I have a small problem.
I have text that looks like:
B.1 My name is John
I want to only obtain:
My name is John
I'm having difficulty leaving out both the B and the 1, at the same time
You can do this with sub and a regular expression.
TestStrings = c("B.1 My name is John", "A.12 This is another sentence")
sub("\\b[A-Z]\\.\\d+\\s+", "", TestStrings)
[1] "My name is John" "This is another sentence"
The \\b indicates a word boundary (to eliminate multiple letters)
[A-Z] will match a single capital letter.
\\. will match a period
\\d+ will match one or more digits
\\s+ will match any training blank space.
The part that is matched will be replaced with the empty string.
If you are sure that all the strings that you need have the same (or similar) initial part you can do
> a<-"B.1 My name is John"
> substr(a, 5, nchar(a))
[1] "My name is John"
Related
I want to remove all words that start with "a" in a string.
Input:
string <- "This is a sentence about nothing."
My attempt:
stringr::str_remove_all(string,"a*\\b")
output I got:
[1] "This is sentence about nothing."
output I want:
[1] "This is sentence nothing."
I am not sure how to detect based on one letter but perform action(e.g., remove, replace) on the whole word. Any input is appreciated!
The a*\b pattern matches zero or more a chars followed with end of string or a word char. It does not match a word unless it is an a word.
You can use
stringr::str_remove_all(string,"\\ba\\w*")
stringr::str_replace_all(string,"\\ba\\w*", "")
gsub("\\ba\\w*", "", string, perl=TRUE) ## ASCII only letters/digits
where \ba\w* matches a word boundary, a, and then zero or more word chars.
If you also want to remove any whitespaces before the word, add \s* at the start:
stringr::str_remove_all(string,"\\s*\\ba\\w*")
stringr::str_replace_all(string,"\\s*\\ba\\w*", "")
gsub("\\s*\\ba\\w*", "", string, perl=TRUE) ## ASCII only letters/digits/whitespaces
If you need to make sure you only remove natural langugage words consisting only of letters, then you can replace \w with \p{L}:
stringr::str_remove_all(string,"\\s*\\ba\\p{L}*")
stringr::str_replace_all(string,"\\s*\\ba\\p{L}*", "")
gsub("(*UCP)\\s*\\ba\\p{L}*", "", string, perl=TRUE) ## any Uncicode letters/digits/whitespaces
I want to extract a name field from a text eg
name = "My name is John Smith"
should return John Smith
My current code is
grep(".^[A-Z][a-z]+\\s[A-Z][a-z]+", name, value = TRUE)
We can use sub to capture the words that start with uppercase, followed by lower case, then a space followed by the word with upper case, lower case letters of the string followed by other characters (.*) and replace with the backreference (\\1) of the captured group
sub(".*([A-Z][a-z]+\\s[A-Z][a-z]+).*", "\\1", name)
#[1] "John Smith"
edit: added #DJack's recommendation
data
name <- c("My name is John Smith")
I have a string that contains a persons name and city. It's formatted like this:
mock <- "Joe Smith (Cleveland, OH)"
I simply want the state abbreviation remaining, so it in this case, the only remaining string would be "OH"
I can get rid of the the parentheses and comma
[(.*?),]
Which gives me:
"Joe Smith Cleveland OH"
But I can't figure out how to combine all of it. For the record, all of the records will look like that, where it ends with ", two letter capital state abbreviation" (ex: ", OH", ", KY", ", MD" etc...)
You may use
mock <- "Joe Smith (Cleveland, OH)"
sub(".+,\\s*([A-Z]{2})\\)$","\\1",mock)
## => [1] "OH"
## With stringr:
str_extract(mock, "[A-Z]{2}(?=\\)$)")
See this R demo
Details
.+,\\s*([A-Z]{2})\\)$ - matches any 1+ chars as many as possible, then ,, 0+ whitespaces, and then captures 2 uppercase ASCII letters into Group 1 (referred to with \1 from the replacement pattern) and then matches ) at the end of string
[A-Z]{2}(?=\)$) - matches 2 uppercase ASCII letters if followed with the ) at the end of the string.
How about this. If they are all formatted the same, then this should work.
mock <- "Joe Smith (Cleveland, OH)"
substr(mock, (nchar(mock) - 2), (nchar(mock) - 1))
If the general case is that the state is in the second and third last characters then match everything, .*, and then a capture group of two characters (..) and then another character . and replace that with the capture group:
sub(".*(..).", "\\1", mock)
## [1] "OH"
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"
>