I'm using a text file in R and using the readLine function and regexs to extract words from it. The file uses special characters around words (such as # sings before and after a word to show it is bolded or # before and after a word to show it should be italicized) to indicate special meanings, which are messing up my regexs.
So far this is my r code which removed all empty lines and then combined my text file into a single vector :
book<-readLines("/Users/Desktop/SAMPLE .txt",encoding="UTF-8")
#remove all empty lines
empty_lines = grepl('^\\s*$', book)
book = book[! empty_lines]
#combine book into one variable
xBook = paste(book, collapse = '')
#remove extra white spaces for a single text of the entire book
updated<-trimws(gsub("\\s+"," ",xBook))
when i run updated, i see the entire file stored in the variable updated but with the special characters:
updated
[1] "It is a truth universally acknowledged, that a #single# man in possession of a good fortune, must be in want of a wife. However little known the feelings or views of such a #man# may be on his first entering a neighbourhood, this truth is so well fixed in the minds of the surrounding families, #that# he is considered the rightful property of some one or other of #their# daughters.
How can I remove all all the leading or trailing # or # from the words in my updated variable?
my desired output is just the plain text, with no indication of words that should be bolded or italicized:
updated
[1] "It is a truth universally acknowledged, that a single man in possession of a good fortune, must be in want of a wife. However little known the feelings or views of such a man may be on his first entering a neighbourhood, this truth is so well fixed in the minds of the surrounding families, that he is considered the rightful property of some one or other of their daughters.
gsub("[##]([a-zA-Z]+)[##]", "\\1", x)
Related
Alright so I have minimal experience with RStudio, I've been googling this for hours now and I'm fed up-- I don't care about the pride of figuring it out on my own anymore, I just want it done. I want to do some stuff with Canterbury Tales-- the Middle English version on Gutenberg.
Downloaded the plaintext, trimmed out the meta data, etc but it's chock-full of "helpful" footnotes and I can't figure out how to cut them out. EX:
"And shortly, whan the sonne was to reste,
So hadde I spoken with hem everichon,
That I was of hir felawshipe anon,
And made forward erly for to ryse,
To take our wey, ther as I yow devyse.
19. Hn. Bifel; E. Bifil. 23. E. were; _rest_ was. 24. E. Hn.
compaignye. 26, 32. E. felaweshipe. Hl. pilgryms; E. pilgrimes.
34. E. oure
But natheles, whyl I have tyme and space,..."
I at least have the vague notion that this is a grep/regex puzzle. Looking at the text in TextEdit, each bundle of footnotes is indented by 4 spaces, and the next verse starts with a capitalized word indented by (edit: 4 spaces as well).
So I tried downloading the package qdap and using the rm_between function to specify removal of text between four spaces and a number; and two spaces and a capital letter (" [0-9]"," "[A-Z]") to no avail.
I mean, this isn't nearly as simple as "make the text lowercase and remove all the numbers dur-hur" which all the tutorials are so helpfully offering. But I'm assuming this is a rather common thing that people have to do when dealing with big texts. Can anyone help me? Or do I have to go into textedit and just manually delete all the footnotes?
EDIT: I restarted the workspace today and all I have is a scan of the file, each line stored in a character vector, with the Gutenburg metadata trimmed out:
text<- scan("thefilepath.txt, what = "character", sep = "\n")
start <-which(text=="GROUP A. THE PROLOGUE.")
end <-which(text==""God bringe us to the Ioye . that ever schal be!")
cant.lines.v <- text[start:end]
And that's it so far. Eventually I will
cant.v<- paste(cant.lines.v, collapse=" ")
And then strsplit and unlist into a vector of individual words-- but I'm assuming, to get rid of the footnotes, I need to gsub and replace with blank space, and that will be easier with each separate line? I just don't know how to encode the pattern I need to cut. I believe it is 4 spaces followed by a number, then continuing on until you get to 4 spaces followed by a capitalized word and a second word w/o numbers and special characters and punctuation.
I hope that I'm providing enough information, I'm not well-versed in this but I am looking to become so...thanks in advance.
I'm trying to extract UK postcodes from address strings in R, using the regular expression provided by the UK government here.
Here is my function:
address_to_postcode <- function(addresses) {
# 1. Convert addresses to upper case
addresses = toupper(addresses)
# 2. Regular expression for UK postcodes:
pcd_regex = "[Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) {0,1}[0-9][A-Za-z]{2})"
# 3. Check if a postcode is present in each address or not (return TRUE if present, else FALSE)
present <- grepl(pcd_regex, addresses)
# 4. Extract postcodes matching the regular expression for a valid UK postcode
postcodes <- regmatches(addresses, regexpr(pcd_regex, addresses))
# 5. Return NA where an address does not contain a (valid format) UK postcode
postcodes_out <- list()
postcodes_out[present] <- postcodes
postcodes_out[!present] <- NA
# 6. Return the results in a vector (should be same length as input vector)
return(do.call(c, postcodes_out))
}
According to the guidance document, the logic this regular expression looks for is as follows:
"GIR 0AA" OR One letter followed by either one or two numbers OR One letter followed by a second letter that must be one of
ABCDEFGHJ KLMNOPQRSTUVWXY (i.e..not I) and then followed by either one
or two numbers OR One letter followed by one number and then another
letter OR A two part post code where the first part must be One letter
followed by a second letter that must be one of ABCDEFGH
JKLMNOPQRSTUVWXY (i.e..not I) and then followed by one number and
optionally a further letter after that AND The second part (separated
by a space from the first part) must be One number followed by two
letters. A combination of upper and lower case characters is allowed.
Note: the length is determined by the regular expression and is
between 2 and 8 characters.
My problem is that this logic is not completely preserved when using the regular expression without the ^ and $ anchors (as I have to do in this scenario because the postcode could be anywhere within the address strings); what I'm struggling with is how to preserve the order and number of characters for each segment in a partial (as opposed to complete) string match.
Consider the following example:
> address_to_postcode("1A noplace road, random city, NR1 2PK, UK")
[1] "NR1 2PK"
According to the logic in the guideline, the second letter in the postcode cannot be 'z' (and there are some other exclusions too); however look what happens when I add a 'z':
> address_to_postcode("1A noplace road, random city, NZ1 2PK, UK")
[1] "Z1 2PK"
... whereas in this case I would expect the output to be NA.
Adding the anchors (for a different usage case) doesn't seem to help as the 'z' is still accepted even though it is in the wrong place:
> grepl("^[Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) {0,1}[0-9][A-Za-z]{2})$", "NZ1 2PK")
[1] TRUE
Two questions:
Have I misunderstood the logic of the regular expression and
If not, how can I correct it (i.e. why aren't the specified letter
and character ranges exclusive to their position within the regular expression)?
Edit
Since posting this answer, I dug deeper into the UK government's regex and found even more problems. I posted another answer here that describes all the issues and provides alternatives to their poorly formatted regex.
Note
Please note that I'm posting the raw regex here. You'll need to escape certain characters (like backslashes \) when porting to r.
Issues
You have many issues here, all of which are caused by whoever created the document you're retrieving your regex from or the coder that created it.
1. The space character
My guess is that when you copied the regular expression from the link you provided it converted the space character into a newline character and you removed it (that's exactly what I did at first). You need to, instead, change it to a space character.
^([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([AZa-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2})$
here ^
2. Boundaries
You need to remove the anchors ^ and $ as these indicate start and end of line. Instead, wrap your regex in (?:) and place a \b (word boundary) on either end as the following shows. In fact, the regex in the documentation is incorrect (see Side note for more information) as it will fail to anchor the pattern properly.
See regex in use here
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([AZa-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^^^^^ ^^^
3. Character class oversight
There's a missing - in the character class as pointed out by #deadcrab in his answer here.
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^
4. They made the wrong character class optional!
In the documentation it clearly states:
A two part post code where the first part must be:
One letter followed by a second letter that must be one of ABCDEFGHJKLMNOPQRSTUVWXY (i.e..not I) and then followed by one number and optionally a further letter after that
They made the wrong character class optional!
\b(?:([Gg][Ii][Rr] 0[Aa]{2})|((([A-Za-z][0-9]{1,2})|(([A-Za-z][A-Ha-hJ-Yj-y][0-9]{1,2})|(([A-Za-z][0-9][A-Za-z])|([A-Za-z][A-Ha-hJ-Yj-y][0-9]?[A-Za-z])))) [0-9][A-Za-z]{2}))\b
^^^^^^
it should be this one ^^^^^^^^
5. The whole thing is just awful...
There are so many things wrong with this regex that I just decided to rewrite it. It can very easily be simplified to perform a fraction of the steps it currently takes to match text.
\b(?:[A-Za-z][A-HJ-Ya-hj-y]?[0-9][0-9A-Za-z]? [0-9][A-Za-z]{2}|[Gg][Ii][Rr] 0[Aa]{2})\b
Answer
As mentioned in the comments below my answer, some postcodes are missing the space character. For missing spaces in the postcodes (e.g. NR12PK), simply add a ? after the spaces as shown in the regex below:
\b(?:[A-Za-z][A-HJ-Ya-hj-y]?[0-9][0-9A-Za-z]? ?[0-9][A-Za-z]{2}|[Gg][Ii][Rr] ?0[Aa]{2})\b
^^ ^^
You may also shorten the regex above with the following and use the case-insensitive flag (ignore.case(pattern) or ignore_case = TRUE in r, depending on the method used.):
\b(?:[A-Z][A-HJ-Y]?[0-9][0-9A-Z]? ?[0-9][A-Z]{2}|GIR ?0A{2})\b
Note
Please note that regular expressions only validate the possible format(s) of a string and cannot actually identify whether or not a postcode legitimately exists. For this, you should use an API. There are also some edge-cases where this regex will not properly match valid postcodes. For a list of these postcodes, please see this Wikipedia article.
The regex below additionally matches the following (make it case-insensitive to match lowercase variants as well):
British Overseas Territories
British Forces Post Office
Although they've recently changed it to align with the British postcode system to BF, followed by a number (starting with BF1), they're considered optional alternative postcodes
Special cases outlined in that article (as well as SAN TA1 - a valid postcode for Santa!)
See this regex in use here.
\b(?:(?:[A-Z][A-HJ-Y]?[0-9][0-9A-Z]?|ASCN|STHL|TDCU|BBND|[BFS]IQ{2}|GX11|PCRN|TKCA) ?[0-9][A-Z]{2}|GIR ?0A{2}|SAN ?TA1|AI-?[0-9]{4}|BFPO[ -]?[0-9]{2,3}|MSR[ -]?1(?:1[12]|[23][135])0|VG[ -]?11[1-6]0|[A-Z]{2} ? [0-9]{2}|KY[1-3][ -]?[0-2][0-9]{3})\b
I would also recommend anyone implementing this answer to read this StackOverflow question titled UK Postcode Regex (Comprehensive).
Side note
The documentation you linked to (Bulk Data Transfer: Additional Validation for CAS Upload - Section 3. UK Postcode Regular Expression) actually has an improperly written regular expression.
As mentioned in the Issues section, they should have:
Wrapped the entire expression in (?:) and placed the anchors around the non-capturing group. Their regular expression, as it stands, will fail in for some cases as seen here.
The regular expression is also missing - in one of the character classes
It also made the wrong character class optional.
here is my regular expression
txt="0288, Bishopsgate, London Borough of Tower Hamlets, London, Greater London, England, EC2M 4QP, United Kingdom"
matches=re.findall(r'[A-Z]{1,2}[0-9][A-Z0-9]? [0-9][ABD-HJLNP-UW-Z]{2}', txt)
(strap in!)
Hi, I'm running into issues involving Unicode encoding in R.
Basically, I'm importing data sets that contain Unicode (UTF-8) characters, and then running grep() searches to match values. For example, say I have:
bigData <- c("foo","αβγ","bar","αβγγ (abgg)", ...)
smallData <- c("αβγ","foo", ...)
What I'm trying to do is take the entries in smallData and match them to entries in bigData. (The actual sets are matrixes with columns of values, so what I'm trying to do is find the indexes of the matches, so I can tell what row to add the values to.) I've been using
matches <- grepl(smallData[i], bigData, fixed=T)
which usually results in a vector of matches. For i=2, it would return 1, since "foo" is element 1 of bigData. This is peachy and all is well. But RStudio seems to not be dealing with unicode characters properly. When I import the sets and view them, they use the character IDs.
dataset <- read_csv("[file].csv", col_names = FALSE, locale = locale())
Using View(dataset) shows "aß<U+03B3>" instead of "αβγ." The same goes for
dataset[1]
A tibble: 1x1 <chr>
[1] aß<U+03B3>
print(dataset[1])
A tibble: 1x1 <chr>
[1] aß<U+03B3>
However, and this is why I'm stuck rather than just adjusting the encoding:
paste(dataset[1])
[1] "αβγ"
Encoding(toString(dataset[1]))
[1] "UTF-8"
So it appears that R is recognizing in certain contexts that it should display Unicode characters, while in others it just sticks to--ASCII? I'm not entirely sure, but certainly a more limited set.
In any case, regardless of how it displays, what I want to do is be able to get
grep("αβγ", bigData)
[1] 2 4
However, none of the following work:
grep("αβ", bigData) #(Searching the two letters that do appear to convert)
grep("<U+03B3>",bigData,fixed=T) #(Searching the code ID itself)
grep("αβ", toString(bigData)) #(converts the whole thing to one string)
grep("\\β", bigData) #(only mentioning because it matches, bizarrely, to ß)
The only solution I've found is:
grep("\u03B3", bigData)
[1] 2 4
Which is not ideal for a couple reasons, most jarringly that it doesn't look like it's possible to just take every <U+####> and replace it with \u####, since not every Unicode character is converted to the <U+####> format, but none of them can be searched. (i.e., α and ß didn't turn into their unicode keys, but they're also not searchable by themselves. So I'd have to turn them into their keys, then alter their keys to a form that grep() can use, then search.)
That means I can't just regex the keys into a searchable format--and even if I could, I have a lot of entries including characters that'd need to be escaped (e.g., () or ), so having to remove the fixed=T term would be its own headache involving nested escapes.
Anyway...I realize that a significant part of the problem is that my set apparently involves every sort of character under the sun, and it seems I have thoroughly entrapped myself in a net of regular expressions.
Is there any way of forcing a search with (arbitrary) unicode characters? Or do I have to find a way of using regular expressions to escape every ( and α in my data set? (coordinate to that second question: is there a method to convert a unicode character to its key? I can't seem to find anything that does that specific function.)
I'm almost certain this has been asked before but due to a certain social media app I drowning in unrelated search results.
So the data set that I'm importing contains actual "#", as in Apartment #404, and I'd like to if possible preserve the character but R thinks it's an end of line or something. At first it would bomb out on the first occurrence, then I set fill=TRUE and now it just ignores the rest of the line after that.
How does one instruct R to treat #'s as regular characters?
If you are not using "#" as a comment symbol in your data, you can use
read.table(..., comment.char="")
That should treat "#" like any other character.
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 9 years ago.
Improve this question
This is my first ever question here and I'm new to R, trying to figure out my first step in how to do data processing, please keep it easy : )
I'm wondering what would be the best function and a useful data structure in R to load unstructured text data for further processing. For example, let's say I have a book stored as a text file, with no new line characters in it.
Is it a good idea to use read.delim() and store the data in a list? Or is a character vector better, and how would I define it?
Thank you in advance.
PN
P.S. If I use "." as my delimeter, it would treat things like "Mr." as a separate sentence. While this is just an example and I'm not concerned about this flaw, just for educational purposes, I'd still be curious how you'd go around this problem.
read.delim reads in data in table format (with rows and columns, as in Excel). It is not very useful for reading a string of text.
To read text from a text file into R you can use readLines(). readLines() creates a character vector with as many elements as lines of text. A line, for this kind of software, is any string of text that ends with a newline. (Read about newline on Wikipedia.) When you write text, you enter your system specific newline character(s) by pressing Return. In effect, a line of text is not defined by the width of your software window, but can run over many visual rows. In effect, a line of text is what in a book would be a a paragraph. So readLines() splits your text at the paragraphs:
> readLines("/path/to/tom_sawyer.txt")
[1] "\"TOM!\""
[2] "No answer."
[3] "\"TOM!\""
[4] "No answer."
[5] "\"What's gone with that boy, I wonder? You TOM!\""
[6] "No answer."
[7] "The old lady pulled her spectacles down and looked over them about the room; then she put them up and looked out under them. She seldom or never looked through them for so small a thing as a boy; they were her state pair, the pride of her heart, and were built for \"style,\" not service—she could have seen through a pair of stove-lids just as well. She looked perplexed for a moment, and then said, not fiercely, but still loud enough for the furniture to hear:"
[8] "\"Well, I lay if I get hold of you I'll—\"
Note that you can scroll long text to the left here in Stackoverflow. That seventh line is longer than this column is wide.
As you can see, readLines() read that long seventh paragraph as one line. And, as you can also see, readLines() added a backslash in front of each quotation mark. Since R holds the individual lines in quotation marks, it needs to distinguish these from those that are part of the original text. Therefore, it "escapes" the original quotation marks. Read about escaping on Wikipedia.
readLines() may output a warning that an "incomplete final line" was found in your file. This only means that there was no newline after the last line. You can suppress this warning with readLines(..., warn = FALSE), but you don't have to, it is not an error, and supressing the warning will do nothing but supress the warning message.
If you don't want to just output your text to the R console but process it further, create an object that holds the output of readLines():
mytext <- readLines("textfile.txt")
Besides readLines(), you can also use scan(), readBin() and other functions to read text from files. Look at the manual by entering ?scan etc. Look at ?connections to learn about many different methods to read files into R.
I would strongly advise you to write your text in a .txt-file in a text editor like Vim, Notepad, TextWrangler etc., and not compose it in a word processor like MS Word. Word files contain more than the text you see on screen or printed, and those will be read by R. You can try and see what you get, but for good results you should either save your file as a .txt-file from Word or compose it in a text editor.
You can also copy-paste your text from a text file open in any other software to R or compose your text in the R console:
myothertext <- c("What did you do?
+ I wrote some text.
+ Ah, interesting.")
> myothertext
[1] "What did you do?\nI wrote some text.\nAh, interesting."
Note how entering Return does not cause R to execute the command before I closed the string with "). R just replies with +, telling me that I can continue to edit. I did not type in those plusses. Try it. Note also that now the newlines are part of your string of text. (I'm on a Mac, so my newline is \n.)
If you input your text manually, I would load the whole text as one string into a vector:
x <- c("The text of your book.")
You could load different chapters into different elements of this vector:
y <- c("Chapter 1", "Chapter 2")
For better reference, you can name the elements:
z <- c(ch1 = "This is the text of the first chapter. It is not long! Why was the author so lazy?", ch2 = "This is the text of the second chapter. It is even shorter.")
Now you can split the elements of any of these vectors:
sentences <- strsplit(z, "[.!?] *")
Enter ?strsplit to read the manual for this function and learn about the attributes it takes. The second attribute takes a regular expression. In this case I told strsplit to split the elements of the vector at any of the three punctuation marks followed by an optional space (if you don't define a space here, the resulting "sentences" will be preceded by a space).
sentences now contains:
> sentences
$ch1
[1] "This is the text of the first chapter" "It is not long"
[3] "Why was the author so lazy"
$ch2
[1] "This is the text of the second chapter" "It is even shorter"
You can access the individual sentences by indexing:
> sentences$ch1[2]
[3] "It is not long"
R will be unable to know that it should not split after "Mr.". You must define exceptions in your regular expression. Explaining this is beyond the scope of this question.
How you would tell R how to recognize subjects or objects, I have no idea.