Say I have text like this:
pattern = "This_is some word/expression I'd like to parse:intelligently(using special symbols-like '.')"
The challenge is how to split it into words, using word separators from the
c(" ","-","/","\\","_",":","(",")",".",",")
family.
Desired result:
"This" "is" "some" "word" "expression" "I'd" "like" "to" "parse" "intelligently" "using" "special" "symbols" "like"
Methods:
I could do sapply or for loop using:
keywords = unlist(strsplit(pattern," "))
keywords = unlist(strsplit(keywords,"-"))
# etc.
Question:
But what's the solution using Reduce(f, x, init, accummulate=TRUE)?
You shouldn't need Reduce here. You should be able to do something like the following:
splitters <- c(" ","/","\\","_",":","(",")",".",",","-") # dash should come last
pattern <- paste0("[", paste(splitters, collapse = ""), "]")
string <- "This_is some word/expression I'd like to parse:intelligently(using special symbols-like '.')"
strsplit(string, pattern)[[1]]
# [1] "This" "is" "some" "word"
# [5] "expression" "I'd" "like" "to"
# [9] "parse" "intelligently" "using" "special"
# [13] "symbols" "like" "'" "'"
Note that a - in a regex character class should come first or last, so I've edited your vector of "splitters" accordingly. Also, you may want to add a + at the end of your "pattern" in case you want to collapse, say, multiple spaces into one.
You can use option perl = TRUE and then split on punctuation or space
> strsplit(pattern, '[[:punct:]]|[[:space:]]', perl = TRUE)
[[1]]
[1] "This" "is" "some" "word" "expression"
[6] "I" "d" "like" "to" "parse"
[11] "intelligently" "using" "special" "symbols" "like"
[16] ""
I'd go with (It will keep "I'd" together)
strsplit(pattern, "[^[:alnum:][:digit:]']")
## [[1]]
## [1] "This" "is" "some" "word" "expression" "I'd" "like" "to" "parse"
## [10] "intelligently" "using" "special" "symbols" "like" "'" "'"
Related
I'm looking at a number of cells in a data frame and am trying to extract any one of several sequences of characters; there's only one of these sequences per per cell.
Here's what I mean:
dF$newColumn = str_extract_all(string = "dF$column1", pattern ="sequence_1|sequence_2")
Am I screwing the syntax up here? Can I pull this sort of thing with stringr? Please rectify my ignorance!
Yes, you can use | since it denotes logical or in regex. Here's an example:
vec <- c("abc text", "text abc", "def text", "text def text")
library(stringr)
str_extract_all(string = vec, pattern = "abc|def")
The result:
[[1]]
[1] "abc"
[[2]]
[1] "abc"
[[3]]
[1] "def"
[[4]]
[1] "def"
However, in your command, you should replace "dF$column1" with dF$column1 (without quotes).
I am trying something I thought would be easy. I'm looking for a single regex solution (though others are welcomed for completeness). I want to split on n occurrences of a delimiter.
Here is some data:
x <- "I like_to see_how_too"
pat <- "_"
Desired outcome
Say I want to split on first occurrence of _:
[1] "I like" "to see_how_too"
Say I want to split on second occurrence of _:
[1] "I like_to see" "how_too"
Ideally, if the solution is a regex one liner generalizable to nth occurrence; the solution will use strsplit with a single regex.
Here's a solution that doesn't fit my parameters of single regex that works with strsplit
x <- "I like_to see_how_too"
y <- "_"
n <- 1
loc <- gregexpr("_", x)[[1]][n]
c(substr(x, 1, loc-1), substr(x, loc + 1, nchar(x)))
Here is another solution using the gsubfn package and some regex-fu. To change the nth occurrence of the delimiter, you can simply swap the number that is placed inside of the range quantifier — {n}.
library(gsubfn)
x <- 'I like_to see_how_too'
strapply(x, '((?:[^_]*_){1})(.*)', c, simplify =~ sub('_$', '', x))
# [1] "I like" "to see_how_too"
If you would like the nth occurrence to be user defined, you could use the following:
n <- 2
re <- paste0('((?:[^_]*_){',n,'})(.*)')
strapply(x, re, c, simplify =~ sub('_$', '', x))
# [1] "I like_to see" "how_too"
Non-Solution
Since R is using PCRE, you can use \K to remove everything that matches the pattern before \K from the main match result.
Below is the regex to split the string at the 3rd _
^[^_]*(?:_[^_]*){2}\K_
If you want to split at the nth occurrence of _, just change 2 to (n - 1).
Demo on regex101
That was the plan. However, strsplit seems to think differently.
Actual execution
Demo on ideone.com
x <- "I like_to see_how_too but_it_seems to_be_impossible"
strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
It still fails to work on a stronger assertion \A
strsplit(x, "\\A[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
Explanation?
This behavior hints at the fact that strsplit find the first match, do a substring to extract the first token and the remainder part, and find the next match in the remainder part.
This removes all the states from the previous matches, and leaves us with a clean state when it tries to match the regex on the remainder. This makes the task of stopping the strsplit function at first match and achieving the task at the same time impossible. There is not even a parameter in strsplit to limit the number of splits.
Rather than split you do match to get your split strings.
Try this regex:
^((?:[^_]*_){1}[^_]*)_(.*)$
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
RegEx Demo
Update: It seems R also supports PCRE and in that case you can do split as well using this PCRE regex:
^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
(*FAIL) behaves like a failing negative assertion and is a synonym for (?!)
(*SKIP) defines a point beyond which the regex engine is not allowed to backtrack when the subpattern fails later
(*SKIP)(*FAIL) together provide a nice alternative of restriction that you cannot have a variable length lookbehind in above regex.
RegEx Demo2
x <- "I like_to see_how_too"
strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## > strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like" "to see" "how" "too"
## > strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like_to see" "how_too"
This uses gsubfn to to preprocess the input string so that strsplit can handle it. The main advantage is that one can specify a vector of numbers, k, indicating which underscores to split on.
It replaces the occurrences of underscore defined by k by a double underscore and then splits on double underscore. In this example we split at the 2nd and 4th underscore:
library(gsubfn)
k <- c(2, 4) # split at 2nd and 4th _
p <- proto(fun = function(., x) if (count %in% k) "__" else "_")
strsplit(gsubfn("_", p, "aa_bb_cc_dd_ee_ff"), "__")
giving:
[[1]]
[1] "aa_bb" "cc_dd" "ee_ff"
If empty fields are allowed then use any other character sequence not in the string, e.g. "\01" in place of the double underscore.
See section 4 of the gusbfn vignette for more info on using gusbfn with proto objects to retain state between matches.
I am doing a bit of NLP with R and am using the stringr package to tokenize some text.
I would like be able to capture contractions, for example, won't so that it is tokenized into "wo" and "n't".
Here is a sample of what I've got:
library(stringr)
s = "won't you buy my raspberries?"
foo = str_extract_all(s, "(n|t)|[[:punct:]]" ) # captures the contraction OK...
foo[[1]]
>[1] "n't" "?"
foo = str_extract_all(s, "(n|t)|\\w+|[[:punct:]]" ) # gets all words,
# but splits the contraction!
foo[[1]]
>[1] "won" "'" "t" "you" "buy" "my" "raspberries" "?"
I am trying to tokenize the above sentence into "wo", "n't", "you", "buy", "my", "raspberries", "?".
I am not too sure if I can do this with the default, extended regular expressions, or if I need to figure out some way to do this a Perl-like pattern.
Does anyone out there know of a way to do tokenization as described above with the stringr package?
EDIT
TO clarify, I am interested in Treebank tokenization
You could do this through lookaheads which was supported by PCRE library.
> s = "won't you buy my raspberries?"
> s
[1] "won't you buy my raspberries?"
> m <- gregexpr("\\w+(?=n[[:punct:]]t)|n?[[:punct:]]t?|\\w+", s, perl=TRUE)
> regmatches(s, m)
[[1]]
[1] "wo" "n't" "you" "buy" "my"
[6] "raspberries" "?"
OR
> m <- gregexpr("\\w+(?=\\w[[:punct:]]\\w)|\\w?[[:punct:]]\\w?|\\w+", s, perl=TRUE)
> regmatches(s, m)
[[1]]
[1] "wo" "n't" "you" "buy" "my"
[6] "raspberries" "?"
OR
Through stringr library,
> s <- "won't you buy my raspberries?"
> str_extract_all(s, perl("\\w+(?=\\w[[:punct:]]\\w)|\\w?[[:punct:]]\\w?|\\w+") )[[1]]
[1] "wo" "n't" "you" "buy" "my"
[6] "raspberries" "?"
You could try the perl wrapper function when working with stringr package functions.
s <- "won't you buy my raspberries?"
pattern <- "(?=[a-z]'[a-z])|(\\s+)|(?=[!?.])"
library(stringr)
str_split(s, perl(pattern))[[1]]
# [1] "wo" "n't" "you" "buy" "my"
# [6] "raspberries" "?"
There are also other wrappers such as fixed and ignore.case
I have a list of city, state data in a data frame. I need to extract only the state abbreviation and store into a new variable column called state. From visual inspection it looks like the state is always the last 2 characters in the string and they are both capitalized. The city, state data looks like the following:
test <- c("Anchorage, AK", "New York City, NY", "Some Place, Another Place, LA")
I tried the following
pattern <- "[, (A-Z){2}]"
strsplit(test, pattern)
The output was:
[[1]]
[1] "Anchorage, "
[[2]]
[1] "New York City, "
[[3]]
[1] "Some Place, Another Place, "
EDI:
I used another regular expresson:
pattern2 <- "([a-z, ])"
sp <- strsplit(test, pattern2)
I get these results:
[[1]]
[1] "A" "" "" "" "" "" "" "" "" "" "AK"
[[2]]
[1] "N" "" "" "Y" "" "" "" "C" "" "" "" "" "NY"
[[3]]
[1] "S" "" "" "" "P" "" "" "" "" "" "A" "" "" "" "" "" ""
[18] "P" "" "" "" "" "" "LA"
So, the abbreviation is there, but when I try to extract using sapply(), I am not sure how to get the last element of a list. I know how to get the first:
sapply(sp, "[[", 1)
I'm not sure you really need a regular expression here. If you always just want the last two characters of the string, just use
substring(test, nchar(test)-1, nchar(test))
[1] "AK" "NY" "LA"
If you really insist on a regular expression, at least consider using regexec rather than strsplit since you're not really interested in splitting, you only want to extract the state.
m <- regexec("[A-Z]+$", test)
unlist(regmatches(test,m))
# [1] "AK" "NY" "LA"
Try:
tt = strsplit(test, ', ')
tt
[[1]]
[1] "Anchorage" "AK"
[[2]]
[1] "New York City" "NY"
[[3]]
[1] "Some Place" "Another Place" "LA"
z = list()
for(i in tt) z[length(z)+1] = i[length(i)]
z
[[1]]
[1] "AK"
[[2]]
[1] "NY"
[[3]]
[1] "LA"
This can work:
regmatches(test, gregexpr("(?<=[,][\\s+])([A-Z]{2})", test, perl = TRUE))
## [[1]]
## [1] "AK"
##
## [[2]]
## [1] "NY"
##
## [[3]]
## [1] "LA"
Explanation compliments of: http://liveforfaith.com/re/explain.pl
(?<= look behind to see if there is:
[,] any character of: ','
[\\s+] any character of: whitespace (\n, \r,
\t, \f, and " "), '+'
) end of look-behind
( group and capture to \1:
[A-Z]{2} any character of: 'A' to 'Z' (2 times)
) end of \1
I think you understood reversely the meaning of '[]' and '()'. '()' means to match a group of characters; '[]' means to match any one character from a class. What you need is
"(, [A-Z]{2})".
library(stringr)
str_extract(test, perl('[A-Z]+(?=\\b$)'))
#[1] "AK" "NY" "LA"
here is a regex for the same
Regex
(?'state'\w{2})(?=")
Test String
"Anchorage, AK", "New York City, NY", "Some Place, Another Place, LA"
Result
MATCH 1
state [12-14] AK
MATCH 2
state [33-35] NY
MATCH 3
state [66-68] LA
live demo here
you may remove the named capture to make it smaller if required
eg
(\w{2})(?=")
I am trying to split a string by the group "%in%" and the character "#". All documentation and everything I can find says that parenthesis are metacharacters used for grouping in R regex. So the code
> strsplit('example%in%aa(bbb)aa#cdef', '[(%in%)#]', perl=TRUE)
SHOULD give me
[[1]]
[1] "example" "aa(bbb)aa" "cdef"
That is, it should leave the parentheses in "aa(bbb)aa" alone, because the parentheses in the matching expression are not escaped. But instead it ACTUALLY gives me
[[1]]
[1] "example" "" "" "" "aa" "bbb" "aa" "cdef"
as if the parentheses were not metacharacters! What is up with this and how can I fix it? Thanks!
This is true with and without the argument perl=TRUE in strsplit.
Not sure what documentation you're reading, but the Extended Regular Expressions section in ?regex says:
Most metacharacters lose their special meaning inside a character class. ...
(Only '^ - \ ]' are special inside character classes.)
You don't need to create a character class. Just use "or" | (you likely don't need to group "%in%" either, but it shouldn't hurt anything):
> strsplit('example%in%aa(bbb)aa#cdef', '(%in%)|#', perl=TRUE)
[[1]]
[1] "example" "aa(bbb)aa" "cdef"
No need to use [ or ( here , just this :
strsplit('example%in%aa(bbb)aa#cdef', '%in%|#')
[[1]]
[1] "example" "aa(bbb)aa" "cdef"
Inside character class [], most of the characters lose their special meaning, including ().
You might want this regex instead:
'%in%|#'