similar to: substring from behind

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()

Hello togehter, i have a litte problem to convert a data.frame. My data.frame looks like this one A 1 R5000 2 R4800 3 R4700 4 3500 5 3800 I need now a command, which outputs all the numbers, without the character in front. The solution look like this one: A 1 5000 2 4800 3 4700 4 3500 5 3800 Thanks. Best regards. Mat -- View this message in context:

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.

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

Dear R People: I am trying to split a character vector into a set of individual letters: Ideal: x3 <- c("dog") "d" "o" "g" I tried the following: > strsplit(x3) Error in strsplit(x3) : argument "split" is missing, with no default > strsplit(x3,1) [[1]] [1] "dog" I know that this is incredibly simple, but what am I doing

Hi, According to its man page substring() "expands (its) arguments cyclically to the length of the longest _provided_ none are of zero length". So, as expected, I get an error here: > substring("abcd", first=2L, last=integer(0)) Error in substring("abcd", first = 2L, last = integer(0)) : invalid substring argument(s) But I don't get one here:

On 2/20/19 7:55 PM, Toby Hocking wrote: > Update: I have observed that stringi::stri_sub is linear time complexity, > and it computes the same thing as base::substring. figure > https://github.com/tdhock/namedCapture-article/blob/master/figure-substring-bug.png > source: > https://github.com/tdhock/namedCapture-article/blob/master/figure-substring-bug.R > > To me this is a

Hi all, (and especially hi to Tomas Kalibera who accepted my patch sent yesterday) I believe that I have found another bug, this time in the substring function. The use case that I am concerned with is when there is a single (character scalar) text/subject, and many substrings to extract. For example substring("AAAA", 1:4, 1:4) or more generally, N=1000

Hi, In R < 3.0.0, we used to get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) A B A B A "bcdef" "2345" "bcd" "23" "b" But in R >= 3.0.0, we get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) [1] "bcdef"

Dear R Users, I am trying to replace a substring in a string with something else. For example: if we have "abc.degg.hijk" I would like to replace all the "." with a SPACE to become "abc degg hijk" I have tried the replace.substring.wild function in the Hmisc package but get this error: > replace.substring.wild("abc.degg.hijk","*.*","

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))",

Hi, How can I use a substring match as a condition in a subset command? Sth like this: subset(input, field1=="blah1" & field2=="blah2") # but now with substring match in field2 subset(input, field1=="blah1" & field3 *substringmatch* "blah3") I've tried with gsub, but it won't work: subset(input, field1=="blah1" &

Hello R users, I have a little problem with a for loop. Below there is an simple example of my problem. I want to delet the commas in the character string. Fore this reason I create a for loop to unpick the string and rebuild him without the commas. The problem is, that "paste" does not work in the loop as I expected. text <- "aaa,bbb,ccc,ddd" characterseq <-

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]]

Hi R users, I am trying generate the indices for locating a in the string it come from. Given the length of the string, it will take too long using the combn() for further comparison. I am wondering if R has any built-in function for this purpose. To make it concrete: this.substring="cc" this.string="ccc" start.location=1,2 end.location=2,3 Thanks in advance, Kevin

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

Dear R-members, Suppose I have a vector with the following strings: V = c("Welfare_Group_1024", "Welfare_Group_1536", "Welfare_Group_160") I want to 'automatically generate a nice y-axis label for this data. A good candidate is something close to "Welfare Group". Is there an easy way to compute something close to the greatest common

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

I have some data that looks like this: > dataP input output corpusFreq pvolOT pvolRatioOT 1 give(my sister, the old book) P 47.0 56016 0.1543651 5 donate(her, the book) P 48.7 68928 0.1899471 9 give(my sister, the book) P 73.4 80136 0.2208333 13 donate(my sister, the old book) P

Last one for you guys: The command: length(gregexpr('cus','hocus pocus')[[1]]) [1] 2 returns the number of times the substring 'cus' appears in 'hocus pocus' (which is two) It's returning the number of **disjoint** matches. So: length(gregexpr('aa','aaa')[[1]]) [1] 1 returns 1. **What I want to do:** I'm looking for a way to count