string<-c("Posted 69 months ago (7/4/2011)")
library(gsubfn)
strapplyc(string, "(.*)", simplify = TRUE)
I apply above function but nothing happens.
In this I want to extract only date part i.e 7/4/2011.
The first one shows how to fix the code in the question to give the desired answer. The next 2 solutions are the same except they use different regular expressions. The fourth solution shows how to do it with gsub. The fifth breaks the gsub into two sub calls and the sixth uses read.table.
1) Escape parens The problem is that ( and ) have special meaning in regular expressions so you must escape them if you want to match them literally. By using "[(]" as we do below (or writing them as "\\(" ) they are matched literally. The inner parentheses define the capture group as we don't want that group to include the literal parentheses themselves:
strapplyc(string, "[(](.*)[)]", simplify = TRUE)
## [1] "7/4/2011"
2) Match content Another way to do it is to match the data itself rather than the surrounding parentheses. Here "\\d+" matches one or more digits:
strapplyc(string, "\\d+/\\d+/\\d+", simplify = TRUE)
## [1] "7/4/2011"
You could specify the number of digits if you want to be even more specific but it seems unnecessary here if the data looks similar to that in the question.
3) Match 8 or more digits and slashes Given that there are no other sequences of 8 or more characters consisting only of slashes and digits in the rest of the string we could just pick out that:
strapplyc(string, "[0-9/]{8,}", simplify = TRUE)
## [1] "7/4/2011"
4) Remove text before and after Another way of doing it is to remove everything up to the ( and after the ) like this:
gsub(".*[(]|[)].*", "", string)
## [1] "7/4/2011"
5) sub This is the same as (4) except it breaks the gsub into two sub invocations, one removing everything up to ( and the other removing ) onwards. The regular expressions are therefore slightly simpler.
sub(".*\\(", "", sub("\\).*", "", string))
6) read.table This solution uses no regular expressions at all. It defines sep and comment.char in read.table so that the second column of the result of read.table is the required date or dates.
read.table(text = string, sep = "(", comment.char = ")", as.is = TRUE)$V2
## [1] "7/4/2011"
Note: Note that you don't need the c in defining string
string <- c("Posted 69 months ago (7/4/2011)")
string2 <- "Posted 69 months ago (7/4/2011)"
identical(string, string2)
## [1] TRUE
We can do this with gsub by matching one or more characters that are not a ( ([^(]+) from the start (^) of the string or | the ) at the end ($) of the string and replace it with ""
gsub("[^[^(]+\\(|\\)$", "", string)
#[1] "7/4/2011"
Or using capture groups
sub("^[^(]+\\(([^)]+).*", "\\1", string)
#[1] "7/4/2011"
Or with str_extract, we match one or more characters that are not a ) ([^)]+) that follows the ( ((?<=[(]))
library(stringr)
str_extract(string, "(?<=[(])[^)]+")
#[1] "7/4/2011"
Related
I have a text string containing digits, letters and spaces. Some of its substrings are month abbreviations. I want to perform a condition-based pattern replacement, namely to enclose a month abbreviation in whitespaces if and only if a given condition is fulfilled. As an example, let the condition be as follows: "preceeded by a digit and succeeded by a letter".
I tried stringr package but I fail to combine the functions str_replace_all() and str_locate_all():
# Input:
txt = "START1SEP2 1DECX JANEND"
# Desired output:
# "START1SEP2 1 DEC X JANEND"
# (A) What I could do without checking the condition:
library(stringr)
patt_month = paste("(", paste(toupper(month.abb), collapse = "|"), ")", sep='')
str_replace_all(string = txt, pattern = patt_month, replacement = " \\1 ")
# "START1 SEP 2 1 DEC X JAN END"
# (B) But I actually only need replacements inside the condition-based bounds:
str_locate_all(string = txt, pattern = paste("[0-9]", patt_month, "[A-Z]", sep=''))[[1]]
# start end
# [1,] 12 16
# To combine (A) and (B), I'm currently using an ugly for() loop not shown here and want to get rid of it
You are looking for lookarounds:
(?<=\d)DEC(?=[A-Z])
See a demo on regex101.com.
Lookarounds make sure a certain position is matched without consuming any characters. They are available in front of sth. (called lookbehind) or to make sure anything that follows is of a certain type (called lookahead). You have positive and negative ones on both sides, thus you have four types (pos./neg. lookbehind/-ahead).
A short memo:
(?=...) is a pos. lookahead
(?!...) is a neg. lookahead
(?<=...) is a pos. lookbehind
(?<!...) is a neg. lookbehind
A Base R version
patt_month <- capture.output(cat(toupper(month.abb),"|"))#concatenate all month.abb with OR
pat <- paste0("(\\s\\d)(", patt_month, ")([A-Z]\\s)")#make it a three group thing
gsub(pattern = pat, replacement = "\\1 \\2 \\3", txt, perl =TRUE)#same result as above
Also works for txt2 <- "START1SEP2 1JANY JANEND" out of the box.
[1] "START1SEP2 1 JAN Y JANEND"
I need to remove the text before the leading period (as well as the leading period) and the text following the last period from a string.
Given this string for example:
"ABCD.EF.GH.IJKL.MN"
I'd like to get the output:
[1] "IJKL"
I have tried the following:
split_string <- sub("^.*?\\.","", string)
split_string <- sub("^\\.+|\\.[^.]*$", "", string)
I believe I have it working for the period and text after for that string output I want. However, the first line needs to be executed multiple times to remove the text before that period in question e.g. '.I'.
One option in base R is to capture as a group ((...)) the word followed by the dot (\\.) and the word (\\w+) till the end ($) of the string. In the replacement, use the backreference (\\1) of the captured word
sub(".*\\.(\\w+)\\.\\w+$", "\\1", str1)
#[1] "IJKL"
Here, we match characters (.*) till the . (\\. - escaped to get the literal value because . is a metacharacter that will match any character if not escaped), followed by the word captured ((\\w+)), followed by a dot and another word at the end ($)of the string. The replacement part is mentioned above
Or another option is regmatches/regexpr from base R
regmatches(str1, regexpr("\\w+(?=\\.\\w+$)", str1, perl = TRUE))
#[1] "IJKL"
Or another option is word from stringr
library(stringr)
word(str1, -2, sep="[.]")
#[1] "IJKL"
data
str1 <- "ABCD.EF.GH.IJKL.MN"
Here is a janky dplyr version in case the other values are of importance and you want to select them later on, just include them in the "select".
df<- data.frame(x=c("ABCD.EF.GH.IJKL.MN"))
df2<-df %>%
separate(x, into=c("var1", "var2","var3","var4","var5")) %>%
select("var4")
Split into groups at period and take the second one from last.
sapply(strsplit(str1, "\\."), function(x) x[length(x) - 1])
#[1] "IJKL"
Get indices of the periods and use substr to extract the relevant portion
sapply(str1, function(x){
ind = gregexpr("\\.", x)[[1]]
substr(x, ind[length(ind) - 1] + 1, ind[length(ind)] - 1)
}, USE.NAMES = FALSE)
#[1] "IJKL"
These alternatives all use no packages or regular expressions.
1) basename/dirname Assuming the test input s shown in the Note at the end convert the dots to slashes and then use dirname and basename.
basename(dirname(chartr(".", "/", s)))
## [1] "IJKL" "IJKL"
2) strsplit Using strsplit split the strings at dot creating a list of character vectors, one vector per input string, and then for each such vector take the last 2 elements using tail and the first of those using indexing.
sapply(strsplit(s, ".", fixed = TRUE), function(x) tail(x, 2)[1])
## [1] "IJKL" "IJKL"
3) read.table It is not clear from the question what the general case is but if all the components of s have the same number of dot separated fields then we can use read.table to create a data.frame with one row per input string and one column per dot-separated component. Then take the column just before the last.
dd <- read.table(text = s, sep = ".", as.is = TRUE)
dd[[ncol(dd)-1]]
## [1] "IJKL" "IJKL"
4) substr Again, the general case is not clear but if the string of interest is always at character positions 12-15 then a simple solution is:
substr(s, 12, 15)
## [1] "IJKL" "IJKL"
Note
s <- c("ABCD.EF.GH.IJKL.MN", "ABCD.EF.GH.IJKL.MN")
I have a string which looks like this:
something-------another--thing
I want to replace the multiple dashes with a single one.
So the expected output would be:
something-another-thing
We can try using sub here:
x <- "something-------another--thing"
gsub("-{2,}", "-", x)
[1] "something-another-thing"
More generally, if we want to replace any sequence of two or more of the same character with just the single character, then use this version:
x <- "something-------another--thing"
gsub("(.)\\1+", "\\1", x)
The second pattern could use an explanation:
(.) match AND capture any single letter
\\1+ then match the same letter, at least one or possibly more times
Then, we replace with just the single captured letter.
you can do it with gsub and using regex.
> text='something-------another--thing'
> gsub('-{2,}','-',text)
[1] "something-another-thing"
t2 <- "something-------another--thing"
library(stringr)
str_replace_all(t2, pattern = "-+", replacement = "-")
which gives:
[1] "something-another-thing"
If you're searching for the right regex to search for a string, you can test it out here https://regexr.com/
In the above, you're just searching for a pattern that is a hyphen, so pattern = "-", but we add the plus so that the search is 'greedy' and can include many hyphens, so we get pattern = "-+"
I would like to insert a colon every five characters starting from the end of the string, preferably using regex and gsub in R.
text <- "My Very Enthusiastic Mother Just Served Us Noodles!"
I have been able to insert a colon every five characters from beginning of the text using:
gsub('(.{5})', "\\1:", text, perl = T)
I have written an inelegant function for achieving this as follows:
library(dplyr)
str_reverse<-function(x){
strsplit(x,split='')[[1]] %>% rev() %>% paste(collapse = "")
}
text2<-str_reverse(text)
text3<-gsub('(.{5})', "\\1:", text2, perl = T)
str_reverse(text3)
to get the desired result
[1] "M:y Ver:y Ent:husia:stic :Mothe:r Jus:t Ser:ved U:s Noo:dles!"
Is there a way this can be achieved directly using regular expressions?
You may use
gsub('(?=(?:.{5})+$)', ":", text, perl = TRUE)
## => [1] "M:y Ver:y Ent:husia:stic :Mothe:r Jus:t Ser:ved U:s Noo:dles!"
See the regex demo
The (?=(?:.{5})+$) pattern matches any location inside the string that is followed with any 5 chars (other than line break chars) 1 or more times up to the end of the string.
If the input string can contain line breaks you need to add (?s) at the start of the pattern (since . in PCRE regex does not match line breaks by default):
'(?s)(?=(?:.{5})+$)'
Example: "example._AL(5)._._4500_GRE/Jan_2018"
I am trying to extract text from the above string containing parentheses. I wanna extract everything starting from AL.
Output should look like: "AL(5)._._4500_GRE/Jan_2018"
There is some question on what we can assume is known but here are a few variations which make various assumptions.
1) word( This removes everything prior to the first word followed by a parenthesis.
"^" matches the start of string
".*?" is the shortest match of anything provided we still match rest of regex
"\\w+" matches a word
"\\(" matches a left paren
(...) forms a capture group which the replacement string can refer to as "\\1"
Code
x <- "example.AL(5)._._4500_GRE/Jan_2018"
sub("^.*?(\\w+\\()", "\\1", x)
## [1] "AL(5)._._4500_GRE/Jan_2018"
1a) or matching a word followed by ( followed by anything and extracting that:
library(gsubfn)
strapplyc(x, "\\w+\\(.*", simplify = TRUE)
## [1] "AL(5)._._4500_GRE/Jan_2018"
2) AL( or if we know that the word is AL then:
sub("^.*?(AL\\(.*)", "\\1", x)
## [1] "AL(5)._._4500_GRE/Jan_2018"
3) remove up to 1st dot or if we know that the part to be removed is the part before and including the first dot:
sub("^.*?\\.", "", x)
## [1] "AL(5)._._4500_GRE/Jan_2018"
4) dot separated fields If the format of the input is dot-separated fields we can parse them all out at once like this:
read.table(text = x, sep = ".", as.is = TRUE)
## V1 V2 V3 V4
## 1 example AL(5) _ _4500_GRE/Jan_2018