I'm dealing with a regular expression in which I has string that has a series of numbers four numbers then name which repeat for multiples.
The text pattern is a series of 4 numbers, then a string. I would like to extract the string after the four numbers. The four numbers must appear before the string. In the example below, I do not want to extract "Not this one", but would like the strings after four numbers.
string_to_inspect <-"Not This One 4586 This one 8888 Another one 8955 PS109 8566 Last One"
My ideal extraction is a character vector that looks like:
"This one" "Another one" "PS109" "Last One"
I have tried
str_extract_all(pattern = "[0-9]{4}(.*?)", string = string_to_inspect)
And it returns a single string that include all the numbers
"4586 This one 8888 Another one 8955 PS109 8566 Last One"
I have tried various combinations but I know I must be missing something critical.
We can split the string by four digits, remove the first one, and then trim the white space.
library(stringr)
str_trim(str_split(string_to_inspect, pattern = "\\s[0-9]{4}\\s")[[1]][-1])
# [1] "This one" "Another one" "PS109" "Last One"
strsplit(string_to_inspect, " [0-9]+ ")
In case you don't want problems with strings mixed with numbers:
string_to_inspect <-"Not This One 4586 This one 8888 Another one 8955 PS109 8566 Last One"
str2insp <- strsplit(string_to_inspect, ' ')[[1]]
str2insp[!gsub('[[:digit:]]', '', str2insp) == '']
outputs:
#[1] "Not" "This" "One" "This" "one" "Another" "one" "PS109" "Last" "One"
Related
I need to extract first 2 words from a string. If the string contains more than 2 words, it should return the first 2 words else if the string contains less than 2 words it should return the string as it is.
I've tried using 'word' function from stringr package but it's not giving the desired output for cases where len(string) < 2.
word(dt$var_containing_strings, 1,2, sep=" ")
Example:
Input String: Auto Loan (Personal)
Output: Auto Loan
Input String: Others
Output: Others
If you want to use stringr::word(), you can do:
ifelse(is.na(word(x, 1, 2)), x, word(x, 1, 2))
[1] "Auto Loan" "Others"
Sample data:
x <- c("Auto Loan (Personal)", "Others")
Something like this?
a <- "this is a character string"
unlist(strsplit(a, " "))[1:2]
[1] "this" "is"
EDIT:
To add the part where original string is returned if number of worlds is less than 2, a simple if-else function can be used:
a <- "this is a character string"
words <- unlist(strsplit(a, " "))
if (length(words) > 2) {
words[1:2]
} else {
a
}
You could use regex in base R using sub
sub("(\\w+\\s+\\w+).*", "\\1", "Auto Loan (Personal)")
#[1] "Auto Loan"
which will also work if you have only one word in the text
sub("(\\w+\\s+\\w+).*", "\\1", "Auto")
#[1] "Auto"
Explanation :
Here we extract the pattern shown inside round brackets which is (\\w+\\s+\\w+) which means :
\\w+ One word followed by \\s+ whitespace followed by \\w+ another word, so in total we extract two words. Extraction is done using backreference \\1 in sub.
(In R) How to split words by title case in a string like "WeLiveInCA" into "We Live In CA" without splitting abbreviations?
I know how to split the string at every uppercase letter, but doing that would split initialisms/abbreviations, like CA or USSR or even U.S.A. and I need to preserve those.
So I'm thinking some type of logical like if a word in a string isn't an initialism then split the word with a space where a lowercase character is followed by an uppercase character.
My snippet of code below splits words with spaces by capital letters, but it breaks initialisms like CA becomes C A undesirably.
s <- "WeLiveInCA"
trimws(gsub('([[:upper:]])', ' \\1', s))
# "We Live In C A"
or another example...
s <- c("IDon'tEatKittensFYI", "YouKnowYourABCs")
trimws(gsub('([[:upper:]])', ' \\1', s))
# "I Don't Eat Kittens F Y I" "You Know Your A B Cs"
The results I'd want would be:
"We Live In CA"
#
"I Don't Eat Kittens FYI" "You Know Your ABCs"
But this needs to be widely applicable (not just for my example)
Try with base R gregexpr/regmatches.
s <- c("WeLiveInCA", "IDon'tEatKittensFYI", "YouKnowYourABCs")
regmatches(s, gregexpr('[[:upper:]]+[^[:upper:]]*', s))
#[[1]]
#[1] "We" "Live" "In" "CA"
#
#[[2]]
#[1] "IDon't" "Eat" "Kittens" "FYI"
#
#[[3]]
#[1] "You" "Know" "Your" "ABCs"
Explanation.
[[:upper:]]+ matches one or more upper case letters;
[^[:upper:]]* matches zero or more occurrences of anything but upper case letters.
In sequence these two regular expressions match words starting with upper case letter(s) followed by something else.
I'm trying to split a vector of strings into two pieces (I only want to keep the first bit) based on the following criteria:
it should split after a full word (i.e. where a space occurs)
it should cut at the space closest to the 12th character
Example:
textvec <- c("this is an example", "I hope someone can help me", "Thank you in advance")
Expected result is a vector like this:
"this is an" , "I hope someone", "Thank you in"
What I tried so far:
I'm able to get the full words that occur before or at the 12th character like this:
t13 <- substr(textvec , 1, 13) #gives me first 13 characters of each string
lastspace <- lapply(gregexpr(" ", t13), FUN=function(x) x[length(x)]) #gives me last space before/at 13th character
result <- substr(t13, start=1, stop=lastspace)
But what I want is to get include the word closest to the 12th character (e.g. "someone" in the example above), not necessarily before or at the 12th character. In case there's a tie, I would like to include the word after the 12th character. I hope I'm explaining myself clearly :)
Using cumsum,
sapply(strsplit(textvec, ' '), function(i) paste(i[cumsum(nchar(i)) <= 12], collapse = ' '))
#[1] "this is an" "I hope someone" "Thank you in"
We can use gregexpr to find the closest space at 12 and then with substr cut the string
substr(textvec, 1, sapply(gregexpr("\\s+", textvec),
function(x) x[which.min(abs(12 - x))])-1)
#[1] "this is an" "I hope someone" "Thank you in"
I have a vector of strings and i want to separate the last sentence from each string in R.
Sentences may end with full stops(.) or even exclamatory marks(!). Hence i am confused as to how to separate the last sentence from a string in R.
You can use strsplit to get the last sentence from each string as shown:-
## paragraph <- "Your vector here"
result <- strsplit(paragraph, "\\.|\\!|\\?")
last.sentences <- sapply(result, function(x) {
trimws((x[length(x)]))
})
Provided that your input is clean enough (in particular, that there are spaces between the sentences), you can use:
sub(".*(\\.|\\?|\\!) ", "", trimws(yourvector))
It finds the longest substring ending with a punctuation mark and a space and removes it.
I added trimws just in case there are trailing spaces in some of your strings.
Example:
u <- c("This is a sentence. And another sentence!",
"By default R regexes are greedy. So only the last sentence is kept. You see ? ",
"Single sentences are not a problem.",
"What if there are no spaces between sentences?It won't work.",
"You know what? Multiple marks don't break my solution!!",
"But if they are separated by spaces, they do ! ! !")
sub(".*(\\.|\\?|\\!) ", "", trimws(u))
# [1] "And another sentence!"
# [2] "You see ?"
# [3] "Single sentences are not a problem."
# [4] "What if there are no spaces between sentences?It won't work."
# [5] "Multiple marks don't break my solution!!"
# [6] "!"
This regex anchors to the end of the string with $, allows an optional '.' or '!' at the end. At the front it finds the closest ". " or "! " as the end of the prior sentence. The negative lookback ?<= ensures the "." or '!' are not matched. Also provides for a single sentence by using ^ for the beginning.
s <- "Sentences may end with full stops(.) or even exclamatory marks(!). Hence i am confused as to how to separate the last sentence from a string in R."
library (stringr)
str_extract(s, "(?<=(\\.\\s|\\!\\s|^)).+(\\.|\\!)?$")
yields
# [1] "Hence i am confused as to how to separate the last sentence from a string in R."
I cannot fully understand why my regular expression does not work to extract the info I want. I have an unlisted vector that looks like this:
text <- c("Senator, 1.4balbal", "rule 46.1, declares",
"Town, 24", "A Town with a Long Name, 23", "THIS IS A DOCUMENT,23)
I would like to create a regular expression to extract only the name of the "Town", even if the town has a long name as the one written in the vector ("A Town with a Long Name"). I have tried this to extract the name of the town:
reg.town <- "[[:alpha:]](.+?)+,(.+?)\\d{2}"
towns<- unlist(str_extract_all(example, reg.prov))
but I extract everything around the ",".
Thanks in advance,
It looks like a town name starts with a capital letter ([[:upper:]]), ends with a comma (or continues to the end of text if there is no comma) ([^,]+) and should be at the start of the input text (^). The corresponding regex in this case would be:
^[[:upper:]][^,]+
Demo: https://regex101.com/r/QXYtyv/1
I have solve the problem thanks to #Dmitry Egorov 's demo post in the comment. the regular expression is this one ([[:upper:]].+?, [[:digit:]])
Thanks for your quick replies!!
You may use the following regex:
> library(stringr)
> text <- c("Senator, 1.4balbal", "rule 46.1, declares", "Town, 24", "A Town with a Long Name, 23", "THIS IS A DOCUMENT,23")
> towns <- unlist(str_extract_all(text, "\\b\\p{Lu}[^,]++(?=, \\d)"))
> towns
[1] "Senator" "Town"
[3] "A Town with a Long Name"
The regex matches:
\\b - a leading word boundary
\\p{Lu} - an uppercase letter
[^,]++ - 1+ chars other than a , (possessively, due to ++ quantifier, with no backtracking into this pattern for a more efficient matching)
(?=, \\d) - a positive lookahead that requires a ,, then a space and then any digit to appear immediately after the last non-, symbol matched with [^,]++.
Note you may get the same results with base R using the same regex with a PCRE option enabled:
> towns_baseR <- unlist(regmatches(text, gregexpr("\\b\\p{Lu}[^,]++(?=, \\d)", text, perl=TRUE)))
> towns_baseR
[1] "Senator" "Town"
[3] "A Town with a Long Name"
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