In R, grep usually matches a vector of multiple strings against one regexp.
Q: Is there a possibility to match a single string against multiple regexps? (without looping through each single regexp pattern)?
Some background:
I have 7000+ keywords as indicators for several categories. I cannot change that keyword dictionary. The dictionary has following structure (keywords in col 1, numbers indicate categories where these keywords belong to):
ab 10 37 41
abbrach* 38
abbreche 39
abbrich* 39
abend* 37
abendessen* 60 63
aber 20 23 45
abermals 37
Concatenating so many keywords with "|" is not a feasible way (and I wouldn't know which of the keywords generated the hit).
Also, just reversing "patterns" and "strings" does not work, as the patterns have truncations, which wouldn't work the other way round.
[related question, other programming language]
What about applying the regexpr function over a vector of keywords?
keywords <- c("dog", "cat", "bird")
strings <- c("Do you have a dog?", "My cat ate by bird.", "Let's get icecream!")
sapply(keywords, regexpr, strings, ignore.case=TRUE)
dog cat bird
[1,] 15 -1 -1
[2,] -1 4 15
[3,] -1 -1 -1
sapply(keywords, regexpr, strings[1], ignore.case=TRUE)
dog cat bird
15 -1 -1
Values returned are the position of the first character in the match, with -1 meaning no match.
If the position of the match is irrelevant, use grepl instead:
sapply(keywords, grepl, strings, ignore.case=TRUE)
dog cat bird
[1,] TRUE FALSE FALSE
[2,] FALSE TRUE TRUE
[3,] FALSE FALSE FALSE
Update: This runs relatively quick on my system, even with a large number of keywords:
# Available on most *nix systems
words <- scan("/usr/share/dict/words", what="")
length(words)
[1] 234936
system.time(matches <- sapply(words, grepl, strings, ignore.case=TRUE))
user system elapsed
7.495 0.155 7.596
dim(matches)
[1] 3 234936
re2r package can match multiple patterns (in parallel). Minimal example:
# compile patterns
re <- re2r::re2(keywords)
# match strings
re2r::re2_detect(strings, re, parallel = TRUE)
To expand on the other answer, to transform the sapply() output into a useful logical vector you need to further use an apply() step.
keywords <- c("dog", "cat", "bird")
strings <- c("Do you have a dog?", "My cat ate by bird.", "Let's get icecream!")
(matches <- sapply(keywords, grepl, strings, ignore.case=TRUE))
# dog cat bird
# [1,] TRUE FALSE FALSE
# [2,] FALSE TRUE TRUE
# [3,] FALSE FALSE FALSE
To know which strings contain any of the keywords (patterns):
apply(matches, 1, any)
# [1] TRUE TRUE FALSE
To know which keywords (patterns) were matched in the supplied strings:
apply(matches, 2, any)
# dog cat bird
# TRUE TRUE TRUE
Related
R newbie here
I have data that looks something like this:
{'id': 19847005, 'profile_sidebar_fill_color': u'http://pbs.foo.com/profile_background', 'profile_text_color': u'333333', 'followers_count': 1105, 'location': u'San Diego, CA', 'profile_background_color': u'9AE4E8', 'listed_count': 43, '009', 'time_zone': u'Pacific Time (US & Canada)', 'protected': False}
I want to extract the location data from this text: San Diego, CA.
I have been trying to use this stringr package to accomplish this, but can't quite get the regex right to capture the city and state. Sometimes state will be present, other times not present.
location_pattern <- "'location':\su'(\w+)'"
rawdata$location <- str_extract(rawdata$user, location_pattern)
You could try
str_extract_all(str1, perl("(?<=location.: u.)[^']+(?=')"))[[1]]
#[1] "San Diego, CA"
Others have given possible solutions, but not explained what likely went wrong with your attempt.
The str_extract function uses POSIX extended regular expressions that do not understand \w and \s, those are specific to Perl regular expressions. You can use the perl function in the stringr package instead and it will then recognize the shortcuts, or you can use [[:space:]] in place of \s and [[:alnum:]_] in place of \w though more likely you will want something like [[:alpha], ] or [^'].
Also, R's string parser gets a shot at the string before it is passed to the matching function, therefore you will need \\s and \\w if you use the perl function (or other regular expressions function in R). the first \ escapes the second so that a single \ remains in the string to be interpreted as part of the regular expression.
It looks like a json string, but if you're not too concerned about that, then perhaps this would help.
library(stringi)
ss <- stri_split_regex(x, "[{}]|u?'|(, '(009')?)|: ", omit=TRUE)[[1]]
(m <- matrix(ss, ncol = 2, byrow = TRUE))
# [,1] [,2]
# [1,] "id" "19847005"
# [2,] "profile_sidebar_fill_color" "http://pbs.foo.com/profile_background"
# [3,] "profile_text_color" "333333"
# [4,] "followers_count" "1105"
# [5,] "location" "San Diego, CA"
# [6,] "profile_background_color" "9AE4E8"
# [7,] "listed_count" "43"
# [8,] "time_zone" "Pacific Time (US & Canada)"
# [9,] "protected" "False"
So now you have the ID names in the left column and the values on the right. It would probably be simple to reassemble the json from this point if need be.
Also, regarding the json-ness, we can coerce m to a data.frame (or leave it as a matrix), and then use jsonlite::toJSON
library(jsonlite)
json <- toJSON(setNames(as.data.frame(m), c("ID", "Value")))
fromJSON(json)
# ID Value
# 1 id 19847005
# 2 profile_sidebar_fill_color http://pbs.foo.com/profile_background
# 3 profile_text_color 333333
# 4 followers_count 1105
# 5 location San Diego, CA
# 6 profile_background_color 9AE4E8
# 7 listed_count 43
# 8 time_zone Pacific Time (US & Canada)
# 9 protected False
I have a problem when I tried to obtain a numeric part in R. The original strings, for example, is "buy 1000 shares of Google at 1100 GBP"
I need to extract the number of the shares (1000) and the price (1100) separately. Besides, I need to extract the number of the stock, which always appears after "shares of".
I know that sub and gsub can replace string, but what commands should I use to extract part of a string?
1) This extracts all numbers in order:
s <- "buy 1000 shares of Google at 1100 GBP"
library(gsubfn)
strapplyc(s, "[0-9.]+", simplify = as.numeric)
giving:
[1] 1000 1100
2) If the numbers can be in any order but if the number of shares is always followed by the word "shares" and the price is always followed by GBP then:
strapplyc(s, "(\\d+) shares", simplify = as.numeric) # 1000
strapplyc(s, "([0-9.]+) GBP", simplify = as.numeric) # 1100
The portion of the string matched by the part of the regular expression within parens is returned.
3) If the string is known to be of the form: X shares of Y at Z GBP then X, Y and Z can be extracted like this:
strapplyc(s, "(\\d+) shares of (.+) at ([0-9.]+) GBP", simplify = c)
ADDED Modified pattern to allow either digits or a dot. Also added (3) above and the following:
strapply(c(s, s), "[0-9.]+", as.numeric)
strapply(c(s, s), "[0-9.]+", as.numeric, simplify = rbind) # if ea has same no of matches
strapply(c(s, s), "(\\d+) shares", as.numeric, simplify = c)
strapply(c(s, s), "([0-9.]+) GBP", as.numeric, simplify = c)
strapplyc(c(s, s), "(\\d+) shares of (.+) at ([0-9.]+) GBP")
strapplyc(c(s, s), "(\\d+) shares of (.+) at ([0-9.]+) GBP", simplify = rbind)
You can use the sub function:
s <- "buy 1000 shares of Google at 1100 GBP"
# the number of shares
sub(".* (\\d+) shares.*", "\\1", s)
# [1] "1000"
# the stock
sub(".*shares of (\\w+) .*", "\\1", s)
# [1] "Google"
# the price
sub(".* at (\\d+) .*", "\\1", s)
# [1] "1100"
You can also use gregexpr and regmatches to extract all substrings at once:
regmatches(s, gregexpr("\\d+(?= shares)|(?<=shares of )\\w+|(?<= at )\\d+",
s, perl = TRUE))
# [[1]]
# [1] "1000" "Google" "1100"
I feel compelled to include the obligatory stringr solution as well.
library(stringr)
s <- "buy 1000 shares of Google at 1100 GBP"
str_match(s, "([0-9]+) shares")[2]
[1] "1000"
str_match(s, "([0-9]+) GBP")[2]
[1] "1100"
If you want to extract all digits from text use this function from stringi package.
"Nd" is the class of decimal digits.
stri_extract_all_charclass(c(123,43,"66ala123","kot"),"\\p{Nd}")
[[1]]
[1] "123"
[[2]]
[1] "43"
[[3]]
[1] "66" "123"
[[4]]
[1] NA
Please note that here 66 and 123 numbers are extracted separatly.
I have a large dataset where all column headers are individual IDS, each 8 characters in length. I would like to split those individual IDs into 2 rows, where the first row of IDs contains the first 7 characters, and the second row contains just the last character.
Current dataset:
ID1: Indiv01A Indiv01B Indiv02A Indiv02B Speci03A Speci03B
Intended dataset:
ID1: Indiv01 Indiv01 Indiv02 Indiv02 Speci03 Speci03
ID2: A B A B A B
I've looked through other posts on splitting data, but they all seem to have a unique way to separate the column name (ie: there's a comma separating the 2 components, or a period).
This is the code I'm thinking would work best, but I just can't figure out how to code for "7 characters" as the split point, rather than a comma:
sapply(strsplit(as.character(d$ID), ",")
Any help would be appreciated.
Here's a regular expression for a solution with strsplit. It splits the string between the 7th and the 8th character:
ID1 <- c("Indiv01A", "Indiv01B", "Indiv02A", "Indiv02B", "Speci03A", "Speci03B")
res <- strsplit(ID1, "(?<=.{7})", perl = TRUE)
# [[1]]
# [1] "Indiv01" "A"
#
# [[2]]
# [1] "Indiv01" "B"
#
# [[3]]
# [1] "Indiv02" "A"
#
# [[4]]
# [1] "Indiv02" "B"
#
# [[5]]
# [1] "Speci03" "A"
#
# [[6]]
# [1] "Speci03" "B"
Now, you can use rbind to create two columns:
do.call(rbind, res)
# [,1] [,2]
# [1,] "Indiv01" "A"
# [2,] "Indiv01" "B"
# [3,] "Indiv02" "A"
# [4,] "Indiv02" "B"
# [5,] "Speci03" "A"
# [6,] "Speci03" "B"
Explanation of the regex pattern:
(?<=.{7})
The (?<=) is a (positive) lookbehind. It matches any position that is preceded by the specified pattern. Here, the pattern is .{7}. The dot (.) matches any character. {7} means 7 times. Hence, the regex matches the position that is preceded by exactly 7 characters.
Here is a gsubfn solution:
library(gsubfn)
strapplyc(ID1, "(.*)(.)", simplify = cbind)
which gives this matrix:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] "Indiv01" "Indiv01" "Indiv02" "Indiv02" "Speci03" "Speci03"
[2,] "A" "B" "A" "B" "A" "B"
or use rbind in place of cbind if you want two columns (rather than two rows).
There are a couple of ways you could go about this.
To extract the final character
First, with substr:
new.vec <- sapply(old.vec, function(x) substr(x, nchar(x), nchar(x)))
or, with sub:
new.vec <- sub('.*(.)', '\\1', old.vec)
where old.vec is the vector of strings that you want to split.
For interest, the latter option uses a regular expression that translates to: "capture (indicating by surrounding with parentheses) the single character (.) that follows zero or more other characters (.*), and replace matches with the captured content (\\1)". For more info, see ?gsub, and here.
The above options allow for varying string lengths. However, if you do always want to split after 7 characters, and the second part of the string always has just a single character, then the following should work:
new.vec <- substr(old.vec, 8, 8)
(Edited to include method for extracting the first part of the string.)
To extract all but the final character
The process is similar.
new.vec <- sapply(old.vec, function(x) substr(x, 1, nchar(x) - 1))
new.vec <- sub('(.*).', '\\1', old.vec)
new.vec <- substr(old.vec, 1, 7)
I match and replace 4-digit numbers preceded and followed by white space with:
str12 <- "coihr 1234 &/()= jngm 34 ljd"
sub("\\s\\d{4}\\s", "", str12)
[1] "coihr&/()= jngm 34 ljd"
but, every try to invert this and extract the number instead fails.
I want:
[1] 1234
does someone has a clue?
ps: I know how to do it with {stringr} but am wondering if it's possible with {base} only..
require(stringr)
gsub("\\s", "", str_extract(str12, "\\s\\d{4}\\s"))
[1] "1234"
regmatches(), only available since R-2.14.0, allows you to "extract or replace matched substrings from match data obtained by regexpr, gregexpr or regexec"
Here are examples of how you could use regmatches() to extract either the first whitespace-cushioned 4-digit substring in your input character string, or all such substrings.
## Example strings and pattern
x <- "coihr 1234 &/()= jngm 34 ljd" # string with 1 matching substring
xx <- "coihr 1234 &/()= jngm 3444 6789 ljd" # string with >1 matching substring
pat <- "(?<=\\s)(\\d{4})(?=\\s)"
## Use regexpr() to extract *1st* matching substring
as.numeric(regmatches(x, regexpr(pat, x, perl=TRUE)))
# [1] 1234
as.numeric(regmatches(xx, regexpr(pat, xx, perl=TRUE)))
# [1] 1234
## Use gregexpr() to extract *all* matching substrings
as.numeric(regmatches(xx, gregexpr(pat, xx, perl=TRUE))[[1]])
# [1] 1234 3444 6789
(Note that this will return numeric(0) for character strings not containing a substring matching your criteria).
It's possible to capture group in regex using (). Taking the same example
str12 <- "coihr 1234 &/()= jngm 34 ljd"
gsub(".*\\s(\\d{4})\\s.*", "\\1", str12)
[1] "1234"
I'm pretty naive about regex in general, but here's an ugly way to do it in base:
# if it's always in the same spot as in your example
unlist(strsplit(str12, split = " "))[2]
# or if it can occur in various places
str13 <- unlist(strsplit(str12, split = " "))
str13[!is.na(as.integer(str13)) & nchar(str13) == 4] # issues warning
I have a bunch of names, and I want to obtain the unique names. However, due to spelling errors and inconsistencies in the data the names might be written down wrong. I am looking for a way to check in a vector of strings if two of them are similair.
For example:
pres <- c(" Obama, B.","Bush, G.W.","Obama, B.H.","Clinton, W.J.")
I want to find that " Obama, B." and "Obama, B.H." are very similar. Is there a way to do this?
This can be done based on eg the Levenshtein distance. There are multiple implementations of this in different packages. Some solutions and packages can be found in the answers of these questions:
agrep: only return best match(es)
In R, how do I replace a string that contains a certain pattern with another string?
Fast Levenshtein distance in R?
But most often agrep will do what you want :
> sapply(pres,agrep,pres)
$` Obama, B.`
[1] 1 3
$`Bush, G.W.`
[1] 2
$`Obama, B.H.`
[1] 1 3
$`Clinton, W.J.`
[1] 4
Maybe agrep is what you want? It searches for approximate matches using the Levenshtein edit distance.
lapply(pres, agrep, pres, value = TRUE)
[[1]]
[1] " Obama, B." "Obama, B.H."
[[2]]
[1] "Bush, G.W."
[[3]]
[1] " Obama, B." "Obama, B.H."
[[4]]
[1] "Clinton, W.J."
Add another duplicate to show it works with more than one duplicate.
pres <- c(" Obama, B.","Bush, G.W.","Obama, B.H.","Clinton, W.J.", "Bush, G.")
adist shows the string distance between 2 character vectors
adist(" Obama, B.", pres)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 9 3 10 7
For example, to select the closest string to " Obama, B." you can take the one which has the minimal distance. To avoid the identical string, I took only distances greater than zero:
d <- adist(" Obama, B.", pres)
pres[min(d[d>0])]
# [1] "Obama, B.H."
To obtain unique names, taking into account spelling errors and inconsistencies, you can compare each string to all previous ones. Then if there is a similar one, remove it. I created a keepunique() function that performs this. keepunique() is then applied to all elements of the vector successively with Reduce().
keepunique <- function(previousones, x){
if(any(adist(x, previousones)<5)){
x <- NULL
}
return(c(previousones, x))
}
Reduce(keepunique, pres)
# [1] " Obama, B." "Bush, G.W." "Clinton, W.J."