I'd like to extract segments of line from a text.
For example:
txt<-"This is some cool text that involves this type of text and not that kind."
extract.context(txt,start="of text",end="that")
"of text and not that"
It kind of depends on what exactly what you will be looking for. If you will be just searching for characters (no punctuation), then this will work nicely.
extract.context<-function(txt, start, end) {
sapply(regmatches(txt, gregexpr(paste0(start,".*",end),txt)), "[", 1)
}
txt<-"This is some cool text that involves this type of text and not that kind."
extract.context(txt,start="of text",end="that")
# [1] "of text and not that"
This method uses a basic regular expression so if you search for character that may be matched by regular expression syntax, it could get confused. Also it's unclear what you want to do should multiple matches occur. Right now i just return the first. But since you didn't provide a lot of context, i'm going to assume that's OK.
Related
I need to concatenate two strings within an R object: one is just regular text; the other is italicized. So, I tried a lot of combinations, e.g.
paste0(" This is Regular", italic( This is Italics))
The desired result should be:
This is Regular This is Italics
Any ideia on how to do it?
Thanks!
In plot labels, you can use expressions, see mathematical annotation :
plot(1,xlab=expression("This is regular"~italic("this is italic")))
To provide an string for which an HTML parser will recognise the need to render the text in Italics, wrap the text in <i> and </i>. For example: "This is plain text, but <i>this is in Italics</i>.".
However, most HTML processors will assume that you want your text to appear as-is and will escape their input by default. This means that the special meanings of certain characters - including < and > will be "turned off". You need to tell the processor not to do this. How you do that will depend on context. I can't tell you that because you haven't given me context.
Are you for example, writing to a raw HTML file? (You need do nothing.) Are you writing to a Markdown file? If so, how? In plain text or in a rendered chunk? Are you writing a caption to a graphic? (Waldi has suggested a solution.) Etc, etc....
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.
In R Markdown, to make a text bold, we just need to do:
**code**
The the word code shows in bold.
I was wondering if there is a way to create a new command, let's say:
***code***
That would make the text highlighted?
Thanks!
It is not easily possible to create new markup, but one can change the way existing markup commands are rendered. Text enclosed by three stars is interpreted as emphasized strong emphasis. So one has to change that interpretation and change it to something else. One way to do so is via pandoc Lua filters. We just have to match on pandoc's internal representation of emphasized strong text and convert it to whatever we want:
function Strong (strong)
-- if this contains only one element, and if that element
-- is emphasized text, convert it to highlighted text.
local element = #strong.content == 1 and strong.content[1]
if element and element.t == 'Emph' then
table.insert(element.content, 1, pandoc.RawInline('html', '<mark>'))
table.insert(element.content, pandoc.RawInline('html', '</mark>'))
return element.content
end
end
The above works for HTML output. One would have to define what "highlighted text" means for each targeted format.
See this and this question for other approaches to the problem, and for details of how to use the filter with R Markdown.
I scrapped tweets from the twitter API and the package rtweet but I don't know how to work with text with emojis because they are in the form '\U0001f600' and all the regex code that I tried failed until now. I can't get anything of it.
For example
text = 'text text. \U0001f600'
grepl('U',text)
Give me FALSE
grepl('000',text)
Also give me FALSE.
Another problem is that they are often sticked to the word before (for example i am here\U0001f600 )
So how can I make R recognize emojis of that format? What can I put in the grepl that will return me TRUE for any emojis of that format?
In R there tends to be a package for most things. And in this case textclean and with it comes the lexicon package which has a lot of dictionaries. Using textclean you have 2 functions you can use, replace_emoji and replace_emoji_identifier
text = c("text text. \U0001f600", "i am here\U0001f600")
# replace emoji with identifier:
textclean::replace_emoji_identifier(text)
[1] "text text. lexiconvygwtlyrpywfarytvfis " "i am here lexiconvygwtlyrpywfarytvfis "
# replace emoji with text representation
textclean::replace_emoji(text)
[1] "text text. grinning face " "i am here grinning face "
Next you could use sentimentr to use sentiment scoring on the emoji's or for text analysis quanteda. If you just want to check the presence as in your expected output:
grepl("lexicon[[:alpha:]]{20}", textclean::replace_emoji_identifier(text))
[1] TRUE TRUE
Your problem is that you use a single character \ in your code:
text = 'text text. \U0001f600'
It really should be \\:
text = 'text text. \\U0001f600'
I had a similar experience using the rtweet library.
In my case the tweets bring some Unicode code points, not just emoji, and with the following format: "some text<U+code-point>". What I did in this case was "convert" that code point to its graphic representation:
library(stringi)
#I use gsub() to replace "<U+code-point>" with "\\ucode-point", the appropriate format
# And stri_unescape_unicode() to un-escape all Unicode sequences
stri_unescape_unicode(gsub("<U\\+(\\S+)>",
"\\\\u\\1", #replace by \\ucode-point
"some text with #COVID<U+30FC>19"))
#[1] "some text with #COVIDー19"
If the Unicode code point is not delimited as in my case (<>), you should change the regular expression from "<U\\+(\\S+)>" to "U(\\S+)" . You should be careful here, because this will work correctly if a space character appears after the code point. In case you have words attached to the code point both before and after, it must be more specific and indicate the number of characters that compose it, example "U(....)".
You can try refining this regular expression using Character Classes, or specifying only hexadecimal digits "U([A-Fa-f0-9]+)".
Note that in the RStudio console, the emoji are not going to be seen, you can apply this function but to see the emoji you must use an R library for this purpose. However other characters can be seen: "#COVID<U+30FC>19" appears in the RStudio console as "#COVIDー19".
Edit: Actually "\\S+" didn't work for me when there were consecutive Unicode code points like "<U+0001F926><U+200D><U+2642>". In this case it only replaced the first occurrence, I didn't delve into that, I just changed my regular expression to "<U\\+([A-Fa-f0-9]+)>".
"[A-Fa-f0-9]" represents hexadecimal digits.
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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.