I have a list of years like this:
2018-
2001–2020
1999-
2005-
I would like to create a regex to match the year with these criteria:
xxxx- matches xxxx
yyyy-nnnn matches nnnn
Can you please help me?
I've tried [[:digit:]]{4}$, or alternatively [[:digit:]]{4}-$, but they only partially work.
To get the last year in the "range," established by - character, the cleanest way
my $year = (split /-/, $range)[-1];
If there isn't anything after the last delimiter then the last returned element by split is what is before it, so the last element in its return list (obtained with index -1) is either the second given year -- as in 2001-2020 -- or the only one, as in other examples. This performs no checking of input.
With a regex, one way is to seek the last number in the string
my ($year) = $range =~ /([0-9]+)[^0-9]*$/;
where if you use [0-9]{4} then there is a small additional measure of checking.
The POSIX character class [[:digit:]] and its negation [[:^digit:]] (or \P{PosixDigit}) can be used instead if desired, but note that these match all manner of Unicode "digit characters," just like \d and \D do (a few hundred), on top of the ascii [0-9] (unless /a modifier is used).
A full test program, for both
use warnings;
use strict;
use feature 'say';
my #ranges = qw(2018- 2001-2020 1999- 2005-);
foreach my $range (#ranges) {
my $year = (split /-/, $range)[-1];
# Or, using regex
# my ($year) = $range =~ /([0-9]+)[^0-9]*$/;
say $year;
}
Prints as desired.
We can capture the 4 digits as group, followed by a - at the end ($) of the string and replace with the backreference (\\1) of the captured group
sub(".*(\\d{4})-?$", "\\1", str1)
#[1] "2018" "2020" "1999" "2005"
data
str1 <- c("2018-", "2001-2020", "1999-", "2005-")
You can split the text on "-" and get the last number.
x <- c("2018-", "2001-2020", "1999-", "2005-")
sapply(strsplit(str1, '-', fixed = TRUE), tail, 1)
#[1] "2018" "2020" "1999" "2005"
Related
Assuming I have the following string:
string = "path/stack/over_flow/Pedro_account"
I am intrested in matching the first 2 characters after the last / and before the first _. So in this case the desired out put is:
Pe
What I have so far is a mix of substr and str_extract:
substr(str_extract(string, "[^/]*$"),1,2)
which of course will give an answer but I belive there is a nice regex for it as well, and that is what I'm looking for.
You can use
library(stringr)
str_extract(string, "(?<=/)[^/]{2}(?=[^/]*$)")
## => [1] "Pe"
See the R demo and the regex demo. Details:
(?<=/) - a location immediately preceded with a / char
[^/]{2} - two chars other than /
(?=[^/]*$) - a location immediately preceded with zero or more chars other than / till the end of string.
Using basename to get the last folder name, then substring:
substr(basename("path/stack/over_flow/Pedro_account"), 1, 2)
# [1] "Pe"
Remove everything till last / and extract first 2 characters.
Base R -
string = "path/stack/over_flow/Pedro_account"
substr(sub('.*/', '', string), 1, 2)
#[1] "Pe"
stringr
substr(stringr::str_remove(string, '.*/'), 1, 2)
You can use str_match with a capture group:
/ Match literally
([^/_]{2}) Capture 2 chars other than / or _ in group 1
[^/]* Match optional chars other than /
$ End of string
See a regex demo and a R demo.
Example
library(stringr)
string = "path/stack/over_flow/Pedro_account"
str_match(string, "/([^/_]{2})[^/]*$")[,2]
Output
[1] "Pe"
For a text field, I would like to expose those that contain invalid characters. The list of invalid characters is unknown; I only know the list of accepted ones.
For example for French language, the accepted list is
A-z, 1-9, [punc::], space, àéèçè, hyphen, etc.
The list of invalid charactersis unknown, yet I want anything unusual to resurface, for example, I would want
This is an 2-piece à-la-carte dessert to pass when
'Ã this Øs an apple' pumps up as an anomalie
The 'not contain' notion in R does not behave as I would like, for example
grep("[^(abc)]",c("abcdef", "defabc", "apple") )
(those that does not contain 'abc') match all three while
grep("(abc)",c("abcdef", "defabc", "apple") )
behaves correctly and match only the first two. Am I missing something
How can we do that in R ? Also, how can we put hypen together in the list of accepted characters ?
[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+
The above regex matches any of the following (one or more times). Note that the parameter ignore.case=T used in the code below allows the following to also match uppercase variants of the letters.
a-z Any lowercase ASCII letter
1-9 Any digit in the range from 1 to 9 (excludes 0)
[:punct:] Any punctuation character
The space character
àâæçéèêëîïôœùûüÿ Any valid French character with a diacritic mark
- The hyphen character
See code in use here
x <- c("This is an 2-piece à-la-carte dessert", "Ã this Øs an apple")
gsub("[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+", "", x, ignore.case=T)
The code above replaces all valid characters with nothing. The result is all invalid characters that exist in the string. The following is the output:
[1] "" "ÃØ"
If by "expose the invalid characters" you mean delete the "accepted" ones, then a regex character class should be helpful. From the ?regex help page we can see that a hyphen is already part of the punctuation character vector;
[:punct:]
Punctuation characters:
! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { | } ~
So the code could be:
x <- 'Ã this Øs an apple'
gsub("[A-z1-9[:punct:] àéèçè]+", "", x)
#[1] "ÃØ"
Note that regex has a predefined, locale-specific "[:alpha:]" named character class that would probably be both safer and more compact than the expression "[A-zàéèçè]" especially since the post from ctwheels suggests that you missed a few. The ?regex page indicates that "[0-9A-Za-z]" might be both locale- and encoding-specific.
If by "expose" you instead meant "identify the postion within the string" then you could use the negation operator "^" within the character class formalism and apply gregexpr:
gregexpr("[^A-z1-9[:punct:] àéèçè]+", x)
[[1]]
[1] 1 8
attr(,"match.length")
[1] 1 1
How do I match the year such that it is general for the following examples.
a <- '"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}'
b <- 'Þegar það gerist (1998/I) (TV)'
I have tried the following, but did not have the biggest success.
gsub('.+\\(([0-9]+.+\\)).?$', '\\1', a)
What I thought it did was to go until it finds a (, then it would make a group of numbers, then any character until it meets a ). And if there are several matches, I want to extract the first group.
Any suggestions to where I go wrong? I have been doing this in R.
You could use
library(stringr)
strings <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)')
years <- str_match(strings, "\\((\\d+(?: B\\.C\\.)?)")[,2]
years
# [1] "1953" "1998"
The expression here is
\( # (
(\d+ # capture 1+ digits
(?: B\.C\.)? # B.C. eventually
)
Note that backslashes need to be escaped in R.
Your pattern contains .+ parts that match 1 or more chars as many as possible, and at best your pattern could grab last 4 digit chunks from the incoming strings.
You may use
^.*?\((\d{4})(?:/[^)]*)?\).*
Replace with \1 to only keep the 4 digit number. See the regex demo.
Details
^ - start of string
.*? - any 0+ chars as few as possible
\( - a (
(\d{4}) - Group 1: four digits
(?: - start of an optional non-capturing group
/ - a /
[^)]* - any 0+ chars other than )
)? - end of the group
\) - a ) (OPTIONAL, MAY BE OMITTED)
.* - the rest of the string.
See the R demo:
a <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)', 'Johannes Passion, BWV. 245 (1725 Version) (1996) (V)')
sub("^.*?\\((\\d{4})(?:/[^)]*)?\\).*", "\\1", a)
# => [1] "1953" "1998" "1996"
Another base R solution is to match the 4 digits after (:
regmatches(a, regexpr("\\(\\K\\d{4}(?=(?:/[^)]*)?\\))", a, perl=TRUE))
# => [1] "1953" "1998" "1996"
The \(\K\d{4} pattern matches ( and then drops it due to \K match reset operator and then a (?=(?:/[^)]*)?\\)) lookahead ensures there is an optional / + 0+ chars other than ) and then a ). Note that regexpr extracts the first match only.
I want to use regex to replace commands or tags around strings. My use case is converting LaTeX commands to bookdown commands, which means doing things like replacing \citep{*} with [#*], \ref{*} with \#ref(*), etc. However, lets stick to the generalized question:
Given a string <begin>somestring<end> where <begin> and <end> are known and somestring is an arbitrary sequence of characters, can we use regex to susbstitute <newbegin> and <newend> to get the string <newbegin>somestring<newend>?
For example, consider the LaTeX command \citep{bonobo2017}, which I want to convert to [#bonobo2017]. For this example:
<begin> = \citep{
somestring = bonobo2017
<end> = }
<newbegin> = [#
<newend> = ]
This question is basically the inverse of this question.
I'm hoping for an R or notepad++ solution.
Additional Examples
Convert \citet{bonobo2017} to #bonobo2017
Convert \ref{myfigure} to \#ref(myfigure)
Convert \section{Some title} to # Some title
Convert \emph{something important} to *something important*
I'm looking for a template regex that I can fill in my <begin>, <end>, <newbegin> and <newend> on a case-by-case basis.
You can try something like this with dplyr + stringr:
string = "\\citep{bonobo2017}"
begin = "\\citep{"
somestring = "bonobo2017"
end = "}"
newbegin = "[#"
newend = "]"
library(stringr)
library(dplyr)
string %>%
str_extract(paste0("(?<=\\Q", begin, "\\E)\\w+(?=\\Q", end, "\\E)")) %>%
paste0(newbegin, ., newend)
or:
string %>%
str_replace_all(paste0("\\Q", begin, "\\E|\\Q", end, "\\E"), "") %>%
paste0(newbegin, ., newend)
You can also make it a function for convenience:
convertLatex = function(string, BEGIN, END, NEWBEGIN, NEWEND){
string %>%
str_replace_all(paste0("\\Q", BEGIN, "\\E|\\Q", END, "\\E"), "") %>%
paste0(NEWBEGIN, ., NEWEND)
}
convertLatex(string, begin, end, newbegin, newend)
# [1] "[#bonobo2017]"
Notes:
Notice that I manually added an additional \ to "\\citep{bonobo2017}", this is because raw strings don't exist in R(I hope they do exist), so a single \ would be treated as an escape character. I need another \ to escape the first \.
The regex in str_extract uses positive lookbehind and positve lookahead to extract the somestring in between begin and end.
str_replace takes another approach of removing begin and end from string.
The "\\Q", "\\E" pair in the regex means "Backslash all nonalphanumeric characters" and "\\E" ends the expression. This is especially useful in your case since you likely have special characters in your Latex command. This expression automatically escapes them for you.
One column of my data.frame looks like the following:
c("BP_1_CSPP", "BP_2_GEGS", "BP_3_AEAG", "BP_4_KPAP", "BP_5_TAKP",
"BP_6_GGDR", "BP_7_MQQP", "BP_8_EEEE", "BP_9_RSDP", "BP_10_APAS",
"BP_11_KRGG", "BP_12_RSQQ", "BP_13_QQLS", "BP_14_EPEV", "BP_15_AAPS",
"BP_16_SDVT", "BP_17_GQQQ", "BP_18_AETP", "BP_19_PPSA", "BP_20_DATP",
"EpQ_1_AYAT", "EpQ_2_HEKL", "EpQ_3_SCSV", "EpQ_4_MAYV", "EpQ_5_LKDP",
"EpQ_6_ERCE", "EpQ_7_DNPA", "EpQ_8_YGIS", "EpQ_9_GMSS", "EpQ_10_AAKK",
"EpQ_11_NIRI", "EpQ_12_ERRR", "EpQ_13_MDRE", "EpQ_14_SRQM", "EpQ_15_DWSI",
"EpQ_16_VLVQ", "EpQ_17_GRTI", "EpQ_18_EKVR", "EpQ_19_PDVA", "EpQ_20_ADVT",
"LbT_1_RPGG", "LbT_2_TQGD", "LbT_3_EVKS", "LbT_4_VIEM", "LbT_5_GSAD",
"LbT_6_VRPI", "LbT_7_CELG", "LbT_8_APQQ", "LbT_9_SAEE", "LbT_10_GEAE",
"LbT_11_EELR", "LbT_12_EWAN", "LbT_13_IKEE", "LbT_14_VSDF", "LbT_15_WEDV",
"LbT_16_SGGA", "LbT_17_KATN", "LbT_18_EREG", "LbT_19_AWAS", "LbT_20_VDRD",
"abc_1_CVTQ", "abc_2_KEAP", "abc_3_TAYI", "abc_4_MITN", "abc_5_MPTV",
"abc_6_TRTG", "abc_7_KSTI", "abc_8_KEAI", "abc_9_HVYS", "abc_10_LGMG",
"abc_11_VAYQ", "abc_12_AGTG", "abc_13_TDSW", "abc_14_HKKS", "abc_15_YGLA",
"abc_16_WEEW", "abc_17_HSTI", "abc_18_EKCI", "abc_19_PAGI", "abc_20_TGTI",
"TcII")
Considering all the numbers < 10, which are located within the strings (e.g. "BP_1_CSPP", "BP_2_GEGS" , I wanted to add a leading zero to them, such that I would have:
"BP_01_CSPP", "BP_02_GEGS", "BP_03_AEAG", "BP_04_KPAP", "BP_05_TAKP",
"BP_06_GGDR"
and so on.
This question almost did the job, yet it does not worked for my data as:
The "0" will not be inserted at the same position all the time (some strings have 3 characters before the 0 to be inserted (e.g. BP_1_CSPP) while others have 4 (e.g. EpQ_3_SCSV)
I will still have some characters after the zero to be inserted i.e. the zero will be inserted at the middle of the string.
We can use sub to match the pattern of _ followed by a single number (([0-9])) captured as a group (inside the brackets) followed by _ and replace it with _ followed by 0, the backreference of the capture group (\\1) followed by _.
v1 <- sub("_([0-9])_", "_0\\1_", v1)
v1
#[1] "BP_01_CSPP" "BP_02_GEGS" "BP_03_AEAG" "BP_04_KPAP" "BP_05_TAKP" "BP_06_GGDR" "BP_07_MQQP" "BP_08_EEEE" "BP_09_RSDP" "BP_10_APAS" "BP_11_KRGG"
#[12] "BP_12_RSQQ" "BP_13_QQLS" "BP_14_EPEV" "BP_15_AAPS" "BP_16_SDVT" "BP_17_GQQQ" "BP_18_AETP" "BP_19_PPSA" "BP_20_DATP" "EpQ_01_AYAT" "EpQ_02_HEKL"
#[23] "EpQ_03_SCSV" "EpQ_04_MAYV" "EpQ_05_LKDP" "EpQ_06_ERCE" "EpQ_07_DNPA" "EpQ_08_YGIS" "EpQ_09_GMSS" "EpQ_10_AAKK" "EpQ_11_NIRI" "EpQ_12_ERRR" "EpQ_13_MDRE"
#[34] "EpQ_14_SRQM" "EpQ_15_DWSI" "EpQ_16_VLVQ" "EpQ_17_GRTI" "EpQ_18_EKVR" "EpQ_19_PDVA" "EpQ_20_ADVT" "LbT_01_RPGG" "LbT_02_TQGD" "LbT_03_EVKS" "LbT_04_VIEM"
#[45] "LbT_05_GSAD" "LbT_06_VRPI" "LbT_07_CELG" "LbT_08_APQQ" "LbT_09_SAEE" "LbT_10_GEAE" "LbT_11_EELR" "LbT_12_EWAN" "LbT_13_IKEE" "LbT_14_VSDF" "LbT_15_WEDV"
#[56] "LbT_16_SGGA" "LbT_17_KATN" "LbT_18_EREG" "LbT_19_AWAS" "LbT_20_VDRD" "abc_01_CVTQ" "abc_02_KEAP" "abc_03_TAYI" "abc_04_MITN" "abc_05_MPTV" "abc_06_TRTG"
#[67] "abc_07_KSTI" "abc_08_KEAI" "abc_09_HVYS" "abc_10_LGMG" "abc_11_VAYQ" "abc_12_AGTG" "abc_13_TDSW" "abc_14_HKKS" "abc_15_YGLA" "abc_16_WEEW" "abc_17_HSTI"
#[78] "abc_18_EKCI" "abc_19_PAGI" "abc_20_TGTI" "TcII"
If we are using strsplit, another option is split by _, replace the numbers by formatting with sprintf and then paste together
sapply(strsplit(v1, "_"), function(x) {
if(length(x)>1) x[2] <- sprintf("%02d", as.numeric(x[2]))
paste(x, collapse="_")})