I am trying to understand how do i extract the string which is between two hyphens.
For example,
node->testtransport-fasttrack-direct
I want the string fasttrack to be extracted and it shouldnt be based on the position of the strings as they might change.
I want the hard code to extract the string present between two hyphens
Thank you in advance.
Here are some approaches. No packages are used.
1) Here we assume that the part between the two minus signs must be all upper case letters so >DHLPAKET is excluded because even though it is between two minus signs it has a character which is not an upper case letter. Match the start (^) and then anything (.*) followed by minus (-) followed by an upper case string which is captured ([A-Z]+) and another minus (-) and everything else and finally the end of string ($). Replace all that with the captured portion (\1)
x <- "WRO2->DHLPAKET-ASCHHEIM-DI"
sub("^.*-([A-Z]+)-.*$", "\\1", x)
## [1] "ASCHHEIM"
2) If the two minus signs surrounding the string of interest are always the second and third minus signs then this would work. It uses read.table picking off the third minus-separated field.
read.table(text = x, sep = "-", as.is = TRUE)$V3
## [1] "ASCHHEIM"
Related
I have a vector that is constructed with numbers and letters. I want to get all the characters before the LAST letter of each value (which is I guess, always the 2nd letter of the vector). Using stringr (preferably)...
Example :
x = c("1H23456789H10", "97845784584H2", "0H987654321H0", "0P45454545A3", "63A00000000000A91")
str_extract_all(string = x, pattern = ????????)
I tried some tricks here : https://evoldyn.gitlab.io/evomics-2018/ref-sheets/R_strings.pdf
The result I want is :
"1H23456789" instead of "1H23456789H10"
"97845784584" instead of "97845784584H2"
"0H987654321", instead of "0H987654321H0"
"0P45454545", instead of "0P45454545A3"
"63A00000000000" instead of "63A00000000000A91"
str_extract(string = x, pattern = "[^A-Z]*[A-Z][^A-Z]*")
# [1] "1H23456789" "0H987654321" "0P45454545" "63A00000000000"
Explanation: we want to extract 1 pattern match per input, so we use str_extract not str_extract_all. Our pattern [^A-Z]*, any number of non-letters, followed by [A-Z] exactly one letter, followed by [^A-Z]* any number of non-letters. I just used capital letters based on your input, but you could change A-Z to A-Za-z inside the brackets to include lower case letters.
I have a vector with some codes. However, for an unknown reason, some of the code start with X# (# being a number 0-9). If my vector item does start with x#, I need to remove the first two letters.
Examples:
codes <- c('x0fa319-432f39-4fre78', '23weq0-4fsf198-417203', 'x2431-5435-1242-qewf')
expectedResult <- c('fa319-432f39-4fre78', '23weq0-4fsf198-417203', '431-5435-1242-qewf')
I tried using str_replace and gsub, but I couldn't get it right:
gsub("X\\d", "", codes)
but this would remove the x# even if it was in the middle of the string.
Any ides?
You can use
codes <- c('x0fa319-432f39-4fre78', '23weq0-4fsf198-417203', 'x2431-5435-1242-qewf')
sub("^x\\d", "", codes, ignore.case=TRUE)
See the R demo.
The ^x\d pattern matches x and any digit at the start of a string.
sub replaces the first occurrence only.
ignore.case=TRUE enables case insensitive matching.
My objective would be replacing a string by a symbol repeated as many characters as have the string, in a way as one can replace letters to capital letters with \\U\\1, if my pattern was "...(*)..." my replacement for what is captured by (*) would be something like x\\q1 or {\\q1}x so I would get so many x as characters captured by *.
Is this possible?
I am thinking mainly in sub,gsub but you can answer with other libraris like stringi,stringr, etc.
You can use perl = TRUE or perl = FALSE and any other options with convenience.
I assume the answer can be negative, since seems to be quite limited options (?gsub):
a replacement for matched pattern in sub and gsub. Coerced to character if possible. For fixed = FALSE this can include backreferences "\1" to "\9" to parenthesized subexpressions of pattern. For perl = TRUE only, it can also contain "\U" or "\L" to convert the rest of the replacement to upper or lower case and "\E" to end case conversion. If a character vector of length 2 or more is supplied, the first element is used with a warning. If NA, all elements in the result corresponding to matches will be set to NA.
Main quantifiers are (?base::regex):
?
The preceding item is optional and will be matched at most once.
*
The preceding item will be matched zero or more times.
+
The preceding item will be matched one or more times.
{n}
The preceding item is matched exactly n times.
{n,}
The preceding item is matched n or more times.
{n,m}
The preceding item is matched at least n times, but not more than m times.
Ok, but it seems to be an option (which is not in PCRE, not sure if in PERL or where...) (*) which captures the number of characters the star quantifier is able to match (I found it at https://www.rexegg.com/regex-quantifier-capture.html) so then it could be used \q1 (same reference) to refer to the first captured quantifier (and \q2, etc.). I also read that (*) is equivalent to {0,} but I'm not sure if this is really the fact for what I'm interested in.
EDIT UPDATE:
Since asked by commenters I update my question with an specific example provide by this interesting question. I modify a bit the example. Let's say we have a <- "I hate extra spaces elephant" so we are interested in keeping the a unique space between words, the 5 first characters of each word (till here as the original question) but then a dot for each other character (not sure if this is what is expected in the original question but doesn't matter) so the resulting string would be "I hate extra space. eleph..." (one . for the last s in spaces and 3 dots for the 3 letters ant in the end of elephant). So I started by keeping the 5 first characters with
gsub("(?<!\\S)(\\S{5})\\S*", "\\1", a, perl = TRUE)
[1] "I hate extra space eleph"
How should I replace the exact number of characters in \\S* by dots or any other symbol?
Quantifiers cannot be used in the replacement pattern, nor the information how many chars they match.
What you need is a \G base PCRE pattern to find consecutive matches after a specific place in the string:
a <- "I hate extra spaces elephant"
gsub("(?:\\G(?!^)|(?<!\\S)\\S{5})\\K\\S", ".", a, perl = TRUE)
See the R demo and the regex demo.
Details
(?:\G(?!^)|(?<!\S)\S{5}) - the end of the previous successful match or five non-whitespace chars not preceded with a non-whitespace char
\K - a match reset operator discarding text matched so far
\S - any non-whitespace char.
gsubfn is like gsub except the replacement string can be a function which inputs the match and outputs the replacement. The function can optionally be expressed a formula as we do here replacing each string of word characters with the output of the function replacing that string. No complex regular expressions are needed.
library(gsubfn)
gsubfn("\\w+", ~ paste0(substr(x, 1, 5), strrep(".", max(0, nchar(x) - 5))), a)
## [1] "I hate extra space. eleph..."
or almost the same except function is slightly different:
gsubfn("\\w+", ~ paste0(substr(x, 1, 5), substring(gsub(".", ".", x), 6)), a)
## [1] "I hate extra space. eleph..."
I have a vector of strings and I want to remove -es from all strings (words) ending in either -ses or -ces at the same time. The reason I want to do it at the same time and not consequitively is that sometimes it happens that after removing one ending, the other ending appears while I don't want to apply this pattern to a single word twice.
I have no idea how to use two patterns at the same time, but this is the best I could:
text <- gsub("[sc]+s$", "[sc]", text)
I know the replacement is not correct, but I wonder how can I show that I want to replace it with the letter I just detected (c or s in this case). Thank you in advance.
To remove es at the end of words, that is preceded with s or c, you may use
gsub("([sc])es\\b", "\\1", text)
gsub("(?<=[sc])es\\b", "", text, perl=TRUE)
To remove them at the end of strings, you can go on using your $ anchor:
gsub("([sc])es$", "\\1", text)
gsub("(?<=[sc])es$", "", text, perl=TRUE)
The first gsub TRE pattern is ([sc])es\b: a capturing group #1 that matches either s or c, and then es is matched, and then \b makes sure the next char is not a letter, digit or _. The \1 in the replacement is the backreference to the value stored in the capturing group #1 memory buffer.
In the second example with the PCRE regex (due to perl=TRUE), (?<=[sc]) positive lookbehind is used instead of the ([sc]) capturing group. Lookbehinds are not consuming text, the text they match does not land in the match value, and thus, there is no need to restore it anyhow. The replacement is an empty string.
Strings ending with "ces" and "ses" follow the same pattern, i.e. "*es$"
If I understand it correctly than you don't need two patterns.
Example:
x = c("ces", "ses", "mes)
gsub( pattern = "*([cs])es$", replacement = "\\1", x)
[1] "c" "s" "mes"
Hope it helps.
M
I have the following codes
"ABC.A.SVN.10.10.390.10.UDGGL"
"XYZ.Z.SVN.11.12.111.99.ASDDL"
and I need to replace the characters that exist between the 2nd and the 3rd dot. In this case it is SVN but it may well be any combination of between A and ZZZ, so really the only way to make this work is by using the dots.
The required outcome would be:
"ABC.A..10.10.390.10.UDGGL"
"XYZ.Z..11.12.111.99.ASDDL"
I tried variants of grep("^.+(\\.\\).$", "ABC.A.SVN.10.10.390.10.UDGGL") but I get an error.
Some examples of what I have tried with no success :
Link 1
Link 2
EDIT
I tried #Onyambu 's first method and I ran into a variant which I had not accounted for: "ABC.A.AB11.1.12.112.1123.UDGGL". In the replacement part, I also have numeric values. The desired outcome is "ABC.A..1.12.112.1123.UDGGL" and I get it using sub("\\.\\w+.\\B.",".",x) per the second part of his answer!
See code in use here
x <- c("ABC.A.SVN.10.10.390.10.UDGGL", "XYZ.Z.SVN.11.12.111.99.ASDDL")
sub("^(?:[^.]*\\.){2}\\K[^.]*", "", x, perl=T)
^ Assert position at the start of the line
(?:[^.]*\.){2} Match the following exactly twice
[^.]*\. Match any character except . any number of times, followed by .
\K Resets the starting point of the pattern. Any previously consumed characters are no longer included in the final match
[^.]* Match any character except . any number of times
Results in [1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
x= "ABC.A.SVN.10.10.390.10.UDGGL" "XYZ.Z.SVN.11.12.111.99.ASDDL"
sub("([A-Z]+)(\\.\\d+)","\\2",x)
[1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
([A-Z]+) Capture any word that has the characters A-Z
(\\.\\d+) The captured word above, must be followed with a dot ie\\..This dot is then followed by numbers ie \\d+. This completes the capture.
so far the captured part of the string "ABC.A.SVN.10.10.390.10.UDGGL" is SVN.10 since this is the part that matches the regular expression. But this part was captured as SVN and .10. we do a backreference ie replace the whole SVN.10 with the 2nd part .10
Another logic that will work:
sub("\\.\\w+.\\B.",".",x)
[1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
Not exactly regex but here is one more approach
#DATA
S = c("ABC.A.SVN.10.10.390.10.UDGGL", "XYZ.Z.SVN.11.12.111.99.ASDDL")
sapply(X = S,
FUN = function(str){
ind = unlist(gregexpr("\\.", str))[2:3]
paste(c(substring(str, 1, ind[1]),
"SUBSTITUTION",
substring(str, ind[2], )), collapse = "")
},
USE.NAMES = FALSE)
#[1] "ABC.A.SUBSTITUTION.10.10.390.10.UDGGL" "XYZ.Z.SUBSTITUTION.11.12.111.99.ASDDL"