search for: substring

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 substring(paste(rep("A", N), collapse=""), 1:N, 1:N) The pro...

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 clear indication of a bug in substring, but again it would > be nice to h...

...8, "" , substring ( hardware, 1,...

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 clear indication of a bug in substring, but again it would be nice to have some feedback/confirmatio...

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)) : inva...

Dear all, I have a string, let's say "testing", and I would like to extract in sequence each letter (character) from it. But when I use substr() I only properly get the first character, the rest is empty (""). What am I getting wrong? For example, I have this code: >>> x <- "testing" k <- nchar(x) for (i in 1:k) { y <- substr(x, i, 1)

Hi, A noobie question: I'm simply trying to run a conditional statement that evaluates if a substring is found within a larger string. I find that if it IS found, my function returns TRUE (great!), but if not, the condition does not evaluate to FALSE. ex): if( grep("hi", "hop", fixed = TRUE) ) print('yes, your substring is in your string') else print('no, yo...

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 there, I'm pretty sure that it's written down somewhere but I cannot find it so far. The little example shows different approaches to replace a substring. Only the last one works. I think it has something to do with the fact that "substr" is used on the left side. Can anybody refer to an explanation for this behaviour? Thanks a lot in advance! Antje values <- factor(c(rep("abc",3), rep("bcd",3), rep("cde...

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 t...

Hi, Do anybody know when and why the below error we get? Error in substr(x, as.integer(start), as.integer(stop)) : invalid substring argument(s) in substr() Thanks in advance, Shubha [[alternative HTML version deleted]]

Hi All, I'm looking for a way to get many substrings from a longer string and then stitch them together. But, since the longer string is really, really long (like 250 MB long), I don't want to do this in a loop and load and re-load the longer string many times. Does anybody have an idea? Maybe I could pass in two vectors (the first would have...

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"...

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried

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() reports the correct size as constructed. # Start of R session: > tmp <- paste(rep("1",8192),sep="",collapse="") > nch...

Hi, Is there an easy way to test if a substring exists in a string? I''m using this check: if mystring.index(substring) > 0 but it doesn''t seem to be working.... Thanks! -- Posted via http://www.ruby-forum.com/.

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(...

...bb,ee=TA2.ee,gg=TA2.gg,tt=TA2.tt,uu=TA2.uu*)* etc for TA3~TA5. Although these generate the output I desire, I would like to use a single statement for producing 5 different arrays (instead of 5 different statements) I have tried the following codes, however the last statement (paste("T", substring(i,3,4), sep="") <- c(bb ......) gives error message reading "Error in paste("T", substring(i, 3, 4), sep = "") <- c(bb = paste(paste("T", : target of assignment expands to non-language object" for (i in unique(a$id)) for (j in unique(a$pro)...

... when the replacement string is shorter than the portion of the string to be replaced? The documentation to substr (in R 1.3.1) gives me: If the portion to be replaced is longer than the replacement string, then only the portion the length of the string is replaced. And so I try: R> x <- "abcdef" R> substr(x,2,3) <- "xy" #ok R> x [1]

... when the replacement string is shorter than the portion of the string to be replaced? The documentation to substr (in R 1.3.1) gives me: If the portion to be replaced is longer than the replacement string, then only the portion the length of the string is replaced. And so I try: R> x <- "abcdef" R> substr(x,2,3) <- "xy" #ok R> x [1]