Prostate_Cancer" I want to create a file, call it: "repeating_functions.R" and place it in the root directory such that I can call these functions from within the sub-folders in each

Hey, I just googled my error and many things came up. I followed the leads and read the ?read.delim page; I tried changing header = TRUE, and row.names = TRUE-- but I've still been having trouble fixing it, so I would greatly appreciate any help you can provide. Here is my code: rm(list=ls()) source("../../functions.R") uncurated <-

Hi guys, My code (next 2 lines below) isn't doing what I'm expecting it to: tmp <- ifelse(uncurated$days_to_tumor_recurrence=="null","norecurrence","recurrence") curated$recurrence_status <- tmp I want the column "recurrence_status" in my output file to have values "recurrence" (if the input value had a number in the

Computer Friends, with the following example lines: [107] "98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1" [108] "99-625: Cell type: S; Surv(months): 21; STATUS(0=alive, 1=dead): 1" i want to be able to isolate the number of months of survival for each row. is there a regular expression that can find the first instance of a ";", delete

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 together, i have a question for "substring". I know i can filter a number like this one: bill$No<-substring(bill$Customer,2,4) in this case i get the 2nd, 3rd and 4th number of my Customer ID. But how can i do this, if i want the 2nd, 3rd and 4th number of a column. Like this one. I have: Mercedes_02352 Audi_03555 and now i want to filter this data.frame to 235 355 can you

All - I have a column of SiteNames: SiteName OYS-PIA2-FL-1 OYS-PIA2-LA-1 OYS-PI-LA-BB-1 OYS-PIA2-LA-10 ... [truncated] and I want to include only the last few digits into a new column. I tried substr(data$SiteName, 13, 20) but because some SiteName values are of a different length, the final hyphen (i.e., "-") was included: "1" "1" "-1" "10"

I have a text file like this 2.5 3.6 7.1 7.9 100 3 4 2 3 200 3.1 4 3 3 300 2.2 3.3 2 4 I used "r <- read.table("a.txt", header=T)" The row names becomes X2.5, X3.6... What I need is the row names are numeric, so I can use the row names as numbers on x-axis for plotting. e.g. "plot(colMeans(r)~names(r))",

When concatenating mbox files like described here https://xaizek.github.io/2013-03-30/merge-mbox-mailboxes/. You will end up with an 'unsorted' mbox file. Is this going to be a problem esspecially when they are large >2GB's and new emails will be written to it? The email client nicely sorts the message from folder A "foldera 5 last" as last, but of course the mbox is

Hey, I have a dataset and I want to identify the records by groups for further use in ggplot. Here is a sample data: ID Value AL1 1 AL2 2 CA1 3 CA4 4 I want to identify all the records that in the same state (AL1 AND A2), group them as "AL", and do the same for CA1 and CA4. How can I have an output like: ID Value State AL1 1 AL AL2 2 AL CA1 3 CA CA4 4

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Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns

Hello, I have a data as follow: ID    Visit xa1 xa2 yb1 yc23 yb33   I want to look at frequency of visit for ID and create a new column as response .  For example my response would be 2 for x and 3 for y. I think I need to write a loop, but I don't know how. I really appreciate your help. Thanks a lot. Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

If I have data (below) and need some help in figuring out how I can change the values of my date column, so that a year will be from September-August? So the year 1990 = September 89-August 90; 1991 = September 90-August 91, etc... I was trying to use the if() function, but am unable to figure it out. I basically need to change the years associated with September-December to the following

Hi, A column of my df looks like A 10/20/30 40/20 60/10/10/5 80/10 I want to split it such that the last column has the last composition and if there are not enough the middle columns get the 0s. That way my df would look like A1 A2 A3 A4 10 20 0 30 40 0 0 20 60 10 10 5 80 0 0 10 How can I do that ?? [[alternative HTML version deleted]]

Hi, I have a char vector with year values. Some cells have single year value '2001-' and some have range like 1996-2007. I need to remove hyphen character '-' from all the values within the character vector named as 'end'. After removing the hyphen I need to get the last number from the cells where there are year range values i.e if the cell has range 1996-2007, the code

Dear All, I have a data frame of vectors of publication names such as 'pub': pub1 <- c('Brown DK, Santos R, Rome DF, Don Juan X') pub2 <- c('Benigni D') pub3 <- c('Arstra SD, Van den Hoops DD, lamarque D') pub <- rbind(pub1, pub2, pub3) I would like to construct a dataframe with only author's last name and each last name in columns and the

I have the following timeseries "tab" ===================================== > str(tab) mts [1:23, 1:2] 79.5 89.1 84.9 75.7 72.8 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr [1:2] "Ipex...I" "Omel...E" - attr(*, "tsp")= num [1:3] 2006 2008 12 - attr(*, "class")= chr [1:2] "mts" "ts" > tab

Hi all, >From a list of strings, I desire to filter out the followings: 1. Digits at the beginning of the strings 2. Character "SPE" following the digits (if it exists) 3. Any characters followed by hyphen The following produces the desired result, but would like to know whether this can be done more efficiently. Any suggestions would be much appreciated. dat <- c("2148