The format I would like to allow in my text boxes are comma delimited lists followed by a line break in between the comma delimited lists. Here is an example of what I want from the user:
1,2,3
1,2,4
1,2,5
1,2,6
So far I have limited the user using this ValidationExpression:
^([1-9][0-9]*[]*[ ]*,[ ]*)*[1-9][0-9]*$
However with that expression, the user is only able to enter one row of comma delimited numbers.
How can proceed to accept multiple rows by accepting line breaks?
It is possible to check if the input has the correct format. I would recommend to use groups and repeat them:
((\d+,)+\d+\n?)+
But to check if the matrix is symmetric you have to use something else then regex.
Check it out here: https://regex101.com/r/GqtOuQ/2/
If you want to be a bit more user friendly it is possible to allow as much horizontal spaces as the user wants to add between the number and comma. This can be done with he regex group \h which allows every whitespace except \n.
The regex code looks now a bit more messy:
((\h*\d+\h*,\h*)+\h*\d+\h*\n?\h*)+
Check this out here: https://regex101.com/r/GqtOuQ/3
Here is the version that should work with .NET:
(([ \t]*\d+[ \t]*,[ \t]*)+[ \t]*\d+[ \t]*\n?[ \t]*)+
Related
I have a question similar to this one but instead of having two specific characters to look between, I want to get the text between a space and a specific character. In my example, I have this string:
myString <- "This is my string I scraped from the web. I want to remove all instances of a picture. picture-file.jpg. The text continues here. picture-file2.jpg"
but if I were to do something like this: str_remove_all(myString, " .*jpg) I end up with
[1] "This"
I know that what's happening is R is finding the first instance of a space and removing everything between that space and ".jpg" but I want it to be the first space immediately before ".jpg". My final result I hope for looks like this:
[1] "This is my string I scraped from the web. I want to remove all instances of a picture. the text continues here.
NOTE: I know that a solution may arise which does what I want, but ends up putting two periods next to each other. I do not mind a solution like that because later in my analysis I am removing punctuation.
You can use
str_remove_all(myString, "\\S*\\.jpg")
Or, if you also want to remove optional whitespace before the "word":
str_remove_all(myString, "\\s*\\S*\\.jpg")
Details:
\s* - zero or more whitespaces
\S* - zero or more non-whitespaces
\.jpg - .jpg substring.
To make it case insensitive, add (?i) at the pattern part: "(?i)\\s*\\S*\\.jpg".
If you need to make sure there is no word char after jpg, add a word boundary: "(?i)\\s*\\S*\\.jpg\\b"
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 question.
My text file contains lines such as:
1.1 Description.
This is the description.
1.1.1 Quality Assurance
Random sentence.
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
I'm trying to find out how to get:
1.1 Description
1.1.1 Quality Assurance
1.6.1 Quality Control
Right now, I have:
txt1 <- readLines("text1.txt")
txt2<-grep("^[0-9.]+", txt1, value = TRUE)
file<-write(txt2, "text3.txt")
which results in:
1.1 Description.
1.1.1 Quality Assurance
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
You are using grep with value=TRUE, which
returns a character vector containing the selected elements of x
(after coercion, preserving names but no other attributes).
This means, that if your regular expression matches anything in the line, the all line will be returned. You managed to build your regular expression to match numbers in the begining of the line. So all the lines which begin with numbers get selected.
It seems that your goal is not to select the all line, but to select only until there is a line break or a period.
So, you need to adjust the regular expression to be more specific, and you need to extract only the matching portion of the line.
A regular expression that matches what you want can be:
"^([0-9]\\.?)+ .+?(\\.|$)"
It selects numbers with dots, followed by a space, followed by anything, and stops matching things when a . comes or the line ends. I recommend the following website to better understand what the regex does: https://regexr.com/
The next step is extracting from the given lines only the matching portion, and not the all line where the regex has a match. For this we'll use the function regexpr, which tells us where the matches are, and the function regmatches, which helps us extract those matches:
txt1 <- readLines("text.txt")
regmatches(txt1, regexpr("^([0-9]\\.?)+ .+?(\\.|$)", txt1))
I've got some kind of logfile I'd like to read and analyse. Unfortunately the files are saved in a pretty "ugly" way (with lots of special characters in between), so I'm not able to read in just the lines with each one being an entry. The only way to separate the different entries is using regular expressions, since the beginning of each entry follows a specified pattern.
My first approach was to identify the pattern in the character vector (I use read_file from the readr-package) and use the corresponding positions to split the vector with strsplit. Unfortunately the positions seem not always to match, since the result doesn't always correspond to the entries (I'd guess that there's a problem with the special characters).
A typical line of the file looks as follows:
16/10/2017, 21:51 - George: This is a typical entry here
The corresponding regular expressions looks as follows:
([[:digit:]]{2})/([[:digit:]]{2})/([[:digit:]]{4}), ([[:digit:]]{2}):([[:digit:]]{2}) - ([[:alpha:]]+):
The first thing I want is a data.frame with each line corresponding to a specific entry (in a next step I'd split the pattern into its different parts).
What I tried so far was the following:
regex.log = "([[:digit:]]{2})/([[:digit:]]{2})/([[:digit:]]{4}), ([[:digit:]]{2}):([[:digit:]]{2}) - ([[:alpha:]]+):"
log.regex = gregexpr(regex.log, file.log)[[1]]
log.splitted = substring(file.log, log.regex, log.regex[2:355]-1)
As can be seen this logfile has 355 entries. The first ones are separated correctly. How can I separate the character vector using a regular expression without loosing the information of the regular expression/pattern?
Use capturing and non-capturing groups to identify the parts you want to keep, and be sure to use anchors:
file.log = "16/10/2017, 21:51 - George: This is a typical entry here"
regex.log = "^((?:[[:digit:]]{2})\\/(?:[[:digit:]]{2})\\/(?:[[:digit:]]{4}), (?:[[:digit:]]{2}):(?:[[:digit:]]{2}) - (?:[[:alpha:]]+)): (.*)$"
gsub(regex.log,"\\1",file.log)
>> "16/10/2017, 21:51 - George"
gsub(regex.log,"\\2",file.log)
>> "This is a typical entry here"
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.