I am trying to remove the decimal points in decimal numbers in R. Please note I want to keep the full stop of strings.
Example:
data= c("It's 6.00pm, and is late.")
I know that I have to use regex for this, but I am struggling. My desired output is:
"It's 6 00pm, and is late."
Thank you in advance.
Try this:
sub("(?<=\\d)\\.(?=\\d)", " ", data, perl = TRUE)
This solution uses lookbehind (?<=...) and lookahead (?=...)to assert that the period you wish to remove be enclosed by digits (thus avoiding matching the period at the sentence end). If you have several such cases within strings, then use gsubinstead of sub.
I suggest using a simple pattern to find the target text, then adding parenthesis to identify the parts of the matching text that you want to retain.
# Test data
data <- c("It's 6.00pm, and is late.")
The target pattern is a literal dot with a string of digits before and after it. \\d+ matches one or more digits and \\. matches a literal dot. Testing the pattern to see if it works:
grepl("\\d+\\.\\d+", data)
Result
TRUE
If we wanted too eliminate the whole thing we could do a simple replacement with an empty string. Testing if this targets the correct text:
sub("\\d+\\.\\d+", "", data)
Result
"It's pm, and is late."
Instead, to discard only a section of matched text we can identify the parts we want to keep, which is done by surrounding them with parenthesis. Once done we can refer to the captured text in the replacement. \\1 refers to the first chunk of text captured and \\2 refers to the second chunk of text, corresponding to the first and second sets of parenthesis
# pattern replacement
sub("(\\d+)\\.(\\d+)", "\\1\\2", data)
Result
[1] "It's 600pm, and is late."
This effectively removes the dot by omitting it from the replacement text.
Related
I have a string ARC GUNNA SPARKYA 2011QUARTER HORSE.
I'd like to extract only the ARC GUNNA SPARKYA part. I.e., everything to the left of the "2011QUARTER."
I will also have valid strings which I want the pattern NOT to match. Valid strings would be "10RUNS FAST" or "QUICKER 1".
Note that the above means I need a pattern which can explicitly pick up just any four numbers followed by the uppercase word "QUARTER."
I tried ([0-9A-Za-z]+( [0-9A-Za-z]+)+) but that pattern matches the part I want to keep too, so I can't use it to do something like gsub.
Can you please help me understand what regex pattern will accomplish this--particularly in R?
Thank you!
You could use sub with a capture group, and use that group in the replacement.
(.*?)\s+\d{4}QUARTER\b.*
Explanation
(.*?) Capture group 1, match any character, as few as possible
\s+ Match 1+ whitespace characters
\d{4}QUARTER\b Match 4 digits followed by the word QUARTER
.* Match the rest of the line
See a regex101 demo.
text <- "ARC GUNNA SPARKYA 2011QUARTER HORSE"
result = sub("(.*?)\\s+\\d{4}QUARTER\\b.*", "\\1", text)
result
Output
[1] "ARC GUNNA SPARKYA"
I have the following string:
x = "marchTextIWantToDisplayWithSpacesmarch"
I would like to delete the 'march' portion at the beginning of the string and then add a space before each uppercase letter in the remainder to yield the following result:
"Text I Want To Display With Spacesmarch"
To insert whitepace, I used gsub("([a-z]?)([A-Z])", "\\1 \\2", x, perl= T) but I have no clue how to modify the pattern so that the first 'march' is excluded from the returned string. I'm trying to get better at this so any help would be greatly appreciated.
An option would be to capture the upper case letter as a group ((...)) and in the replacement create a space followed by the backreference (\\1) of the captured group
gsub("([A-Z])", " \\1", x)
#[1] "march Text I Want To Display With Spacesmarch"
If we need to remove the 'march'
sub("\\b[a-z]\\w+\\s+", "", gsub("([A-Z])", " \\1", x))
[#1] "Text I Want To Display With Spacesmarch"
data
x <- "marchTextIWantToDisplayWithSpacesmarch"
No, you can't achieve your replacement using single gsub because in one of your requirement, you want to remove all lowercase letters starting from the beginning, and your second requirement is to introduce a space before every capital letter except the first capital letter of the resultant string after removing all lowercase letters from the beginning of text.
Doing it in single gsub call would have been possible in cases where somehow we can re-use some of the existing characters to make the conditional replace which can't be the case here. So in first step, you can use ^[a-z]+ regex to get rid of all lowercase letters only from the beginning of string,
sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch")
leaving you with this,
[1] "TextIWantToDisplayWithSpacesmarch"
And next step you can use this (?<!^)(?=[A-Z]) regex to insert a space before every capital letter except the first one as you might not want an extra space before your sentence. But you can combine both and write them as this,
gsub('(?<!^)(?=[A-Z])', ' ', sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch"), perl=TRUE)
which will give you your desired string,
[1] "Text I Want To Display With Spacesmarch"
Edit:
Explanation of (?<!^)(?=[A-Z]) pattern
First, let's just take (?=[A-Z]) pattern,
See the pink markers in this demo
As you can see, in the demo, every capital letter is preceded by a pink mark which is the place where a space will get inserted. But we don't want space to be inserted before the very first letter as that is not needed. Hence we need a condition in regex, which will not select the first capital letter which appears at the start of string. And for that, we need to use a negative look behind (?<!^) which means that Do not select the position which is preceded by start of string and hence this (?<!^) helps in discarding the upper case letter that is preceded by just start of string.
See this demo where the pink marker is gone from the very first uppercase letter
Hope this clarifies how every other capital letter is selected but not the very first. Let me know if you have any queries further.
You may use a single regex call to gsub coupled with trimws to trim the resulting string:
trimws(gsub("^\\p{Ll}+|(?<=.)(?=\\p{Lu})", " ", x, perl=TRUE))
## => [1] "Text I Want To Display With Spacesmarch"
It also supports all Unicode lowercase (\p{Ll}) and uppercase (\p{Lu}) letters.
See the R demo online and the regex demo.
Details
^\\p{Ll}+ - 1 or more lowercase letters at the string start
| - or
(?<=.)(?=\\p{Lu}) - any location between any char but linebreak chars and an uppercase letter.
Here is an altenative with a single call to gsubfn regex with some ifelse logic:
> gsubfn("^\\p{Ll}*(\\p{L})|(?<=.)(?=\\p{Lu})", function(n) ifelse(nchar(n)>0,n," "), x, perl=TRUE,backref=-1)
[1] "Text I Want To Display With Spacesmarch"
Here, the ^\\p{Ll}*(\\p{L}) part matches 0+ lowercase letters and captures the next uppercase into Group 1 that will be accessed by passing n argument to the anonymous function. If n length is non-zero, this alternative matched and the we need to replace with this value. Else, we replace with a space.
Since this is tagged perl, my 2 cents:
Can you chain together the substitutions inside sub() and gsub()? In newer perl versions an /r option can be added to the s/// substitution so the matched string can be returned "non-destructively" and then matched again. This allows hackish match/substitution/rematches without mastering advanced syntax, e.g.:
perl -E '
say "marchTextIWantToDisplayWithSpacesmarch" =~
s/\Amarch//r =~ s/([[:upper:]])/ $1/gr =~ s/\A\s//r;'
Output
Text I Want To Display With Spacesmarch
This seems to be what #pushpesh-kumar-rajwanshi and #akrun are doing by wrapping gsub inside sub() (and vice versa). In general I don't thinkperl = T captures the full magnificently advanced madness of perl regexps ;-) but gsub/sub must be fast operating on vectors, no?
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 4 years ago.
I'm trying to understand a regular expression someone has written in the gsub() function.
I've never used regular expressions before seeing this code, and i have tried to work out how it's getting the final result with some googling, but i have hit a wall so to speak.
gsub('.*(.{2}$)', '\\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'm trying to use stringr or R base calls to conditionally add a white-space for instances in a large vector where there is a numeric value then a special character - in this case a $ sign without a space. str_pad doesn't appear to allow for a reference vectors.
For example, for:
$6.88$7.34
I'd like to add a whitespace after the last number and before the next dollar sign:
$6.88 $7.34
Thanks!
If there is only one instance, then use sub to capture digit and the $ separately and in the replacement add the space between the backreferences of the captured group
sub("([0-9])([$])", "\\1 \\2", v1)
#[1] "$6.88 $7.34"
Or with a regex lookaround
gsub("(?<=[0-9])(?=[$])", " ", v1, perl = TRUE)
data
v1 <- "$6.88$7.34"
This will work if you are working with a vectored string:
mystring<-as.vector('$6.88$7.34 $8.34$4.31')
gsub("(?<=\\d)\\$", " $", mystring, perl=T)
[1] "$6.88 $7.34 $8.34 $4.31"
This includes cases where there is already space as well.
Regarding the question asked in the comments:
mystring2<-as.vector('Regular_Distribution_Type† Income Only" "Distribution_Rate 5.34%" "Distribution_Amount $0.0295" "Distribution_Frequency Monthly')
gsub("(?<=[[:alpha:]])\\s(?=[[:alpha:]]+)", "_", mystring2, perl=T)
[1] "Regular_Distribution_Type<U+2020> Income_Only\" \"Distribution_Rate 5.34%\" \"Distribution_Amount $0.0295\" \"Distribution_Frequency_Monthly"
Note that the \ appears due to nested quotes in the vector, should not make a difference. Also <U+2020> appears due to encoding the special character.
Explanation of regex:
(?<=[[:alpha:]]) This first part is a positive look-behind created by ?<=, this basically looks behind anything we are trying to match to make sure what we define in the look behind is there. In this case we are looking for [[:alpha:]] which matches a alphabetic character.
We then check for a blank space with \s, in R we have to use a double escape so \\s, this is what we are trying to match.
Finally we use (?=[[:alpha:]]+), which is a positive look-ahead defined by ?= that checks to make sure our match is followed by another letter as explained above.
The logic is to find a blank space between letters, and match the space, which then is replaced by gsub, with a _
See all the regex here
I would like to split strings like the following:
x <- "abc-1230-xyz-[def-ghu-jkl---]-[adsasa7asda12]-s-[klas-bst-asdas foo]"
by dash (-) on the condition that those dashes must not be contained inside a pair of []. The expected result would be
c("abc", "1230", "xyz", "[def-ghu-jkl---]", "[adsasa7asda12]", "s",
"[klas-bst-asdas foo]")
Notes:
There is no nesting of square brackets inside each other.
The square brackets can contain any characters / numbers / symbols except square brackets.
The other parts of the string are also variable so that we can only assume that we split by - whenever it's not inside [].
There's a similar question for python (How to split a string by commas positioned outside of parenthesis?) but I haven't yet been able to accurately adjust that to my scenario.
You could use look ahead to verify that there is no ] following sooner than a [:
-(?![^[]*\])
So in R:
strsplit(x, "-(?![^[]*\\])", perl=TRUE)
Explanation:
-: match the hyphen
(?! ): negative look ahead: if that part is found after the previously matched hyphen, it invalidates the match of the hyphen.
[^[]: match any character that is not a [
*: match any number of the previous
\]: match a literal ]. If this matches, it means we found a ] before finding a [. As all this happens in a negative look ahead, a match here means the hyphen is not a match. Note that a ] is a special character in regular expressions, so it must be escaped with a backslash (although it does work without escape, as the engine knows there is no matching [ preceding it -- but I prefer to be clear about it being a literal). And as backslashes have a special meaning in string literals (they also denote an escape), that backslash itself must be escaped again in this string, so it appears as \\].
Instead of splitting, extract the parts:
library(stringr)
str_extract_all(x, "(\\[[^\\[]*\\]|[^-])+")
I am not familiar with r language, but I believe it can do regex based search and replace. Instead of struggling with one single regex split function, I would go in 3 steps:
replace - in all [....] parts by a invisible char, like \x99
split by -
for each element in the above split result(array/list), replace \x99 back to -
For the first step, you can find the parts by \[[^]]