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How can I get the last n characters from a string in R?
Is there a function like SQL's RIGHT?
I'm not aware of anything in base R, but it's straight-forward to make a function to do this using substr and nchar:
x <- "some text in a string"
substrRight <- function(x, n){
substr(x, nchar(x)-n+1, nchar(x))
}
substrRight(x, 6)
[1] "string"
substrRight(x, 8)
[1] "a string"
This is vectorised, as #mdsumner points out. Consider:
x <- c("some text in a string", "I really need to learn how to count")
substrRight(x, 6)
[1] "string" " count"
If you don't mind using the stringr package, str_sub is handy because you can use negatives to count backward:
x <- "some text in a string"
str_sub(x,-6,-1)
[1] "string"
Or, as Max points out in a comment to this answer,
str_sub(x, start= -6)
[1] "string"
Use stri_sub function from stringi package.
To get substring from the end, use negative numbers.
Look below for the examples:
stri_sub("abcde",1,3)
[1] "abc"
stri_sub("abcde",1,1)
[1] "a"
stri_sub("abcde",-3,-1)
[1] "cde"
You can install this package from github: https://github.com/Rexamine/stringi
It is available on CRAN now, simply type
install.packages("stringi")
to install this package.
str = 'This is an example'
n = 7
result = substr(str,(nchar(str)+1)-n,nchar(str))
print(result)
> [1] "example"
>
Another reasonably straightforward way is to use regular expressions and sub:
sub('.*(?=.$)', '', string, perl=T)
So, "get rid of everything followed by one character". To grab more characters off the end, add however many dots in the lookahead assertion:
sub('.*(?=.{2}$)', '', string, perl=T)
where .{2} means .., or "any two characters", so meaning "get rid of everything followed by two characters".
sub('.*(?=.{3}$)', '', string, perl=T)
for three characters, etc. You can set the number of characters to grab with a variable, but you'll have to paste the variable value into the regular expression string:
n = 3
sub(paste('.+(?=.{', n, '})', sep=''), '', string, perl=T)
UPDATE: as noted by mdsumner, the original code is already vectorised because substr is. Should have been more careful.
And if you want a vectorised version (based on Andrie's code)
substrRight <- function(x, n){
sapply(x, function(xx)
substr(xx, (nchar(xx)-n+1), nchar(xx))
)
}
> substrRight(c("12345","ABCDE"),2)
12345 ABCDE
"45" "DE"
Note that I have changed (nchar(x)-n) to (nchar(x)-n+1) to get n characters.
A simple base R solution using the substring() function (who knew this function even existed?):
RIGHT = function(x,n){
substring(x,nchar(x)-n+1)
}
This takes advantage of basically being substr() underneath but has a default end value of 1,000,000.
Examples:
> RIGHT('Hello World!',2)
[1] "d!"
> RIGHT('Hello World!',8)
[1] "o World!"
Try this:
x <- "some text in a string"
n <- 5
substr(x, nchar(x)-n, nchar(x))
It shoudl give:
[1] "string"
An alternative to substr is to split the string into a list of single characters and process that:
N <- 2
sapply(strsplit(x, ""), function(x, n) paste(tail(x, n), collapse = ""), N)
I use substr too, but in a different way. I want to extract the last 6 characters of "Give me your food." Here are the steps:
(1) Split the characters
splits <- strsplit("Give me your food.", split = "")
(2) Extract the last 6 characters
tail(splits[[1]], n=6)
Output:
[1] " " "f" "o" "o" "d" "."
Each of the character can be accessed by splits[[1]][x], where x is 1 to 6.
someone before uses a similar solution to mine, but I find it easier to think as below:
> text<-"some text in a string" # we want to have only the last word "string" with 6 letter
> n<-5 #as the last character will be counted with nchar(), here we discount 1
> substr(x=text,start=nchar(text)-n,stop=nchar(text))
This will bring the last characters as desired.
For those coming from Microsoft Excel or Google Sheets, you would have seen functions like LEFT(), RIGHT(), and MID(). I have created a package known as forstringr and its development version is currently on Github.
if(!require("devtools")){
install.packages("devtools")
}
devtools::install_github("gbganalyst/forstringr")
library(forstringr)
the str_left(): This counts from the left and then extract n characters
the str_right()- This counts from the right and then extract n characters
the str_mid()- This extract characters from the middle
Examples:
x <- "some text in a string"
str_left(x, 4)
[1] "some"
str_right(x, 6)
[1] "string"
str_mid(x, 6, 4)
[1] "text"
I used the following code to get the last character of a string.
substr(output, nchar(stringOfInterest), nchar(stringOfInterest))
You can play with the nchar(stringOfInterest) to figure out how to get last few characters.
A little modification on #Andrie solution gives also the complement:
substrR <- function(x, n) {
if(n > 0) substr(x, (nchar(x)-n+1), nchar(x)) else substr(x, 1, (nchar(x)+n))
}
x <- "moSvmC20F.5.rda"
substrR(x,-4)
[1] "moSvmC20F.5"
That was what I was looking for. And it invites to the left side:
substrL <- function(x, n){
if(n > 0) substr(x, 1, n) else substr(x, -n+1, nchar(x))
}
substrL(substrR(x,-4),-2)
[1] "SvmC20F.5"
Just in case if a range of characters need to be picked:
# For example, to get the date part from the string
substrRightRange <- function(x, m, n){substr(x, nchar(x)-m+1, nchar(x)-m+n)}
value <- "REGNDATE:20170526RN"
substrRightRange(value, 10, 8)
[1] "20170526"
I have a list filled with strings:
string<- c("SPG_L_subgenual_ACC_R", "SPG_R_MTG_L_pole", "MTG_L_pole_CerebellumGM_L")
I need to split the strings so they appear like:
"SPG_L", "subgenual_ACC_R", "SPG_R", "MTG_L_pole", "MTG_L_pole", "CerebellumGM_L"
I tried using the following regex expression to split the strings:
str_split(string,'(?<=[[RL]|pole])_')
But this leads to:
"SPG_L", "subgenual" "ACC_R", "SPG_R", "MTG_L", "pole", "MTG_L", "pole", "CerebellumGM_L"
How do I edit the regex expression so it splits each string element at the "_" after the first occurrence of "R", "L" unless the first occurrence of "R" or "L" is followed by "pole", then it splits the string element after the first occurrence of "pole" and only splits each string element once?
I suggest a matching approach using
^(.*?[RL](?:_pole)?)_(.*)
See the regex demo
Details
^ - start of string
(.*?[RL](?:_pole)?) - Group 1:
.*? - any zero or more chars other than line break chars as few as possible
[RL](?:_pole)? - R or L optionally followed with _pole
_ - an underscore
(.*) - Group 2: any zero or more chars other than line break chars as many as possible
See the R demo:
library(stringr)
x <- c("SPG_L_subgenual_ACC_R", "SPG_R_MTG_L_pole", "MTG_L_pole_CerebellumGM_L", "SFG_pole_R_IFG_triangularis_L", "SFG_pole_R_IFG_opercularis_L" )
res <- str_match_all(x, "^(.*?[RL](?:_pole)?)_(.*)")
lapply(res, function(x) x[-1])
Output:
[[1]]
[1] "SPG_L" "subgenual_ACC_R"
[[2]]
[1] "SPG_R" "MTG_L_pole"
[[3]]
[1] "MTG_L_pole" "CerebellumGM_L"
[[4]]
[1] "SFG_pole_R" "IFG_triangularis_L"
[[5]]
[1] "SFG_pole_R" "IFG_opercularis_L"
split_again = function(x){
if(length(x) > 1){
return(x)
}
else{
str_split(
string = x,
pattern = '(?<=[R|L])_',
n = 2)
}
}
str_split(
string = string,
pattern = '(?<=pole)_',
n = 2) %>%
lapply(split_again) %>%
unlist()
you could use sub then strsplit as shown:
strsplit(sub("^.*?[LR](?:_pole)?\\K_",":",string,perl=TRUE),":")
[[1]]
[1] "SPG_L" "subgenual_ACC_R"
[[2]]
[1] "SPG_R" "MTG_L_pole"
[[3]]
[1] "MTG_L_pole" "CerebellumGM_L"
I have a vector of URLs and need to extract a certain part of it. I've tried using a regex tester to see if my attempts worked, but they were no good.
The URLs I have are in this format: https://www.baseball-reference.com/teams/MIL/1976.shtml
I ned to extract the three letters after "teams/" (so for the example above, I need "MIL")
Does anyone have any idea how to get the correct regular expression to get this working? Thanks.
1) basename/dirname Try this:
u <- "https://www.baseball-reference.com/teams/MIL/1976.shtml" # input data
basename(dirname(u))
## [1] "MIL"
2) sub or with a regular expression:
sub(".*teams/(.*?)/.*", "\\1", u)
## [1] "MIL"
3) strsplit Split the string on / and take the second last component.
s <- strsplit(u, "/")[[1]]
s[length(s) - 1]
## [1] "MIL"
4) gsub Since the required substring is all upper case and no other characters in the input are this gsub which removes all characters that are not upper case letters would work:
gsub("[^A-Z]", "", u)
## [1] "MIL"
Many different ways to achieve this using regexp's. Here's one:
url <- "https://www.baseball-reference.com/teams/MIL/1976.shtml"
gsub(".+teams/(\\w{3}).+$", "\\1", url);
#[1] "MIL"
Or
x <- c('https://www.baseball-reference.com/teams/MIL/1976.shtml')
pattern <- "/teams/([^/]+)"
m <- regexec(pattern, x)
res = regmatches(x, m)[[1]]
res[2]
which yields
[1] "MIL"
Consider using the stringr package to simplify your code when handling strings.
Use a regular expression with positive lookbehind to catch alphanumeric codes following the string "teams\":
stringr::str_extract(url, "(?<=teams\\/)[A-Z]*")
In your case, if the URLs literally all begin with the same string https://www.baseball-reference.com/teams/ then you can avoid regex entirely and use a simple substring to get the three-letter code which follows:
stringr::str_sub(url, 42, 44)
Here are the results:
> url <- "https://www.baseball-reference.com/teams/MIL/1976.shtml"
>
> stringr::str_extract(url, "(?<=teams\\/)[A-Z]*")
[1] "MIL"
>
> stringr::str_sub(url, 42, 44)
[1] "MIL"
Suppose I have a long string:
"XOVEWVJIEWNIGOIWENVOIWEWVWEW"
How do I split this to get every 5 characters followed by a space?
"XOVEW VJIEW NIGOI WENVO IWEWV WEW"
Note that the last one is shorter.
I can do a loop where I constantly count and build a new string character by character but surely there must be something better no?
Using regular expressions:
gsub("(.{5})", "\\1 ", "XOVEWVJIEWNIGOIWENVOIWEWVWEW")
# [1] "XOVEW VJIEW NIGOI WENVO IWEWV WEW"
Using sapply
> string <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
> sapply(seq(from=1, to=nchar(string), by=5), function(i) substr(string, i, i+4))
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
You can try something like the following:
s <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW" # Original string
l <- seq(from=5, to=nchar(s), by=5) # Calculate the location where to chop
# Add sentinels 0 (beginning of string) and nchar(s) (end of string)
# and take substrings. (Thanks to #flodel for the condense expression)
mapply(substr, list(s), c(0, l) + 1, c(l, nchar(s)))
Output:
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
Now you can paste the resulting vector (with collapse=' ') to obtain a single string with spaces.
No *apply stringi solution:
x <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
stri_sub(x, seq(1, stri_length(x),by=5), length=5)
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
This extracts substrings just like in #Jilber answer, but stri_sub function is vectorized se we don't need to use *apply here.
You can also use a sub-string without a loop. substring is the vectorized substr
x <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
n <- seq(1, nc <- nchar(x), by = 5)
paste(substring(x, n, c(n[-1]-1, nc)), collapse = " ")
# [1] "XOVEW VJIEW NIGOI WENVO IWEWV WEW"
I have a vector from which I just need the first word. The words have different lengths. Words are separated by a symbol (. and _) How can I use the substr() function to get a new vector with just the first word?
I was thinking of something like this
x <- c("wooombel.ab","mugran.cd","friendly_ef.ab","hungry_kd.xy")
y <- substr(x,0, ???)
I think sub with some regular expressions would be the easiest solution:
sub(pattern = "[._].*", replacement = "", x = x)
# [1] "wooombel" "mugran" "friendly" "hungry"
Try:
sapply(strsplit(x,'[._]'), function(x) x[1])
[1] "wooombel" "mugran" "friendly" "hungry"
You could also use package stringr. It has some really handy functions for string manipulation.
One that comes to mind for this problem is word. It has a sep argument that allows the use of a regular expression.
> x <- c("wooombel.ab","mugran.cd","friendly_ef.ab","hungry_kd.xy")
> library(stringr)
> word(x, sep = "[._]")
# [1] "wooombel" "mugran" "friendly" "hungry"
Another option that allows you to continue to use substr is str_locate. So if we just subtract 1 from its result, we can get the desired first words.
> substr(x, 1, str_locate(x, "[._]")-1)
# [1] "wooombel" "mugran" "friendly" "hungry"
An extraction approach with stringi:
library(stringi)
stri_extract_first_regex(x, "[a-z]+(?=[._])")
## [1] "wooombel" "mugran" "friendly" "hungry"
Though "[^a-z]+(?=[._])" may be more explicit.
Regex explanation:
[^a-z]+ any character except: 'a' to 'z' (1 or
more times)
(?= look ahead to see if there is:
[._] any character of: '.', '_'
) end of look-ahead