similar to: substr gives empty output

In y <- substr(x, i, 1) your third integer needs to be the location not the number of digits, so change it to y <- substr(x, i, i) and you should get what you want. Cheers, Tim > Date: Sun, 21 Jan 2018 10:50:31 -0500 > From: Ek Esawi <esawiek at gmail.com> > To: Luigi Marongiu <marongiu.luigi at gmail.com>, r-help at r-project.org > Subject: Re: [R] substr

Hi, (First, apology for my earlier incorrectly addressed "subscribe" post.) Can somebody tell me what exactly is going on below. Basically, I am running into some kind of "string truncation" problem when I try to get a substring starting past the 8192nd character (see sample session below). There doesn't appear to be any problem creating the string, and nchar()

Dear R People: Here is a toy example: > x <- c("2E","5W","12H") > substr(x,2,2) [1] "E" "W" "2" > Sometimes x has 3 elements, sometimes 2. I want to extract the last element, and then extract the other 1 or 2 elements. How can I do this, please? TIA, Sincerely, Erin -- Erin Hodgess Associate Professor Department of

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

I think there is a memory bug in `substr` that is triggered by a UTF-8 corner case: an incomplete UTF-8 byte sequence at the end of a string.? With a valgrind level 2 instrumented build of R-devel I get: > string <- "abc\xEE"??? # \xEE indicates the start of a 3 byte UTF-8 sequence > Encoding(string) <- "UTF-8" > substr(string, 1, 10) ==15375== Invalid read of

Hi all, I'm getting the following error from substring: > substr("<I>Jens Oehlschl\xe4gel-Akiyoshi", 1, 100) Error in substr("<I>Jens Oehlschl\xe4gel-Akiyoshi", 1, 100) : invalid multibyte string at '<e4>gel-A<6b>iyoshi' Is that normal / intended? I've tried setting the Encoding/locale to Latin-1/UTF-8 but that does not help. nchar

Thanks for the quick response Ivan. readLines with encoding='latin1' works for me (on Ubuntu). However I was more concerned with the inconsistency in results between substr and regexpr. I was expecting that if one of them errors because of an unknown encoding then the other should as well. Even better, if regexpr works, why shouldn't substr work as well? Incidentally the analogous

A useful extension of ALTREP is having two new string methods which return the number of characters of a given string element and to return a substring of an element. Having these methods would allow retrieving these values without needing to create a CHARSXP for the full element data, which could potentially be costly for long elements. For example say you have an ALTREP altstring vector where

How to substring the following vector (in data frame b) >b a 11 12 1234 1245 124567 126786 145769 such that: a1??? a2 1???? 1 1???? 2 12??? 23 12??? 34 124?? 567 126???787 145???769 ? I tried logical commands with substr() did not work out.

Hi, i have a vector filled with names: [1] Alvaro Adela ... [25] Beatriz Berta ... ... [100000] ... I would like to drop last character in every name. I use the next program: for (i in 1:100000) { ? ? ? ? ? ? ? ? ? ? ? ? ? largo <- nchar(names[i]-1) ? ? ? ? ? ? ? ? ? ? ? ? ? names[i] <- substring (names[i],1,largo] ? ? ? ? ? ? ? ? ? ? ? ? ?} Is another and faster way of do it? Thanks,

Hi, I am trying to create the assignment function: "substring<-" <- function(text, first, last=100000, sub) { if(is.character(first)) { if(!missing(last)) stop('wrong # arguments') return(sedit(text, first, sub)) } lf <- length(first) if(length(text)==1 && lf > 1) { if(missing(last)) last <- nchar(text) last <- rep(last,

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

Help-Rs,   I'm doing some string manipulation in a file where I converted a string date in mm/dd/yyyy format and returned the date yyyy.   I've used regexpr (hat tip to Gabor G for a very nice earlier post on this function) in steps (I've un-nested the code and provided it and an example of what I did below.  My question is: is there a more efficient way to do this.  Specifically is

Thank you, but this split the area into two and distorts the shape of the plot. (compared to ``` p + geom_abline(slope = slope_1, intercept = intercept_1 - 1/w[2], linetype = "dashed", col = "royalblue") + geom_abline(slope = slope_1, intercept = intercept_1 + 1/w[2], linetype = "dashed", col = "royalblue") ``` Why there

Hi all, I have a numeric vector with 1 decimal place, and I'd like to extract the last 3 characters, including the decimal point. The vector ranges from 0 to 20. x <- round(runif(100)*20, digits=1) Some of numbers have 3 characters, and some have 4. I've read up on the substr() function but that extracts characters based on exact positions. How can I extract just the last 3

Hi, everyone When I ran this cript, There is Error in substring(tmp.subject, tmp.end[ex] + 1, tmp.start[ex + 1] - 1) : invalid substring argument(s) Could someone figure out what the problem is? for(i in 1:length(genebody[,1])){ tmp.id<-as.vector(genebody[i,1]) # get gene id tmp.subject<-as.vector(genebody[i,2]) # get gene sequence

Hi Put fill outside aes p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2]), fill = "blue", alpha=0.1) The "hole" is because you have two levels of data (red and blue). To get rid of this you should put new data in ribbon call. Something like newdat <- trainset newdat$z <- factor(0) p+geom_ribbon(data=newdat, aes(ymin =

[R 1.7.1 on Windows XP Pro] Since R allows missing values for character variables, why are NA's not propagated by character manipulation functions? For example: > temp <- c("a", NA) > temp [1] "a" NA > is.na(temp) [1] FALSE TRUE > paste(temp[1], temp[2]) [1] "a NA" > substr(temp, 1, 1) [1] "a" "N" >

also from this site: https://plotly.com/ggplot2/geom_ribbon/ I get the answer is geom_ribbon but I am still missing something ``` #! plot p = ggplot(data = trainset, aes(x=x, y=y, color=z)) + geom_point() + scale_color_manual(values = c("red", "blue")) # show support vectors df_sv = trainset[svm_model$index, ] p = p + geom_point(data = df_sv, aes(x=x, y=y),

As usual, a web search ("find statistical mode in R") brought up something that is possibly useful -- Did you try this before posting? If not, please do so in future and let us know what your results were if you subsequently post here. Here's what SO suggested: Mode <- function(x) { ux <- unique(x) ux[which.max(tabulate(match(x, ux)))] } # ergo: