Hello, again. See comments inline
Em 29-05-2012 16:28, Sabina Arndt escreveu:
Hello, thank you very much for your reply, Rui Barradas. OK, I did what you said:x<- readLines("sabina.txt") s<- strsplit(x, ";[[:space:]]\\[") r<- lapply(s, function(x) sapply(strsplit(x, "[[:blank:]]"), tail, 1)) length(r)[1] 20 I don't know why your result here was 21 since the file consists of only 20 lines.
Don't worry. When I copied the file it probably included some junk character in the end.
r[[21]]<- NULL r[[20]]<- r[[20]][ -length(r[[20]]) ] r1<- lapply(r, function(x) x[nchar(x)> 0]) country.list<- r1[ -which(sapply(r1, function(x) is.null(x))) ] rm(s, r, r1) country.listlist() I also tried this:r[[20]]<- NULL r[[19]]<- r[[19]][ -length(r[[19]]) ] r1<- lapply(r, function(x) x[nchar(x)> 0]) country.list<- r1[ -which(sapply(r1, function(x) is.null(x))) ] rm(s, r, r1) country.listlist() But the result was the same. For some reason this seems to be empty. But if I try this before "country.list<- r1[ -which(sapply(r1, function(x) is.null(x))) ]":
It should be. The error is that I've made some experiences with the data, since 'r' has some empty strings in its elements. In my workspace everything was converted either to non-empty strings or NULLs.
This is how to do it. r1 <- lapply(r, function(x) x[nchar(x) > 0]) r1 <- lapply(r1, function(x) if(length(x)) x else NULL) # second pass country.list <- r1[ -which(sapply(r1, is.null)) ] country.list
r[[18]][1] "England" "Scotland" "Germany" "Germany" "England" "WOS:000296579800006" This is almost correct. But the last country name is missing of this record and replaced with the value of the very last column / field of this record. Do you know how to correct this?
After removing the nulls, in my workspace the list numbers are different, but you could remove unwanted values along the lines of
bad <- -length(r[[18]]) r[[18]] <- r[[18]][ -bad ] Note that you could do this to 'country.list', it might be simpler.
In addition to that there are some additional adjustments I need to apply to the country names before output since there are many different versions of US addresses, e.g. (See 000296579800006.). I'm not sure I understand your function correctly, do you think the edits I mentioned could be fit in there as well?
If it all works correctly, adjustments can be made, if not it might be premature. I don't know.
See how it goes, so far.
Thank you very much for bearing with me! I swear I ususally am not that dumb! Faithfully yours, Sabina Arndt
You're welcome, Rui Barradas
Date: Tue, 29 May 2012 13:00:36 +0100 From: ruipbarra...@sapo.pt To: sabina.ar...@hotmail.de CC: r-help@r-project.org Subject: Re: [R] Relist strings? [Was: How to remove square brackets, etc. from address strings?] Hello, The error message means that 'x' is not a character vector. Can't you try it only with the text in the link you've posted, http://pastebin.com/mYZNDXg6 ? I'm asking this because I've just checked it and it doesn't give any eror. Em 29-05-2012 12:39, Sabina Arndt escreveu:Hello r-help members, thank you very much for your reply, Rui Barradas.Your data file has more than one line.Yes, each line is a new record and I read several such data files into one data.frame.This is problably why it gives you that error. Process just one file, like I've said, then say something. (Moreover, it makes sense to solve the problems with a smaller set then move on to the larger one.) Rui BarradasI've called it "sabrina.txt" and then processed with: x<- readLines("sabrina.txt") s<- strsplit(x, ";[[:space:]]\\[")Thank you; but this gives me an error message: Error in strsplit(x, ";[[:space:]]\\[") : non-character argument So I cannot check the rest of your suggestion, unfortunately.Do you happen to have any idea on how I could put the country names back into their original lines / order, though?...As far as I can tell they're in the original order. But what do you mean by "back into their original lines"?Each line of my data.frame represents a record - except for the first one which is the header. Each record has different addresses in the field / column I'm analyzing. In fact, the records vary in the number of addresses they feature (The first has eight, the second only one, etc.). I don't want a simple list of all the country names but a new field in my data.frame which contains for each record the country name(s) extracted from the addresses of that very same record. I'd like to measure the number of elements after applying strsplit() to each string. I tried: ... results<- strsplit(results, ";") numbers<- sapply(results, length) results<- unlist(results) ... But this doesn't seem to work, because:numbers[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ... Does anybody know how I would achieve these results instead:numbers[1][1] 8numbers[2][1] 1results[1][1] "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY"results[2][1] "GERMANY" Thank you very much in advance! Faithfully yours, Sabina Arndt PS: I updated the subject of my message to reflect the progress I've made thanks to your replies. I hope this is appropriate and clearer this way.Am 27.05.2012 19:04, schrieb Rui Barradas:Hello, Though I've not been following this thread, it seems like a regular expressions problem. In the code below, I've created a 'testdata' variable based on your post. # create a vector with two elements. x<- "[Engel, Kathrin M. Y.; Schroeck, ... etc ... y<- gsub("Germany", "Portugal", x) testdata<- c(x, y) # 's' is a list of character vectors, each element's final word is a country s<- strsplit(testdata, ";[[:space:]]+\\[") lapply(s, function(x) sapply(strsplit(x, "[[:blank:]]"), tail, 1)) If this isn't it, sorry for the intrusion. Rui Barradas Em 27-05-2012 17:29, Sabina Arndt escreveu:Hello r-help members, I'm very grateful for the reply which Sarah Goslee sent to me in such a prompt and helpful manner. It took me some time, but with a few amendments her suggestion now works not only for an example but for my entire data file as well:results[1] "GERMANY" "GERMANY" "GERMANY" "GERMANY" [5] "GERMANY" "GERMANY" "GERMANY" "GERMANY" ... Thank you very much for that, dear Sarah! All these names actually belong to the very first record, though, which contains eight addresses instead of only one:testdata[1][1] "[Engel, Kathrin M. Y.; Schroeck, Kristin; Schoeneberg, Torsten; Schulz, Angela] Univ Leipzig, Fac Med, Inst Biochem, Leipzig, Germany; [Teupser, Daniel; Holdt, Lesca Miriam; Thiery, Joachim] Univ Leipzig, Fac Med, Inst Lab Med Clin Chem& Mol Diagnost, Leipzig, Germany; [Toenjes, Anke; Kern, Matthias; Blueher, Matthias; Stumvoll, Michael] Univ Leipzig, Fac Med, Dept Internal Med, Leipzig, Germany; [Dietrich, Kerstin; Kovacs, Peter] Univ Leipzig, Fac Med, Interdisciplinary Ctr Clin Res, Leipzig, Germany; [Kruegel, Ute] Univ Leipzig, Fac Med, Rudolf Boehm Inst Pharmacol& Toxicol, Leipzig, Germany; [Scheidt, Holger A.; Schiller, Juergen; Huster, Daniel] Univ Leipzig, Fac Med, Inst Med Phys& Biophys, Leipzig, Germany; [Brockmann, Gudrun A.] Humboldt Univ, Inst Anim Sci, D-10099 Berlin, Germany; [Augustin, Martin] Ingenium Pharmaceut AG, Martinsried, Germany"results[1][1] "GERMANY" How can I put the country names back into their original lines / order? This is an example of the correct result I'd like to receive:results[1][1] "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" "GERMANY" How can I achieve this result? I think counting the semicolons outside square brackets - i.e. the ones before a "[" but behind a "]" would be helpful in this regard, but I'm not sure how to do that, unfortunately. These semicolons directly follow the country names, like this, e.g.: "... Germany; [..." If I add "+ 1" to their number it results in the number of addresses for each record / line. Thank you very much in advance! Faithfully yours, Sabina Arndt Am 26.05.2012 00:19, schrieb Sarah Goslee:Part of your problem is that your regexes have spaces in them, so that's what you're matching. A small reproducible example would be more useful. I'm not feeling inclined to wade through all your linked files on Friday evening, but see if this helps:testdata<- "[Engel, Kathrin M. Y.; Schroeck, Kristin; Schoeneberg, Torsten; Schulz, Angela] Univ Leipzig, Fac Med, Inst Biochem, Leipzig, New Zealand; [Teupser, Daniel; Holdt, Lesca Miriam; Thiery, Joachim] Univ Leipzig, Fac Med, Inst Lab Med Clin Chem& Mol Diagnost, Leipzig, USA; [Toenjes, Anke; Kern, Matthias; Blueher, Matthias; Stumvoll, Michael] Univ Leipzig, Fac Med, Dept Internal Med, Leipzig, Germany; [Dietrich, Kerstin; Kovacs, Peter] Univ Leipzig, Fac Med, Interdisciplinary Ctr Clin Res, Leipzig, Germany; [Kruegel, Ute] Univ Leipzig, Fac Med, Rudolf Boehm Inst Pharmacol& Toxicol, Leipzig, Germany; [Scheidt, Holger A.; Schiller, Juergen; Huster, Daniel] Univ Leipzig, Fac Med, Inst Med Phys& Biophys, Leipzig, Germany; [Brockmann, Gudrun A.] Humboldt Univ, Inst Anim Sci, D-10099 Berlin, Germany; [Augustin, Martin] Ingenium Pharmaceut AG, Martinsried, Germany" results<- gsub("\\[.*?\\]", "", testdata) results<- unlist(strsplit(results, ";")) results<- sapply(results, function(x)sub("^.*, ([A-Za-z ]*)$", "\\1", x)) names(results)<- NULL results[1] "New Zealand" "USA" "Germany" "Germany" "Germany" "Germany" "Germany" "Germany" Sarah On Fri, May 25, 2012 at 4:31 PM, Sabina Arndt<sabina.ar...@hotmail.de> wrote:Hello r-help members, the solutions which Sarah Goslee and arun sent to me in such a prompt and helpful manner work well with the examples I cut from the data.frame I'm analyzing. Thank you very much for that! I incorporated them into my R-script and discovered that it still doesn't work properly, unfortunately. I have no idea why that's the case. You see, I want to extract country names from the contents of tab-delimited text files. This is an example of the data I'm using: http://pastebin.com/mYZNDXg6 This is the script I'm using to import the data: http://pastebin.com/Z10UUH3z (It requires the text files to be in a folder which doesn't contain any other .txt files.) This is the script I'm using to extract the country names: http://pastebin.com/G37fuPba This is the string that's in the relevant field of the first record I'm working on: [Engel, Kathrin M. Y.; Schroeck, Kristin; Schoeneberg, Torsten; Schulz, Angela] Univ Leipzig, Fac Med, Inst Biochem, Leipzig, Germany; [Teupser, Daniel; Holdt, Lesca Miriam; Thiery, Joachim] Univ Leipzig, Fac Med, Inst Lab Med Clin Chem& Mol Diagnost, Leipzig, Germany; [Toenjes, Anke; Kern, Matthias; Blueher, Matthias; Stumvoll, Michael] Univ Leipzig, Fac Med, Dept Internal Med, Leipzig, Germany; [Dietrich, Kerstin; Kovacs, Peter] Univ Leipzig, Fac Med, Interdisciplinary Ctr Clin Res, Leipzig, Germany; [Kruegel, Ute] Univ Leipzig, Fac Med, Rudolf Boehm Inst Pharmacol& Toxicol, Leipzig, Germany; [Scheidt, Holger A.; Schiller, Juergen; Huster, Daniel] Univ Leipzig, Fac Med, Inst Med Phys& Biophys, Leipzig, Germany; [Brockmann, Gudrun A.] Humboldt Univ, Inst Anim Sci, D-10099 Berlin, Germany; [Augustin, Martin] Ingenium Pharmaceut AG, Martinsried, Germany This is the incorrect result my extraction script gives me for the first record:C1s[1][1] "[ENGEL, KATHRIN M. Y." "KRISTIN" "TORSTEN" [4] "GERMANY" "DANIEL" "LESCA MIRIAM" [7] "GERMANY" "ANKE" "MATTHIAS" [10] "MATTHIAS" "GERMANY" "KERSTIN" [13] "GERMANY" "GERMANY" "[SCHEIDT, HOLGER A." [16] "JUERGEN" "GERMANY" "HUMBOLDT" [19] "GERMANY" For some reason the first and sixth pair of the eight square brackets are not removed ... Do you understand why? Instead I'd like to get this result, though:C1s[1][1] "GERMANY" "GERMANY" "GERMANY" [4] "GERMANY" "GERMANY" "GERMANY" [7] "HUMBOLDT" "GERMANY" What am I doing wrong? What are the errors in my R-script? Would anybody be so kind as to take a look and help me out, please? Thank you very much in advance! Faithfully yours, Sabina Arndt
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