In a linear model, I have some splines, using the bs function from the splines package (like lm(y ~ bs(x, 3))).
In the model summary and model data frame (colnames(fit$model)) these terms appear as, e.g., bs(efc17age, 3).
Now I would like to extract the variable name using regular expressions. However,, I just don't understand regex syntax.
This is how far I came:
x <- "bs(e17age, 3)1"
sub("bs\\((*?)", "", x)
> [1] "e17age, 3)1"
I just want to have "e17age"... It must be so easy, if you understand regex...
You can use the following snippet:
x <- "bs(e17age, 3)1"
sub("^bs\\(([^,]*).*", "\\1", x)
Regex ^bs\\(([^,]*).* matches bs( at the start of the string, then captures any number of characters other than , with ([^,]*), and then matches any character up to the end. With the replacement string \\1, we get our captured text back.
See IDEONE demo
Related
I'm trying to extract a substring in R, using stringr. Some timeago I wrote a script that did the job, however it does not work anymore. Probably due to an update, but I don't know.
My string looks like(!) this: myStr <- " layout = (3,3); //lala".
The string will always contain the layout keyword, the equal sign and the two braces (open ... close). However, the number of arguments in between can vary: (1,23,455,22) would also be possible. After the part after ) can varying as well.
I like to obtain the substring starting form ( and ending with ). Thus this example must give: (3,3). Others may give e.g. (1,23,455,22).
Up to now I used this:
library(stringr)
str_extract(" layout = (3,3); //lala", "*\\(.*\\)")
However this does not work anymore. It gives me this error:
Error in stri_extract_first_regex(string, pattern, opts_regex = attr(pattern, :
Syntax error in regexp pattern. (U_REGEX_RULE_SYNTAX)
It used to work in the past. What is wrong this the regular expression?
EDIT:
If by string contains two pair of braces, the substring should select the left pair (the other is commented-out with //):
Str <- "layout = (1,2,3,4) //lala(huhu)"
gsub(".*([(])(.*)([)]).*", "\\1\\2\\3", Str)
#gives "(huhu)" which is not good; should be (1,2,3,4)
Your regex "*\\(.*\\)" is not correct as it starts with *, a quantifier, and that causes the incorrect regex syntax issue as you cannot have multiple string start positions (it is a logical error checked by the regex engine when parsing the expression).
The substring should select the left pair
Use lazy matching in the left part - .*?:
myStr <- "layout = (1,2,3,4) //lala(huhu)"
gsub(".*?(\\([^()]*\\)).*", "\\1", myStr)
## ^^^
See IDEONE demo
Result: [1] "(1,2,3,4)"
Lazy matching will ensure we match as few characters as possible before the first occurrence of the subsequent pattern.
Note that if you want to extract multiple (number,number....) values, you need to use
library(stringr)
str_extract(Str,"\\(\\d+(\\s*,\\d+)*\\)")
See the regex demo here.
If the numbers and commas along with the parentheses needs to be extracted, use those patterns,
str_extract(Str,"\\([0-9,]+\\)")
#[1] "(1,2,3,4)"
str_extract(myStr,"\\([0-9,]+\\)")
#[1] "(3,3)"
I have a bunch of strings that contain the word "radius" followed by one or two digits. They also contain a lot of other letters, digits, and underscores. For example, one is "inflow100_radius6_distance12". I want a regex that will just return the one or two digits following "radius." If R recognized \K, then I would just use this:
radius\K[0-9]{1,2}
and be done. But R doesn't allow \K, so I ended up with this instead (which selects radius and the following numbers, and then cuts off "radius"):
result <- regmatches(input_string, gregexpr("radius[0-9]{1,2}", input_string))
result <- unlist(substr(result, 7, 8)))
I'm pretty new to regex, so I'm sure there's a better way. Any ideas?
\K is recognized. You can solve the problem by turning on the perl = TRUE parameter.
result <- regmatches(x, gregexpr('radius\\K\\d+', x, perl=T))
1) Match the entire string replacing it with the digits after radius:
sub(".*radius(\\d+).*", "\\1", "inflow100_radius6_distance12")
## [1] "6"
The regular expression can be visualized as follows:
.*radius(\d+).*
Debuggex Demo
2) This also works, involves a simpler regular expression and converts it to numeric at the same time:
library(gsubfn)
strapply("inflow100_radius6_distance12", "radius(\\d+)", as.numeric, simplify = TRUE)
## [1] 6
Here is a visualization of the regular expression:
radius(\d+)
Debuggex Demo
I am trying to extract all of the words in the string below contained within the brackets following the word 'tokens' only if the 'tokens' occurs after 'tag(noun)'.
For example, I have the string:
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),
inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),
inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),
head([lexmatch([department]),inputmatch(['Department']),tag(noun),
tokens([department])])],0/29,[])."
I want to get a list of all of the words that occur within the brackets after the word 'tokens' only when the word tokens occurs after 'tag(noun)'.
Therefore, I want my output to be a vector of the following:
[1] new, york, state, department
How do I do this? I'm assuming I have to use a regular expression, but I'm lost on how to write this in R.
Thanks!
Remove newlines and then extract the portion matched to the part between parentheses in pattern pat. Then split apart such strings by commas and simplify into a character vector:
library(gsubfn)
pat <- "tag.noun.,tokens..(.*?)\\]"
strapply(gsub("\\n", "", m), pat, ~ unlist(strsplit(x, ",")), simplify = c)
giving:
[1] "new" "york" "state" "department"
Visualization: Here is the debuggex representation of the regular expression in pat. (Note that we need to double the backslash when put within R's double quotes):
tag.noun.,tokens..(.*?)\]
Debuggex Demo
Note that .*? means match the shortetst string of any characters such that the entire pattern matches - without the ? it would try to match the longest string.
How about something like this. Here i'll use the regcatputedmatches helper function to make it easier to extract the captured matches.
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),head([lexmatch([department]),inputmatch(['Department']),tag(noun),tokens([department])])],0/29,[])."
rx <- gregexpr("tag\\(noun\\),tokens\\(\\[([^]]+)\\]\\)", m, perl=T)
lapply(regcapturedmatches(m,rx), function(x) {
unlist(strsplit(c(x),","))
})
# [[1]]
# [1] "new" "york" "state" "department"
The regular expression is a bit messy because your desired match contains many special regular expression symbols so we need to properly escape them.
Here is a one liner if you like:
paste(unlist(regmatches(m, gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T))), collapse=",")
[1] "new,york,state,department"
Broken down:
# Get match indices
indices <- gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T)
# Extract the matches
matches <- regmatches(m, indices)
# unlist and paste together
paste(unlist(matches), collapse=",")
[1] "new,york,state,department"
Given a regular expression containing capture groups (parentheses) and a string, how can I obtain all the substrings matching the capture groups, i.e., the substrings usually referenced by "\1", "\2"?
Example: consider a regex capturing digits preceded by "xy":
s <- "xy1234wz98xy567"
r <- "xy(\\d+)"
Desired result:
[1] "1234" "567"
First attempt: gregexpr:
regmatches(s,gregexpr(r,s))
#[[1]]
#[1] "xy1234" "xy567"
Not what I want because it returns the substrings matching the entire pattern.
Second try: regexec:
regmatches(s,regexec("xy(\\d+)",s))
#[[1]]
#[1] "xy1234" "1234"
Not what I want because it returns only the first occurence of a matching for the entire pattern and the capture group.
If there was a gregexec function, extending regexec as gregexpr extends regexpr, my problem would be solved.
So the question is: how to retrieve all substrings (or indices that can be passed to regmatches as in the examples above) matching capture groups in an arbitrary regular expression?
Note: the pattern for r given above is just a silly example, it must remain arbitrary.
For a base R solution, what about just using gsub() to finish processing the strings extracted by gregexpr() and regmatches()?
s <- "xy1234wz98xy567"
r <- "xy(\\d+)"
gsub(r, "\\1", regmatches(s,gregexpr(r,s))[[1]])
# [1] "1234" "567"
Not sure about doing this in base, but here's a package for your needs:
library(stringr)
str_match_all(s, r)
#[[1]]
# [,1] [,2]
#[1,] "xy1234" "1234"
#[2,] "xy567" "567"
Many stringr functions also have parallels in base R, so you can also achieve this without using stringr.
For instance, here's a simplified version of how the above works, using base R:
sapply(regmatches(s,gregexpr(r,s))[[1]], function(m) regmatches(m,regexec(r,m)))
strapplyc in the gsubfn package does that:
> library(gsubfn)
>
> strapplyc(s, r)
[[1]]
[1] "1234" "567"
Try ?strapplyc for additional info and examples.
Related Functions
1) A generalization of strapplyc is strapply in the same package. It takes a function which inputs the captured portions of each match and returns the output of the function. When the function is c it reduces to strapplyc. For example, suppose we wish to return results as numeric:
> strapply(s, r, as.numeric)
[[1]]
[1] 1234 567
2) gsubfn is another related function in the same package. It is like gsub except the replacement string can be a replacement function (or a replacement list or a replacement proto object). The replacement function inputs the captured portions and outputs the replacement. The replacement replaces the match in the input string. If a formula is used, as in this example, the right hand side of the formula is regarded as the function body. In this example we replace the match with XY{#} where # is twice the matched input number.
> gsubfn(r, ~ paste0("XY{", 2 * as.numeric(x), "}"), s)
[1] "XY{2468}wz98XY{1134}"
UPDATE: Added strapply and gsubfn examples.
Since R 4.1.0, there is gregexec:
regmatches(s,gregexec(r,s))[[1]][2, ]
[1] "1234" "567"
In R, is there a better/simpler way than the following of finding the location of the last dot in a string?
x <- "hello.world.123.456"
g <- gregexpr(".", x, fixed=TRUE)
loc <- g[[1]]
loc[length(loc)] # returns 16
This finds all the dots in the string and then returns the last one, but it seems rather clumsy. I tried using regular expressions, but didn't get very far.
Does this work for you?
x <- "hello.world.123.456"
g <- regexpr("\\.[^\\.]*$", x)
g
\. matches a dot
[^\.] matches everything but a dot
* specifies that the previous expression (everything but a dot) may occur between 0 and unlimited times
$ marks the end of the string.
Taking everything together: find a dot that is followed by anything but a dot until the string ends. R requires \ to be escaped, hence \\ in the expression above. See regex101.com to experiment with regex.
How about a minor syntax improvement?
This will work for your literal example where the input vector is of length 1. Use escapes to get a literal "." search, and reverse the result to get the last index as the "first":
rev(gregexpr("\\.", x)[[1]])[1]
A more proper vectorized version (in case x is longer than 1):
sapply(gregexpr("\\.", x), function(x) rev(x)[1])
and another tidier option to use tail instead:
sapply(gregexpr("\\.", x), tail, 1)
Someone posted the following answer which I really liked, but I notice that they've deleted it:
regexpr("\\.[^\\.]*$", x)
I like it because it directly produces the desired location, without having to search through the results. The regexp is also fairly clean, which is a bit of an exception where regexps are concerned :)
There is a slick stri_locate_last function in the stringi package, that can accept both literal strings and regular expressions.
To just find a dot, no regex is required, and it is as easy as
stringi::stri_locate_last_fixed(x, ".")[,1]
If you need to use this function with a regex, to find the location of the last regex match in the string, you should replace _fixed with _regex:
stringi::stri_locate_last_regex(x, "\\.")[,1]
Note the . is a special regex metacharacter and should be escaped when used in a regex to match a literal dot char.
See an R demo online:
x <- "hello.world.123.456"
stringi::stri_locate_last_fixed(x, ".")[,1]
stringi::stri_locate_last_regex(x, "\\.")[,1]