Dear Andrew,
Thank you for the fast reply. I forgot about strwrap. Though my problem is slightly different.
I do have the actual vector. Of course, I could first join the strings - but this then involves both a join and a split (done by strwrap). Maybe its possible to avoid the join and the split. My 2nd approach may be also fine, but I have not tested it thoroughly (and I may miss an existing solution).
Sincerely, Leonard On 10/29/2022 12:51 AM, Andrew Simmons wrote:
I would suggest using strwrap(), the documentation at ?strwrap has plenty of details and examples. For paragraphs, I would usually do something like: strwrap(x = , width = 80, indent = 4) On Fri, Oct 28, 2022 at 5:42 PM Leonard Mada via R-help <r-help@r-project.org> wrote:Dear R-Users, text = " What is the best way to split/cut a vector of strings into lines of preferred width? I have come up with a simple solution, albeit naive, as it involves many arithmetic divisions. I have an alternative idea which avoids this problem. But I may miss some existing functionality!" # Long vector of strings: str = strsplit(text, " |(?<=\n)", perl=TRUE)[[1]]; lenWords = nchar(str); # simple, but naive solution: # - it involves many divisions; cut.character.int = function(n, w) { ncm = cumsum(n); nwd = ncm %/% w; count = rle(nwd)$lengths; pos = cumsum(count); posS = pos[ - length(pos)] + 1; posS = c(1, posS); pos = rbind(posS, pos); return(pos); } npos = cut.character.int(lenWords, w=30); # lets print the results; for(id in seq(ncol(npos))) { len = npos[2, id] - npos[1, id]; cat(str[seq(npos[1, id], npos[2, id])], c(rep(" ", len), "\n")); } The first solution performs an arithmetic division on all string lengths. It is possible to find out the total length and divide only the last element of the cumsum. Something like this should work (although it is not properly tested). w = 30; cumlen = cumsum(lenWords); max = tail(cumlen, 1) %/% w + 1; pos = cut(cumlen, seq(0, max) * w); count = rle(as.numeric(pos))$lengths; # everything else is the same; pos = cumsum(count); posS = pos[ - length(pos)] + 1; posS = c(1, posS); pos = rbind(posS, pos); npos = pos; # then print The cut() may be optimized as well, as the cumsum is sorted ascending. I did not evaluate the efficiency of the code either. But do I miss some existing functionality? Note: - technically, the cut() function should probably return a vector of indices (something like: rep(seq_along(count), count)), but it was more practical to have both the start and end positions. Many thanks, Leonard ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
