In R I need to search a character vector as shown below. I need to return "AB" separately from "ABC" so I am using word-boundaries. However, I also need to find "AB-C" as something distinct from "AB"; there are some questions along these lines here but I can't get the proper invocation. Put another way, I need each of these strings to be found uniquely as I loop over them, and my grep expression needs to always return a single answer.
vec <- c("AB", "ABC", "AB-C")
grep("\\bAB\\b", vec) # 1 + 3, but only want 1
We just specify the start (^) and end ($) of the string
grep("^AB$", vec)
#[1] 1
Related
Is there a way to replace a character only if it is not repeating, or repeating a certain number of times?
str = c("ddaabb", "daabb", "aaddbb", "aadbb")
gsub("d{1}", "c", str)
[1] "ccaabb" "caabb" "aaccbb" "aacbb"
#Expected output
[1] "ddaabb" "caabb" "aaddbb" "aacbb"
You can use negative lookarounds in your regex to exclude cases where d is preceeded or followed by another d:
gsub("(?<!d)d(?!d)", "c", str, perl=TRUE)
Edit: adding perl=TRUE as suggested by OP. For more info about regex engine in R see this question
Now that you've added "or repeating a specified number of times," the regex-based approaches may get messy. Thus I submit my wacky code from a previous comment.
foo <- unlist(strsplit(str, '')
bar <- rle(foo)
and then look for instances of bar$lengths == desired_length and use the returned indices to locate (by summing all bar$lengths[1:k] ) the position in the original sequence. If you only want to replace a specific character, check the corresponding value of bar$values[k] and selectively replace as desired.
This might sound quite silly but it's driving me nuts.
I have a matrix that has alphanumeric values and I'm struggling to test if some elements of that matrix match only the initial and final letters. As I don't care the middle character, I'm trying (withouth success) to use a wildcard.
As an example, consider this matrix:
m <- matrix(nrow=3,ncol=3)
m[1,]=c("NCF","NBB","FGF")
m[2,]=c("MCF","N2B","CCD")
m[3,]=c("A3B","N4F","MCP")
I want to evaluate if m[2,2] starts with "N" and ends with "B", regardless of the 2nd letter in the string. I've tried something like
grep("N.B",m)
and it works, but still I want to know if there is a more compact way of doing it, like:
m[2,2]="N.B"
which ovbiously didn't work!
Thanks
You can use grepl with the subseted m like:
grepl("^N.B$", m[2,2])
#[1] TRUE
or use startsWith and endsWith:
startsWith(m[2,2], "N") & endsWith(m[2,2], "B")
#[1] TRUE
Minimal Reprex
Suppose I have the string as1das2das3D. I want to extract everything from the letter a to the letter D. There are three different substrings that match this - I want the shortest / right-most match, i.e. as3D.
One solution I know to make this work is stringr::str_extract("as1das2das3D", "a[^a]+D")
Real Example
Unfortunately, I can't get this to work on my real data. In my real data I have string with (potentially) two URLs and I'm trying to extract the one that's immediately followed by rel=\"next\". So, in the below example string, I'd like to extract the URL https://abc.myshopify.com/ZifQ.
foo <- "<https://abc.myshopify.com/YifQ>; rel=\"previous\", <https://abc.myshopify.com/ZifQ>; rel=\"next\""
# what I've tried
stringr::str_extract(foo, '(?<=\\<)https://.*(?=\\>; rel\\="next)') # wrong output
stringr::str_extract(foo, '(?<=\\<)https://(?!https)+(?=\\>; rel\\="next)') # error
You could do:
stringr::str_extract(foo,"https:[^;]+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
or even
stringr::str_extract(foo,"https(?:(?!https).)+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
Would this be an option?
Splitting string on ; or , comparing it with target string and take url from its previous index.
urls <- strsplit(foo, ";\\s+|,\\s+")[[1]]
urls[which(urls == "rel=\"next\"") - 1]
#[1] "<https://abc.myshopify.com/ZifQ>"
Here may be an option.
gsub(".+\\, <(.+)>; rel=\"next\"", "\\1", foo, perl = T)
#[1] "https://abc.myshopify.com/ZifQ"
I have a character vector that contains text similar to the following:
text <- c("ABc.def.xYz", "ge", "lmo.qrstu")
I would like to remove everything before a .:
> "xYz" "ge" "qrstu"
However, the grep function seems to be treating . as a letter:
pattern <- "([A-Z]|[a-z])+$"
grep(pattern, text, value = T)
> "ABc.def.xYz" "ge" "lmo.qrstu"
The pattern works elsewhere, such as on regexpal.
How can I get grep to behave as expected?
grep is for finding the pattern. It returns the index of the vector that matches a pattern. If, value=TRUE is specified, it returns the value. From the description, it seems that you want to remove substring instead of returning a subset of the initial vector.
If you need to remove the substring, you can use sub
sub('.*\\.', '', text)
#[1] "xYz" "ge" "qrstu"
As the first argument, we match a pattern i.e. '.*\\.'. It matches one of more characters (.*) followed by a dot (\\.). The \\ is needed to escape the . to treat it as that symbol instead of any character. This will match until the last . character in the string. We replace that matched pattern with a '' as the replacement argument and thereby remove the substring.
grep doesn't do any replacements. It searches for matches and returns the indices (or the value if you specify value=T) that give a match. The results you're getting are just saying that those meet your criteria at some point in the string. If you added something that doesn't meet the criteria anywhere into your text vector (for example: "9", "#$%23", ...) then it wouldn't return those when you called grep on it.
If you want it just to return the matched portion you should look at the regmatches function. However for your purposes it seems like sub or gsub should do what you want.
gsub(".*\\.", "", text)
I would suggest reading the help page for regexs ?regex. The wikipedia page is a decent read as well but note that R's regexs are a little different than some others. https://en.wikipedia.org/wiki/Regular_expression
You may try str_extract function from stringr package.
str_extract(text, "[^.]*$")
This would match all the non-dot characters exists at the last.
Your pattern does work, the problem is that grep does something different than what you are thinking it does.
Let's first use your pattern with str_extract_all from the package stringr.
library(stringr)
str_extract_all(text, pattern ="([A-Z]|[a-z])+$")
[[1]]
[1] "xYz"
[[2]]
[1] "ge"
[[3]]
[1] "qrstu"
Notice that the results came as you expected!
The problem you are having is that grep will give you the complete element that matches you regular expression and not only the matching part of the element. For example, in the example below, grep will return you the first element because it matches "a":
grep(pattern = "a", x = c("abcdef", "bcdf"), value = TRUE)
[1] "abcdef"
I have a vector of strings:
str.vect<-c ("abcR.1", "abcL.1", "abcR.2", "abcL.2")
str.vect
[1] "abcR.1" "abcL.1" "abcR.2" "abcL.2"
How can I remove the third character from the right in each vector element?
Here is the desired result:
"abc.1" "abc.1" "abc.2" "abc.2"
Thank you very much in advance
You can use nchar to find the length of each element of the vector
> nchar(str.vect)
[1] 6 6 6 6
Then you combine this with strtrim to get the beginning of each string
> strtrim(str.vect, nchar(str.vect)-3)
[1] "abc" "abc" "abc" "abc"
To get the end of the word you can then use substr (actually, you could use substr to get the beginning too...)
> substr(str.vect, nchar(str.vect)-1, nchar(str.vect))
[1] ".1" ".1" ".2" ".2"
And finally you use paste0 (which is paste with sep="") to stick them together
> paste0(strtrim(str.vect, nchar(str.vect)-3), # Beginning
substr(str.vect, nchar(str.vect)-1, nchar(str.vect))) # End
[1] "abc.1" "abc.1" "abc.2" "abc.2"
There are easier ways if you know your strings have some special characteristics
For instance, if the length is always 6 you can directly substitute the nchar calls with the appropriate value.
EDIT: alternatively, R also supports regular expressions, which make this task much easier.
> gsub(".(..)$", "\\1", str.vect)
[1] "abc.1" "abc.1" "abc.2" "abc.2"
The syntax is a bit more obscure, but not that difficult once you know what you are looking at.
The first parameter (".(..)$") is what you want to match
. matches any character, $ denotes the end of the string.
So ...$ indicates the last 3 characters in the string.
We put the last two in parenthesis, so that we can store them in memory.
The second parameter tells us what you want to substitute the matched substring with. In our case we put \\1 which means "whatever was in the first pair of parenthesis".
So essentially this command means: "find the last three characters in the string and change them with the last two".
The solution provided by #nico seems fine, but a simpler alternative might be to use sub:
sub('.(.{2})$', '\\1', str.vect)
This searches for the pattern of: "any character (represented by .) followed by 2 of any character (represented by .{2}), followed by the end of the string (represented by $)". By wrapping the .{2} in parentheses, R captures whatever those last two characters were. The second argument is the string to replace the matched substrings with. In this case, we refer to the first string captured in the matched pattern. This is represented by \\1. (If you captured multiple parts of the pattern, with multiple sets of parentheses, you would refer to subsequent captured regions with, e.g. \\2, \\3, etc.)
str.vect<-c ("abcR.1", "abcL.1", "abcR.2", "abcL.2")
a <- strsplit(str.vect,split="")
a <- strsplit(str.vect,split="")
b <- unlist(lapply(a,FUN=function(x) {x[4] <- ""
paste(x,collapse="")}
))
If you want to parameterize it further change 4 to a variable and put the index of the character you want to remove there.
Not sure how general or efficient this is, but it seems to work with your example string:
(This seems very similar to nico's answer although I am not using the strtrim function.)
my.string <- c("abcR.1", "abcL.1", "abcR.2", "abcL.2")
n.char <- nchar(my.string)
the.beginning <- substr(my.string, n.char-(n.char-1), n.char-3)
the.end <- substr(my.string, n.char-1, n.char)
new.string <- paste0(the.beginning, the.end)
new.string
# [1] "abc.1" "abc.1" "abc.2" "abc.2"
The 3rd character from the right of each element is removed.
sapply(str.vec, function(x) gsub(substr(x, nchar(x)-2,nchar(x)-2), "", x))
This is a very quick and dirty answer, but thats what is needed sometimes:
#Define vector
str.vect <- c("abcR.1", "abcL.1", "abcR.2", "abcL.2")
#Use gsub to remove both 'R' and 'L' independently.
str.vect2 <- gsub("R", '', str.vect )
str.vect_final <- gsub("L", '', str.vect2 )
>str.vect_final
[1] "abc.1" "abc.1" "abc.2" "abc.2"