I'm working with a lot of text files I have loaded into R and I'm trying to replace every instance (or tag) of </SPEAKER> with a certain string found earlier in the text file.
Example:
"<BOB> Lots of text here </SPEAKER> <HARRY> More text here by a different speaker </SPEAKER>"
I'd like to replace every instance of "</SPEAKER>" with the name of, say "<BOB>" and "<HARRY>" based on the NAME that has been found earlier, so I'd get this at the end:
"<BOB> Lots of text here </BOB> <HARRY> More text here by a different speaker </HARRY>"
I was thinking of looping through the vector text but as I only have limited experience with R, I wouldn't know how to tackle this.
If anyone has any suggestions for how to do this, possibly even outside of R using Notepad++ or another text/tag editor, I'd most appreciate any help.
Thanks!
Match
<,
word characters (capturing them in capture group 1),
>,
the shortest string (capturing it in capture group 2) until
</SPEAKER>
and then replace that with the
<,
capture group 1,
>,
capture group 2 and
</ followed by
capture group 1 and
>
This gives
x <- "<BOB> Lots of text here </SPEAKER> <HARRY> More text here by a different speaker </SPEAKER>"
gsub("<(\\w+)>(.*?)</SPEAKER>", "<\\1>\\2</\\1>", x)
## [1] "<BOB> Lots of text here </BOB> <HARRY> More text here by a different speaker </HARRY>"
Related
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Regular expression to stop at first match
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In long passages using bookdown, I have inserted numerous images. Having combined the passages into a single character string (in a data frame) I want to remove the markdown text associated with inserting images, but not any text in between those inserted images. Here is a toy example.
text.string <- "writing  writing to keep  more writing"
str_remove_all(string = text.string, pattern = "!\\[.+\\)")
[1] "writing more writing"
The regex expression doesn't stop at the first closed parenthesis, it continues until the last one and deletes the "writing to keep" in between.
I tried to apply String manipulation in R: remove specific pattern in multiple places without removing text in between instances of the pattern, which uses gsubfn and gsub but was unable to get the solutions to work.
Please point me in the right direction to solve this problem of a regex removal of designated strings, but not the characters in between the strings. I would prefer a stringr solution, but whatever works. Thank you
You have to use the following regex
"!\\[[^\\)]+\\)"
alternatively you can also use this:
"!\\[.*?\\)"
both solution offer a lazy match rather than a greedy one, which is the key to your question
I think you could use the following solution too:
gsub("!\\[[^][]*\\]\\([^()]*\\)", "", text.string)
[1] "writing writing to keep more writing"
I have this code that is meant to add highlights to some numbers in a text stored in "lines"
stringr::str_replace_all(lines, nums, function(x) {paste0("<<", x, ">>")})
where nums is the following pattern being deteced
nums<-(Zero|One|Two|Three|Four|Five|Six|Seven|Eight|Nine)+\\s?(Hundred|Thousand|Million|Billion|Trillion)?'
The problem I'm having is that the line of code above also leads to numbers embedded in words also being detected. In the following text this happens:
Get <<ten>> eggs. That is what is writ<<ten>>. I am <<one>> and al<<one>>.
when it should be:
Get <<ten>> eggs. That is what is written. I am <<one>> and alone.
I don't want to remove the question mark after the \s because I want to detect both numbers like "One" followed by no space and "One Hundred" which has a space in between.
Does anyone know how to do this?
Surround (Zero|One|Two|Three|Four|Five|Six|Seven|Eight|Nine)+ with \b.
\b matches word boundaries, so this expression will newer match inside a word.
I have a (large) dataset that initially consists of an identifier and associated text (in raw HTML). Oftentimes the text will include one or more embedded links. Here's a sample dataset:
id text
1 <p>I love dogs!</p>
2 <p>My <strong>favorite</strong> dog is this kind.</p>
3 <p>I've had both Labs and Huskies in my life.</p>
What I'd like as output (with the text column included in the same spot, but I removed it for visibility here) is:
id link1 link2
1
2 doge.com
3 labs.com huskies.com
I've tried using str_extract_all() paired with <a\s+(?:[^>]*?\s+)?href=(["'])(.*?)\1, but even when I double escape the backslashes I either get an "unexpected" error OR it keeps asking me for more and I have to Escape out. I feel like this method is the one I want and SHOULD work, but I can't seem to get the regex to play nicely. Here are my results so far:
> str_extract_all(text, "<a\s+(?:[^>]*?\s+)?href=(["'])(.*?)\1")
Error: '\s' is an unrecognized escape in character string starting ""<a\s"
> str_extract_all(text, perl(<a\s+(?:[^>]*?\s+)?href=(["'])(.*?)\1))
Error: unexpected '<' in "str_extract_all(text, perl(<"
> str_extract_all(text, "<a\\s+(?:[^>]*?\\s+)?href=(["'])(.*?)\\1")
+
> str_extract_all(text, perl(<a\\s+(?:[^>]*?\\s+)?href=(["'])(.*?)\\1))
Error: unexpected '<' in "str_extract_all(text, perl(<"
I've also tried parseURI from the XML package and for whatever reason it crashes my R session.
The other solutions I've found to date either only deal with single links, or return items in a list or vector altogether. I want to keep things separated by their identifier and in a dataset.
If needed, I could tolerate generating a separate dataset and merging them together, but there will be cases where there are no links, so I'd want to avoid any pitfalls of rows being deleted due to not having a value in any of the link columns.
R does not like quotes within strings so in your example above R is considering the string ongoing:
str_extract_all(text, "<a\\s+(?:[^>]*?\\s+)?href=(["'])(.*?)\\1")
R is still looking for the end of the string since it was not escaped in the regex. R has special cases in which as single \ can be used for escaping, (e.g \n for new line), see this. \' escapes a single quote and \" escapes a double quote in R regex:
str_extract_all(text, "<a\\s+(?:[^>]*?\\s+)?href=([\"])(.*?)\\1", text, perl=T)
"\ itself is a special character that needs escape, e.g. \\d. Do not
confuse these regular expressions with R escape sequences such as
\t."
or in your case \"
Please any advice will be appreciated.. This is time sensitive. I have PDF reports that are mostly blocks of text. They are long reports (~50-100 pages). I'm trying to write an R script that is capable of extracting specific sections of these PDF reports using start/stop positional strings. NOTE: Reports vary in length. Short example:
DOCUMENT TITLE
01. SECTION 1
This is a test section that I DONT want to extract.
This text would normally be much longer... Over 100 words.
Sample Text Text Text Text Text Text Text Text
02. SECTION 2
This is a test section that I do want to extract.
This text would normally be much longer... Over 100 words.
Sample Text Text Text Text Text Text Text Text
...
11. SECTION 11
This is a test section that I do want to extract.
This text would normally be much longer... Over 100 words.
Sample Text Text Text Text Text Text Text Text
...
12. SECTION 12
This is a test section that I DONT want to extract.
This text would normally be much longer... Over 100 words.
Sample Text Text Text Text Text Text Text Text
...
So the goal in this example, is to extract the paragraph below Section 2 and store it as a field/data point. I also want to store Section 11 as a field/data point. Note the document is in PDF format
I have tried used pdftools, tm, stringr, I've literally spent 20+ hours searching for solutions and tutorials on how to do this. I know it is possible as I have done it using SAS before...
Please see code below, I added comments with questions. I believe RegEx will be part of the solution but i'm so lost.
# Init Step
libs <- c("tm","class","stringr","testthat",
"pdftools")
lapply(libs, require, character.only= TRUE)
# File name & location
filename = "~/pdf_test/test.pdf"
# converting PDF to text
textFile <- pdf_text(filename)
cat(textFile[1]) # Text of pg. 1 of PDF
cat(textFile[2]) # Text of pg. 2 of PDF
# I'm at a loss of how to parse the values I want. I have seen things
like:
sectionxyz <- str_extract_all(textFile, #??? )
rm_between()
# 1) How do I loop through each page of PDF file?
# 2) How do I identify start/stop positions for section to be extracted?
# 3) How do I add logic to extract text between start/stop positions
# and then add the result to a data field?
# 4) Sections in PDF will be long sections of text (i.e. 100+ words into a field)
NEW------
So I have been able to:
-Prep doc correctly
-Identify the correct start/stop patterns:
length(grep("^11\\. LIMITS OF LIABILITY( +){1}$",source_main2))
length(grep("Applicable\\s+[Ll]imits\\s+[Oo]f",source_main2))
pat_st_lol <- "^11\\. LIMITS OF LIABILITY( +){1}$"
pat_ed_lol <- "Applicable\\s+[Ll]imits\\s+[Oo]f"
The length(grep()) statements verify only 1 instance is being found. From here I am kind of lost based on how to use gsub or similar to extract the portion of data I want. I tried:
pat <- paste0(".*",pat_st_lol,"(.*)",pat_ed_lol,".*")
test <- gsub(".*^11\\. LIMITS OF LIABILITY( +){1}$(.*)\n",
"Applicable\\s+[Ll]imits\\s+[Oo]f", source_main2)
test2 <-gsub(".*pat_st_lol(.*)\npat_ed_lol.*")
So far, little progress, but progress anyways.
Provided you can come with a systematic to identify the sections you need, you could, as you indicated, use Regex to extract the text you want.
In your above example, something like gsub(".*SECTION 11(.*)\n12\\..*","\\1",string) ought to work.
Now you could define patterns dynamically using paste and iterate through all files. Each result can then be saved in your data.frame, list,....
Here is a brief more detailed explanation of the pattern:
Firstly, .* is way of matching "anything". If you want to match digits you can use \\d or equivalently [0-9]. Here is a short intro to Regex in R (which I found to be quite useful) where you can find several character classes.
.* at the edges of the pattern means that there can be text before/after
(.*) denotes the content we want (so here matching any content as .* is used). Basically it means extract "anything" between SECTION 11 and 12.
\\. means the dot and \n is the "newline" metacharacter (as before "12.", a new line is started)
In Regex you can create groupings within your pattern using the brackets, i.e. gsub(".*(\\d{2}\\:\\d{2})", "\\1","18.05.2018, 21:37") will return 21:37, or gsub("([A-z]) \\d+","\\1","hello 123") will give hello.
Now the second argument in gsub can and is often used to provide a substitute, i.e. something to replace to matched pattern with. Here however, we do not want any substitue, we want to extract something. \\1 means extract the first grouping, i.e. what it inside the first brackets (you could have multiple groupings).
Finally, string is the string from which we want to extract, i.e. the PDF file
Now if you want to perform something similar in a loop you could do the following:
# we are in the loop
# first is your starting point in the extraction, i.e. "SECTION 11"
# last is your end point, i.e. "12."
first <- "SECTION 11" # first and last can be dynamically assigned
last <- "12\\." # "\\" is added before the dot as "." is a Regex metachar
# If last doesn't systematically contain a dot
# you could use gsub to add "\\" before the dot when needed:
# gsub("\\.","\\\\.",".") returns "\\."
# so gsub("\\.","\\\\.","12.") returns "12\\."
pat <- paste0(".*",first,"(.*)","\n",last,".*") #"\n" is added to stop before the newline, but it could be omitted (then "\n" might appear in the extraction)
gsub(pat,"\\1",string) # returns the same as above
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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.