I'm trying to search for a pattern and if it repeats 2nd time, delete to end of line..., please help.
For that using I'm using :%s/my_character.*//g, but this will work for the 1st occurrence of the character in a line, but I need it from 2nd occurrence in the line...
Not sure if I understand it well (you should give a plain example, it always makes it clearer)
I would do it like this:
:s/^\(.\{-}my_character.\{-}\)my_character.*$/\1/
This will search for:
As few character as possible before my_character
my_character
As few character as possible before the second my_character
my_character
Any character
And replace it with characters captured from 1 to 3 in the steps above.
Example:
Input:
werklj z sdkl Azlksd er.
search and replace:
:s/^\(.\{-}z.\{-}\)z.*$/\1/
Output:
werklj z sdkl A
Related
This question already has an answer here:
Reference - What does this regex mean?
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Closed 4 years ago.
I'm trying to understand a regular expression someone has written in the gsub() function.
I've never used regular expressions before seeing this code, and i have tried to work out how it's getting the final result with some googling, but i have hit a wall so to speak.
gsub('.*(.{2}$)', '\\1',"my big fluffy cat")
This code returns the last two characters in the given string. In the above example it would return "at". This is the expected result but from my brief foray into regular expressions i don't understand why this code does what it does.
What i understand is the '.*' means look for any character 0 or more times. So it's going to look at the entire string and this is what will be replaced.
The part in brackets looks for any two characters at the end of the string. It would make more sense to me if this part in brackets was in place of the '\1'. To me it would then read look at the entire string and replace it with the last two characters of that string.
All that does though is output the actual code as the replacement e.g ".{2}$".
Finally i don't understand why '\1' is in the replace part of the function. To me this is just saying replace the entire string with a single backslash and the number one. I say a single backslash because it's my understanding the first backslash is just there to make the second backslash a none special character.
For gsub there are two ways of using the function. The most common way is probably.
gsub("-","TEST","This is a - ")
which would return
This is a TEST
What this does is simply finds the matches in the regular expression and replaces it with the replacement string.
The second way to use gsub is the method in which you described. using \\1, \\2 or \\3...
What this does is looks at the first, second or third capture group in your regular expression.
A capture group is defined by anything inside the circular brackets ex: (capture_group_1)(capture_group_2)...
Explanation
Your analysis is correct.
What i understand is the '.*' means look for any character 0 or more times. So it's going to look at the entire string and this is what will be replaced.
The part in brackets looks for any two characters at the end of the string
The last two characters are placed in a capture group and we are simply replace the whole string with this capture group. Not replacing them with anything.
if it helps, check out the result of this expression.
gsub('(.*)(.{2}$)', 'Group 1: \\1, Group 2: \\2',"my big fluffy cat")
hope the examples can help you to understand it better:
Say we have a string foobarabcabcdef
.* matches whole string.
.*abc it matches: from the beginning matches any chars till the last abc (greedy matching), thus, it matches foobarabcabc
.*(...)$ matches the whole string as well, however, the last 3 chars were groupped. Without the () , the matched string will have a default group, group0, the () will be group1, 2, 3.... think about .*(...)(...)(...)$ so we have:
group 0 : whole string
group 1 : "abc" the first "abc"
group 2 : "abc" the 2nd "abc"
group 3 : "def" the last 3 chars
So back to your example, the \\1 is a reference to group. What it does is: "replace the whole string by the matched text in group1" That is, the .{2}$ part is the replacement.
If you don't understand the backslashs, you have to reference the syntax of r, I cannot tell more. It is all about escaping.
Important part of that regular expression are brackets, that's something called "capturing group".
Regular expression .*(.{2}$) says - match anything and capture last 2 characters at the line. Replacement \\1 is referencing to that group, so it will replace whole match with captured group, which are last two characters in this case.
I have a question.
My text file contains lines such as:
1.1 Description.
This is the description.
1.1.1 Quality Assurance
Random sentence.
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
I'm trying to find out how to get:
1.1 Description
1.1.1 Quality Assurance
1.6.1 Quality Control
Right now, I have:
txt1 <- readLines("text1.txt")
txt2<-grep("^[0-9.]+", txt1, value = TRUE)
file<-write(txt2, "text3.txt")
which results in:
1.1 Description.
1.1.1 Quality Assurance
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
You are using grep with value=TRUE, which
returns a character vector containing the selected elements of x
(after coercion, preserving names but no other attributes).
This means, that if your regular expression matches anything in the line, the all line will be returned. You managed to build your regular expression to match numbers in the begining of the line. So all the lines which begin with numbers get selected.
It seems that your goal is not to select the all line, but to select only until there is a line break or a period.
So, you need to adjust the regular expression to be more specific, and you need to extract only the matching portion of the line.
A regular expression that matches what you want can be:
"^([0-9]\\.?)+ .+?(\\.|$)"
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 am trying to pick patterns within a specific string and their respective location. I have explained below with an example:
String = "Web_797-Web_797-Web_797-Web_797-PCP_IM_PAR-Pharm_1-Pharm_1-
Web_797-PCP_IM_PAR-Prior_OP-Web_797-Prior_OP-Event_0-"
pattern = "Web_797-*Web_797" (Web_797 followed by Web_797 with anything in between)
I used the following function:
str_locate_all(String,pattern)[[1]]
I am getting the following result:
start end
[1,] 1 15
[2,] 17 31
which is what I need partially. However I the pattern is not able to pick the following combination (highlighted in black).
String = "Web_797-Web_797-Web_797-Web_797-PCP_IM_PAR-Pharm_1-Pharm_1-
Web_797-PCP_IM_PAR-Prior_OP-Web_797-Prior_OP-Event_0-"
I would appreciate if anyone could help with this. I believe there is something wrong with the way I am defining the pattern but not able to fix it.
The problem with your pattern pattern = "Web_797-*Web_797" is the -* part. That means zero or more dashes (-). I believe what you wanted was a dash followed by any characters. So a first (incorrect) attempt would be
pattern = "Web_797-.*Web_797" Where the . means "any character". But that is not quite right. You only want to collect characters until the next time you see Web_797, not all the way until the last time you see Web_797. By default, the matches are "greedy" taking the biggest possible match. If we use
pattern = "Web_797-.*?Web_797" the ? turns off greedy matching so that it only matches to the next Web_797.
I am trying to figure out how I can put together a find and replace command with wildcards or figure out a way to find and replace the following example:
I would like to find terms that contain double quotes in front of them with a single quote at the end:
Example:
find "joe' and replace with 'joe'
Basically, I'm trying to find all terms with terms having "in front and at the end.'
Check the [x] Regular expression checkbox in textpad's replace dialog and enter the following values:
Find what:
"([^'"]*)'
Replace with:
'\1'
Explanation:
In a regular expression, square brackets are used to indicate character classes. A character class beginning with a caret will match anything not in the class.
Thus [^'"] will match any character except ' and ". The following * indicates that any number of these characters can follow. The ( and ) mark a group. And the group we're looking for starts with " and ends with '. Finally in the replace string we can refer to any group via \n where n is the nth group. In our case it is the first and only group and that is why we used \1.
I want to introduce a backspace character at the beginning of the line where a particular pattern is not found. Please advise.
Thanks,
Sagar
If you mean that you want to "remove the first character" then you can do this:
1) Write your regex pattern of what you want to find. For example, if you want to match Remove me at the start of the line, use:
^R\(emove me\)
Here we use ^ to assert the position to the start of the string. We also capture everything apart from the string we wish to keep in a backreference so it can be used later.
2) Replace the matches we find with whatever we grabbed in our backreference, in this case emove me, in effect backspacing the first character.
3) Make sure regular expression is checked and the cursor is at the start of the file, and hit Replace All.
Before
After: