similar to: Print occurrence / positions of words

Hi, I have been using R for a few months and I have this working code. Don't seen any problem but this takes a long time. So if I have about 30000 rows it takes a few minutes. If I have 100000 it does not seem to complete. Original Data: Proto Recv-Q Send-Q Local-Address Foreign-Address State tcp 0 0 172.20.100.2:60255

Emilio Ahora cuando quiero instalar los paquetes pdftools, magick y otros más me salen el siguiente error WARNING: Rtools is required to build R packages but is not currently installed. Please download and install the appropriate version of Rtools before proceeding: https://cran.rstudio.com/bin/windows/Rtools/ Installing package into ?C:/Users/bdominguez/Documents/R/win-library/3.6? (as ?lib?

Buenas tarde a todo en s: Tenia la versión de R 3.6 y utilizaba la paquetería de pdftools para extraer información de archivos en pdf actualice la versión 3.6.1 y ya no reconoce la paquetería alguien que me pueda ayudar. Prácticamente no reconoce las funciones de pdftools library(pdftools) library(stringr)? library(NLP)? library(tm)? library(tesseract)? library(magick)?

Hello again: Here is a solution to the dates without leading zeros: pou1 <- function(x) { #Note: x is a data frame #Assume that Column 1 has the date #Column 2 has station #Column 3 has min #Column 4 has max library(stringr) w <- character(length=nrow(x)) z <- str_split(x[,1],"/") for(i in 1:nrow(x)) { u <-

Hello, I'm using the eclat function of the arules package (1.0-6) for the identification of frequent itemsets. I need the tidLists, but if I set in the function tidLists=TRUE R crashes (Windows XP Professional SP3, 32 bit, R version 2.12.1 (2010-12-16), reproducible on two different computers) with two different error messages or non at all. Minimum examples are: library(arules)

Dear All, I'm dealing with a case, where 'manhattan' distance of each of 100 vectors is calculated from 10000 other vectors. For achieving this, following 4 scenarios are tested: 1) scenario 1: > x<-read.table("query.vec") > v<-read.table("query.vec2") > d<-matrix(nrow=nrow(v),ncol=nrow(x)) > for (i in 1:nrow(v)){ + d[i,]<-

Hello! I have a column in my data frame that I have to split: I have to distill the numbers from the text. Below is my example and my solution. x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1")) x library(stringr) out<-as.data.frame(str_split_fixed(x$x,"aaa",2)) out2<-as.data.frame(str_split_fixed(out$V2,"_bbb",2))

Hello Wise Ones... I need a clever way around a problem with findInterval. Consider: vec1 <- 1:10 vec2 <- seq(1, 10, by = 0.1) x1 <- c(2:3) a1 <- findInterval(x1, vec1); a1 # example 1 a2 <- findInterval(x1, vec2); a2 # example 2 In the problem I'm working on, vec* may be either integer or numeric, like vec1 and vec2. I need to remove one or more sections of this vector;

You can use na.strings="" in read.table() or read.csv() library(stringr) vec1<-unlist(str_split(readLines(textConnection("3,7,11,,12,14,15,,17,18,19")),",")) ?vec1[vec1==""]<- NA ?vec1 # [1] "3"? "7"? "11" NA?? "12" "14" "15" NA?? "17" "18" "19" #or

Dear List, I have this column/vector: vec <- c("function", "missing", "string") and want to compute a second column/vector: - value if the pattern "unc" is found: 1 - value if the pattern "iss" is found: 2 - value if none of the patterns is found: 0 This should be the result: > vec2 [1] 1 2 0 Any help? Tried it with grep, but the output

Hello list, I want to make a large rulebased algorithm, to provide decision support for drug prescriptions. I have defined the algorithm in a function, with a for loop and many if statements. The structure should be as follows: 1. Iterate over a list of drug names. For each drug: 2. Get some drug related data (external dataset). Row of a dataframe. 3. Check if adaptions should be made to

I want to create a number of vectors like : vec1 <- rnorm(1) vec2 <- rnorm(2) vec3 <- rnorm(3) and so on........... Here I tried following : for (i in 1:10) paste("vec", i, sep="") <- rnorm(i) However obviously that is not working. Here vectors I need to be seperated i.e I do not want to create a "list". How to modify above code? -- View this

# stringr Strings are not glamorous, high-profile components of R, but they do play a big role in many data cleaning and preparations tasks. R provides a solid set of string operations, but because they have grown organically over time, they can be inconsistent and a little hard to learn. Additionally, they lag behind the string operations in other programming languages, so that some things that

# stringr Strings are not glamorous, high-profile components of R, but they do play a big role in many data cleaning and preparations tasks. R provides a solid set of string operations, but because they have grown organically over time, they can be inconsistent and a little hard to learn. Additionally, they lag behind the string operations in other programming languages, so that some things that

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

Hi guys, I'm a new to R and following along with Tutorials using this book: http://www.amazon.com/Practical-Statistical-Analysis-Non-structured-Applications/dp/012386979X In one of them, they use the twitteR package and describe the following function (see below). From what I can tell from the documentation (R), there's a method to call it directly in an interactive session. The way

Dear R forum I have a data.frame df = data.frame(currency_type = c("EURO_o_n", "EURO_o_n", "EURO_1w", "EURO_1w", "USD_o_n", "USD_o_n", "USD_1w", "USD_1w"), rates = c(0.47, 0.475, 0.461, 0.464, 1.21, 1.19, 1.41, 1.43))   currency_type     rates 1      EURO_o_n   0.470 2      EURO_o_n   0.475 3       EURO_1w   0.461

Hi, there I'm currently trying to figure out how to keep my "factor" levels for a variable when moving it from one data frame or matrix to another. Example below: vec1<-(rep("10",5)) vec2<-(rep("30",5)) vec3<-(rep("80",5)) vecs<-c(vec1, vec2, vec3) resp<-rnorm(2,15) dat<-as.data.frame(cbind(resp, vecs))

Simple one, have read and googled, still no luck! I want to create several empty vectors all of the same length. I would like multiple empty vectors (vec1, vec2, vec3) and want to create them all in one line. I've tried vec1,vec2,vec3 <- vector(length=5) and c(vec1,vec2,vec3) <- vector(length=5) and several other attempts but nothing seems to work ... suggestions? Thanks Jim

Ivan and Bert, thank you so much for your help. Ivan, your solution worked perfectly. I didn't really understand how to do string processing on a vector of strings, and your solution demonstrated it for me. I modified it to work with the tidyverses' stringr library in this way: bg3_race_sum <- bg3_race %>% left_join(pl_vars, by=c("variable" = "name"))