I have had a good morning of trying to tidy this up but cannot find a more elegant solution.
I have the following as a value:
TEST <- Pia1.2016-10-08.1103+N2353.tif
and from this I need the date and the time 'extracted', I have the following (which works but I am 100% sure there is a better way to do it)
DATEDIR <- sub("[P][i][a][1]\\.","",TEST)
DATEDIR <- sub("\\...............","", DATEDIR)
DATEDIR # to check
I have not got round ot extracting the time bit yet as I thought I would clear this up first, although I would like the time variable to be called
TIMEDIR <-
Many thanks!
You may use
TEST <- 'Pia1.2016-10-08.1103+N2353_hc.tif'
date <- sub('.*?\\.(\\d{4}-\\d{2}-\\d{2})\\..*', '\\1', TEST)
time <- sub('.*?\\.\\d{4}-\\d{2}-\\d{2}\\.(\\d{2})(\\d{2}).*', '\\1:\\2', TEST)
# => [1] "2016-10-08"
# [1] "11:03"
See the R demo online. See the Regex 1 and Regex 2.
The first pattern matches
.*? - any 0+ chars, as few as possilbe
\\. - a dot
(\\d{4}-\\d{2}-\\d{2}) - Capturing group 1 (referred to with \1 from the replacement pattern): 4 digits, -, 2 digits, - and 2 digits
\\. - a .
.* - any 0+ chars, as many as possilbe.
The second pattern matches and captures the subsequent two digits into Group 1 and the next two digits into Group 2, and the \1:\2 replacement formats the time into a HH:mm string.
Related
I've solved 2022 advent of code day 6, but was wondering if there was a regex way to find the first occurance of 4 non-repeating characters:
From the question:
bvwbjplbgvbhsrlpgdmjqwftvncz
bvwbjplbgvbhsrlpgdmjqwftvncz
# discard as repeating letter b
bvwbjplbgvbhsrlpgdmjqwftvncz
# match the 5th character, which signifies the end of the first four character block with no repeating characters
in R I've tried:
txt <- "bvwbjplbgvbhsrlpgdmjqwftvncz"
str_match("(.*)\1", txt)
But I'm having no luck
You can use
stringr::str_extract(txt, "(.)(?!\\1)(.)(?!\\1|\\2)(.)(?!\\1|\\2|\\3)(.)")
See the regex demo. Here, (.) captures any char into consequently numbered groups and the (?!...) negative lookaheads make sure each subsequent . does not match the already captured char(s).
See the R demo:
library(stringr)
txt <- "bvwbjplbgvbhsrlpgdmjqwftvncz"
str_extract(txt, "(.)(?!\\1)(.)(?!\\1|\\2)(.)(?!\\1|\\2|\\3)(.)")
## => [1] "vwbj"
Note that the stringr::str_match (as stringr::str_extract) takes the input as the first argument and the regex as the second argument.
I have a field which contains two charecters, some digits and potentially a single letter. For example
QU1Y
ZL002
FX16
TD8
BF007P
VV1395
HM18743
JK0001
I would like to consistently return all letters in their original position, but digits as follows.
for 1 to 3 digits :
return all digits OR the digits left padded with zeros
For 4 or more digits :
it must not begin with a zero and return the 4 first digits OR if the first is a zero then truncate to three digits
example from the data above
QU001Y
ZL002
FX016
TD008
BF007P
VV1395
HM1874
JK001
The implementation will be in R but I'm interested in a straight regex solution, I'll work out the R side of things. It may not be possible in straight regex which is why I can't get my head round it.
This identifies the correct ones, but I'm hoping to correct those which are not
right.
"[A-Z]{2}[1-9]{0,1}[0-9]{1,3}[F,Y,P]{0,1}"
For the curious, they are flight numbers but entered by a human. Hence the variety...
You may use
> library(gsubfn)
> l <- c("QU1Y", "ZL002", "FX16", "TD8", "BF007P", "VV1395", "HM18743", "JK0001")
> gsubfn('^[A-Z]{2}\\K0*(\\d{1,4})\\d*', ~ sprintf("%03d",as.numeric(x)), l, perl=TRUE)
[1] "QU001Y" "ZL002" "FX016" "TD008" "BF007P" "VV1395" "HM1874" "JK001"
The pattern matches
^ - start of string
[A-Z]{2} - two uppercase letters
\\K - the text matched so far is removed from the match
0* - 0 or more zeros
(\\d{1,4}) - Capturing group 1: one to four digits
\\d* - 0+ digits.
Group 1 is passed to the callback function where sprintf("%03d",as.numeric(x)) pads the value with the necessary amount of digits.
I trying to extract a substring by pattern using gsub() R function.
# Example: extracting "7 years" substring.
string <- "Psychologist - 7 years on the website, online"
gsub(pattern="[0-9]+\\s+\\w+", replacement="", string)`
`[1] "Psychologist - on the website, online"
As you can see, it's easy to exlude needed substring using gsub(), but I need to inverse the result and getting "7 years" only.
I think about using "^", something like that:
gsub(pattern="[^[0-9]+\\s+\\w+]", replacement="", string)
Please, could anyone help me with correct regexp pattern?
You may use
sub(pattern=".*?([0-9]+\\s+\\w+).*", replacement="\\1", string)
See this R demo.
Details
.*? - any 0+ chars, as few as possible
([0-9]+\\s+\\w+) - Capturing group 1:
[0-9]+ - one or more digits
\\s+ - 1 or more whitespaces
\\w+ - 1 or more word chars
.* - the rest of the string (any 0+ chars, as many as possible)
The \1 in the replacement replaces with the contents of Group 1.
You could use the opposite of \d, which is \D in R:
string <- "Psychologist - 7 years on the website, online"
sub(pattern = "\\D*(\\d+\\s+\\w+).*", replacement = "\\1", string)
# [1] "7 years"
\D* means: no digits as long as possible, the rest is captured in a group and then replaces the complete string.
See a demo on regex101.com.
In my data (which is text), there are abbreviations.
Is there any functions or code that search for abbreviations in text? For example, detecting 3-4-5 capital letter abbreviations and letting me count how often they happen.
Much appreciated!
detecting 3-4-5 capital letter abbreviations
You may use
\b[A-Z]{3,5}\b
See the regex demo
Details:
\b - a word boundary
[A-Z]{3,5} - 3, 4 or 5 capital letters (use [[:upper:]] to match letters other than ASCII, too)
\b - a word boundary.
R demo online (leveraging the regex occurrence count code from #TheComeOnMan)
abbrev_regex <- "\\b[A-Z]{3,5}\\b";
x <- "XYZ was seen at WXYZ with VWXYZ and did ABCDEFGH."
sum(gregexpr(abbrev_regex,x)[[1]] > 0)
## => [1] 3
regmatches(x, gregexpr(abbrev_regex, x))[[1]]
## => [1] "XYZ" "WXYZ" "VWXYZ"
You can use the regular expression [A-Z] to match any ocurrence of acapital letter. If you want this pattern to be repeated 3 times you can add \1{3} to your regex. Consider using variables and a loop to get the job done for 3 to 5 repetition times.
I am trying to add parentheses around certain book titles character strings and I want to be able to paste with the paste0 function. I want to take this string:
a <- c("I Like What I Know 1959 02e pdfDrama (amazon.com)", "My Liffe 1993 07e pdfDrama (amazon.com)")
wrap certain strings in parentheses:
a
[1] “I Like What I Know (1959) (02e) (pdfDrama) (amazon.com)”
[2] ”My Life (1993) (07e) (pdfDrama) (amazon.com)”
I have tried but can't figure out a way to replace them within the string:
paste0("(",str_extract(a, "\\d{4}"),")")
paste0("(",str_extract(a, ”[0-9]+.e”),”)”)
Help?
I can suggest a regex for a fixed number of words of specific type:
a <- c("I Like What I Know 1959 02e pdfDrama (amazon.com)","My Life 1993 07e pdfDrama (amazon.com)")
sub("\\b(\\d{4})(\\s+)(\\d+e)(\\s+)([a-zA-Z]+)(\\s+\\([^()]*\\))", "(\\1)\\2(\\3)\\4(\\5)\\6", a)
See the R demo
And here is the regex demo. In short,
\\b(\\d{4}) - captures 4 digits as a whole word into Group 1
(\\s+) - Group 2: one or more whitespaces
(\\d+e) - Group 3: one or more digits and e
(\\s+) - Group 4: ibid
([a-zA-Z]+) - Group 5: one or more letters
(\\s+\\([^()]*\\)) - Group 6: one or more whitespaces, (, zero or more chars other than ( and ), ).
The contents of the groups are inserted back into the result with the help of backreferences.
If there are more words, and you need to wrap words starting with a letter/digit/underscore after a 4-digit word in the string, use
gsub("(?:(?=\\b\\d{4}\\b)|\\G(?!\\A))\\s*\\K\\b(\\S+)", "(\\1)", a, perl=TRUE)
See the R demo and a regex demo
Details:
(?:(?=\\b\\d{4}\\b)|\\G(?!\\A)) - either the location before a 4-digit whole word (see the positive lookahead (?=\\b\\d{4}\\b)) or the end of the previous successful match
\\s* - 0+ whitespaces
\\K - omitting the text matched so far
\\b(\\S+) - Group 1 capturing 1 or more non-whitespace symbols that are preceded with a word boundary.