Using the XML package and XPath to scrape addresses from websites, I sometimes can get only a string that has embedded in it the zip code I want. It is straightforward to extract the zip code, but sometimes there are other five-digit strings that show up.
Here are some variations on the problem in a df.
zips <- data.frame(id = seq(1, 5), address = c("Company, 18540 Main Ave., City, ST 12345", "Company 18540 Main Ave. City ST 12345-0000", "Company 18540 Main Ave. City State 12345", "Company, 18540 Main Ave., City, ST 12345 USA", "Company, One Main Ave Suite 18540, City, ST 12345"))
The R statement to extract zip codes (both 5 digit and plus 4 digits) is below, but it is tricked by the faux zip codes of the street number and the suite number (and there may be other possibilities in other address strings).
regmatches(zips$address, gregexpr("\\d{5}([-]?\\d{4})?", zips$address, perl = TRUE))
An answer to a previous SO question suggested that a "regex will return the last consecutive five digit string. It uses a negative look-ahead to ensure the absence of 5-digit strings after the one being returned."
Extracting a zip code from an address string
\b\d{5}\b(?!.*\b\d{5}\b)
But that question and answer deals with PHP and offers an if loop with preg_matches()` I am not familiar with those languages and tools, but the idea might be right.
My question: what R code will find real zip codes and ignore false lookalikes?
This is my first regex answer (I am still learning) so hopefully I don't say anything wrong to lead you in the wrong direction.
Basically, this regex looks for, as you hinted in your question, the last string that looks like a zip code which is not followed by a string that looks like a zip code
the basic syntax is pattern(?!.*pattern) which says to match pattern only if it is not followed (a negative look-ahead assertion, syntax: (?! )) by anything .* and pattern
so we can replace pattern with what you are interested in finding:
[0-9]{5}(-[0-9]{4})?
that is, a digit string [0-9] of exactly 5 characters {5} (which may optionally be followed ? by another group defined as a hyphen and another digit string of length four (-[0-9]{4})
put it all together with gregexpr to search for the matches and regmatches to interpret the results for me, I get:
zips <- data.frame(id = seq(1, 5), address = c("Company, 18540 Main Ave., City, ST 12345", "Company 18540 Main Ave. City ST 12345-0000", "Company 18540 Main Ave. City State 12345", "Company, 18540 Main Ave., City, ST 12345 USA", "Company, One Main Ave Suite 18540, City, ST 12345"))
regmatches(zips$address,
gregexpr('[0-9]{5}(-[0-9]{4})?(?!.*[0-9]{5}(-[0-9]{4})?)', zips$address, perl = TRUE))
# [[1]]
# [1] "12345"
#
# [[2]]
# [1] "12345-0000"
#
# [[3]]
# [1] "12345"
#
# [[4]]
# [1] "12345"
#
# [[5]]
# [1] "12345"
The qdapRegex package has the rm_zip function for this:
zips <- data.frame(id = seq(1, 5),
address = c("Company, 18540 Main Ave., City, ST 12345",
"Company 18540 Main Ave. City ST 12345-0000",
"Company 18540 Main Ave. City State 12345",
"Company, 18540 Main Ave., City, ST 12345 USA",
"Company, One Main Ave Suite 18540, City, ST 12345")
)
lapply(rm_zip(zips$address, extract=TRUE), tail, 1)
## [[1]]
## [1] "12345"
##
## [[2]]
## [1] "12345-0000"
##
## [[3]]
## [1] "12345"
##
## [[4]]
## [1] "12345"
##
## [[5]]
## [1] "12345"
EDIT Per #lawyeR's comments:
I think that you want some regex that is more specific than the dictionary system used by qdapRegex. The current implementation of rm_zip allows for validation purposes and thus I wouldn't alter the regular expression it uses to be more flexible. I also wouldn't alter the function rm_zip to have additional parameters/arguments as qdapRegex attempts to have consistently operating functions.
That being said you could create your own function using the rm_ function and supply your own regular expression. I have done this using both of the parameters specified in your comment:
More complex data set:
zips <- data.frame(id = seq(1, 6),
address = c("Company, 18540 Main Ave., City, ST 12345",
"Company 18540 Main Ave. City ST 12345-0000",
"Company 18540 Main Ave. City State 12345",
"Company, 18540 Main Ave., City, ST 12345 USA",
"Company, One Main Ave Suite 18540m, City, ST 12345",
"company 12345678")
)
Function to grab even if a character follows the zip
## paste together a more flexible regular expression
pat <- pastex(
"#rm_zip",
"(?<!\\d)\\d{5}(?!\\d)",
"(?<!\\d)\\d{5}-\\d{4}(?!\\d)"
)
## Create your own function that extract is set to TRUE
rm_zip2 <- rm_(pattern=pat, extract=TRUE)
rm_zip2(zips$address)
## [[1]]
## [1] "18540" "12345"
##
## [[2]]
## [1] "18540" "12345-0000"
##
## [[3]]
## [1] "18540" "12345"
##
## [[4]]
## [1] "18540" "12345"
##
## [[5]]
## [1] "18540" "12345"
##
## [[6]]
## [1] NA
Function to extract just 5 digit zips
rm_zip3 <- rm_(pattern="(?<!\\d)\\d{5}(?!\\d)", extract=TRUE)
rm_zip3(zips$address)
## [[1]]
## [1] "18540" "12345"
##
## [[2]]
## [1] "18540" "12345"
##
## [[3]]
## [1] "18540" "12345"
##
## [[4]]
## [1] "18540" "12345"
##
## [[5]]
## [1] "18540" "12345"
##
## [[6]]
## [1] NA
Related
I can capture repeated words pretty easily using:
"(?i)\\b(\\w+)(((\\.{3}\\s*|,\\s+)*|\\s+)\\1)+\\b" but this regex does not seem to extend to mutipe words (and why should it in its current state). How could I find repeated phrases using regex?
Here I extract repeated terms (regardless of case) but the same regex doesn't word to extract a repeated phrase:
library(qdapRegex)
rm_default("this is a big Big deal", pattern = "(?i)\\b(\\w+)(((\\.{3}\\s*|,\\s+)*|\\s+)\\1)+\\b", extract=TRUE)
rm_default("this is a big is a Big deal", pattern = "(?i)\\b(\\w+)(((\\.{3}\\s*|,\\s+)*|\\s+)\\1)+\\b", extract=TRUE)
I would hope for a regex that would return:
"is a big is a Big"
for:
x <- "this is a big is a Big deal"
To cover corner cases here's a larger desired test and output...
"this is a big is a Big deal",
"I want want to see",
"I want, want to see",
"I want...want to see see how",
"this is a big is a Big deal for those of, those of you who are.",
"I like it. It is cool",
)
[[1]]
[1] "is a big is a Big"
[[2]]
[1] "want want"
[[3]]
[1] "want, want"
[[4]]
[1] "want...want" "see see"
[[5]]
[1] "is a big is a Big" "those of, those of"
[[6]]
[1] NA
My current regex only gets me to:
rm_default(y, pattern = "(?i)\\b(\\w+)(((\\.{3}\\s*|,\\s+)*|\\s+)\\1)+\\b", extract=TRUE)
## [[1]]
## [1] NA
##
## [[2]]
## [1] "want want"
##
## [[3]]
## [1] "want, want"
##
## [[4]]
## [1] "want...want" "see see"
##
## [[5]]
## [1] NA
I think this does what you want (note we only allow a single space, ..., or , as separators, but you should be able to tweak that easily):
pattern <- "(?i)\\b(\\w.*)((?:\\s|\\.{3}|,)+\\1)+\\b"
rm_default(x, pattern = pattern, extract=TRUE)
Produces:
[[1]]
[1] "is a big is a Big"
[[2]]
[1] "want want"
[[3]]
[1] "want, want"
[[4]]
[1] "want...want" "see see"
[[5]]
[1] "is a big is a Big" "those of, those of"
Try this:
> regmatches(x, gregexpr("(?i)\\b(\\S.*\\S)[ ,.]*\\b(\\1)", x, perl = TRUE))
[[1]]
[1] "is a big is a Big"
[[2]]
[1] "want want"
[[3]]
[1] "want, want"
[[4]]
[1] "want...want" "see see"
[[5]]
[1] "is a big is a Big" "those of, those of"
Here is a visualization (except there is an error in the visualization - the \S parts should be within the group.
(?i)\b(\S.*\S)[ ,.]*\b(\1)
Debuggex Demo
You might want to replace [ ,.] with [ [:punct:]]. I did not do that since debuggex does not support POSIX character groups.
I have a string vector that looks like this and I'd like to split it up:
str <- c("Fruit LoopsJalapeno Sandwich", "Red Bagel", "Basil LeafBarbeque SauceFried Beef")
str_split(str, '[a-z][A-Z]', n = 3)
[[1]]
[1] "Fruit Loop" "alapeno Sandwich"
[[2]]
[1] "Red Bagel"
[[3]]
[1] "Basil Lea" "arbeque Sauc" "ried Beef"
But I need to keep those letters at the end and start of the words.
Here's 2 approaches in base (you can generalize to stringr if you want).
This one subs out this place with a placeholder and then splits on that.
strsplit(gsub("([a-z])([A-Z])", "\\1SPLITHERE\\2", str), "SPLITHERE")
## [[1]]
## [1] "Fruit Loops" "Jalapeno Sandwich"
##
## [[2]]
## [1] "Red Bagel"
##
## [[3]]
## [1] "Basil Leaf" "Barbeque Sauce" "Fried Beef"
This method uses lookaheads and lookbehinds:
strsplit(str, "(?<=[a-z])(?=[A-Z])", perl=TRUE)
## [[1]]
## [1] "Fruit Loops" "Jalapeno Sandwich"
##
## [[2]]
## [1] "Red Bagel"
##
## [[3]]
## [1] "Basil Leaf" "Barbeque Sauce" "Fried Beef"
EDIT Generalized to stringr so you can grab 3 pieces if you want
stringr::str_split(gsub("([a-z])([A-Z])", "\\1SPLITHERE\\2", str), "SPLITHERE", 3)
You could also match instead of splitting based off your string.
unlist(regmatches(str, gregexpr('[A-Z][a-z]+ [A-Z][a-z]+', str)))
# [1] "Fruit Loops" "Jalapeno Sandwich" "Red Bagel"
# [4] "Basil Leaf" "Barbeque Sauce" "Fried Beef"
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.
Need to create an text sparce matrix (DTM) for classification. To prepare the text, first I need to eliminate (separate) the POS-tags the text. My guess was to do it like below. I'm new to R and don't now how to negate a REGEX (see below NOT!).
text <- c("wenn/KOUS ausläuft/VVFIN ./$.", "Kommt/VVFIN vor/PTKVZ ;/$.", "-RRB-/TRUNC Durch/APPR und/KON", "man/PIS zügig/ADJD ./$.", "empfehlung/NN !!!/NE")
My guess how it could work:
(POSs <- regmatches(text, gregexpr('[[:punct:]]*/[[:alpha:][:punct:]]*', text)))
[[1]]
[1] "/KOUS" "/VVFIN" "./$."
[[2]]
[1] "/VVFIN" "/PTKVZ" ";/$."
[[3]]
[1] "-/TRUNC" "/APPR" "/KON"
[[4]]
[1] "/PIS" "/ADJD" "./$."
[[5]]
[1] "/NN" "!!!/NE"
But don't konw how to negate the expression like:
# VVV
(texts <- regmatches(text, NOT!(gregexpr('[[:punct:]]*/[[:alpha:][:punct:]]*', text))))
[[1]]
[1] "wenn" "ausläuft"
[[2]]
[1] "Kommt" "vor"
[[3]]
[1] "Durch" "und"
[[4]]
[1] "man" "zügig"
[[5]]
[1] "empfehlung"
One possibility is to eliminate the tags by, searching for POS-tags and replacing them with '' (i.e. empty text):
text <- c("wenn/KOUS ausläuft/VVFIN ./$.", "Kommt/VVFIN vor/PTKVZ ;/$.", "-RRB-/TRUNC Durch/APPR und/KON", "man/PIS zügig/ADJD ./$.", "empfehlung/NN !!!/NE")
(textlist <- strsplit(paste(gsub('[[:punct:]]*/[[:alpha:][:punct:]]*','', text), sep=' '), " "))
[[1]]
[1] "wenn" "ausläuft"
[[2]]
[1] "Kommt" "vor"
[[3]]
[1] "-RRB" "Durch" "und"
[[4]]
[1] "man" "zügig"
[[5]]
[1] "empfehlung"
With the friendly help of rawr
How can I extract phone numbers from a text file?
x <- c(" Mr. Bean bought 2 tickets 2-613-213-4567 or 5555555555 call either one",
"43 Butter Rd, Brossard QC K0A 3P0 – 613 213 4567",
"Please contact Mr. Bean (613)2134567",
"1.575.555.5555 is his #1 number",
"7164347566"
)
This is a question that's been answered for other languages (see php abd general regex) but doesn't seem to have been tackled on SO for R.
I have searched and found what appears to be possible regexes to find phone numbers (In addition to the regexes from other languages above): http://regexlib.com/Search.aspx?k=phone but have not been able to use gsub within R with these to extract all of these numbers in the example.
Ideally, we'd get something like:
[[1]]
[1] "2-613-213-4567" "5555555555"
[[2]]
[1] "613 213 4567"
[[3]]
[1] "(613)2134567"
[[4]]
[1] "1.575.555.5555"
[[5]]
[1] "7164347566"
This is the best I've been able to do- you have a pretty wide range of formats, including some with spaces, so the regex is pretty general. It just says "look for a string of at least 5 characters made up entirely of digits, periods, brackets, hyphens or spaces":
library(stringr)
str_extract_all(x, "(^| )[0-9.() -]{5,}( |$)")
Output:
[[1]]
[1] " 2-613-213-4567 " " 5555555555 "
[[2]]
[1] " 613 213 4567"
[[3]]
[1] " (613)2134567"
[[4]]
[1] "1.575.555.5555 "
[[5]]
[1] "7164347566"
The leading/trailing spaces could probably be fixed with some additional complexity, or you could just fix it in post.
Update: a bit of searching lead me to this answer, which I slightly modified to allow periods. A bit stricter in terms of requiring a valid (US?) phone number, but seems to cover all your examples:
str_extract_all(x, "\\(?\\d{3}\\)?[.-]? *\\d{3}[.-]? *[.-]?\\d{4}")
Output:
[[1]]
[1] "613-213-4567" "5555555555"
[[2]]
[1] "613 213 4567"
[[3]]
[1] "(613)2134567"
[[4]]
[1] "575.555.5555"
[[5]]
[1] "7164347566"
The monstrosity found here also works once you take out the ^ and $ at either end. Use only if you really need it:
huge_regex = "(?:(?:\\+?1\\s*(?:[.-]\\s*)?)?(?:\\(\\s*([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9])\\s*\\)|([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9]))\\s*(?:[.-]\\s*)?)?([2-9]1[02-9]|[2-9][02-9]1|[2-9][02-9]{2})\\s*(?:[.-]\\s*)?([0-9]{4})(?:\\s*(?:#|x\\.?|ext\\.?|extension)\\s*(\\d+))?"
The qdapRegex now has the rm_phone specifically designed for this task:
x <- c(" Mr. Bean bought 2 tickets 2-613-213-4567 or 5555555555 call either one",
"43 Butter Rd, Brossard QC K0A 3P0 – 613 213 4567",
"Please contact Mr. Bean (613)2134567",
"1.575.555.5555 is his #1 number",
"7164347566"
)
library(qdapRegex)
ex_phone(x)
## [[1]]
## [1] "613-213-4567" "5555555555"
##
## [[2]]
## [1] "613 213 4567"
##
## [[3]]
## [1] "(613)2134567"
##
## [[4]]
## [1] "1.575.555.5555"
##
## [[5]]
## [1] "7164347566"
You would need a complex regex to cover all rules for matching phone numbers, but to cover your examples.
> library(stringi)
> unlist(stri_extract_all_regex(x, '(\\d[.-])?\\(?\\d{3}\\)?[-. ]?\\d{3}[-. ]?\\d{4}\\b'))
# [1] "2-613-213-4567" "5555555555" "613 213 4567" "(613)2134567"
# [5] "1.575.555.5555" "7164347566"