How to remove everything contained after the first number of a string?
x <- c("Hubert 208 apt 1", "Mass Av 300, block 3")
After this question, I succeeded in removing everything before the first number, the first number inclusive:
gsub( "^\\D*\\d+", "", x )
[1] " apt 1" ", block 3"
But the desired output looks like this:
[1] "Hubert 208" "Mass Av 300"
>
In the OP's current code, a minor change can make it work i.e. to capture the matching pattern as a group ((...)) and replace with backreference (\\1)
sub("^(\\D*\\d+).*", "\\1", x)
#[1] "Hubert 208" "Mass Av 300"
Here, the pattern from OP implies ("^\\D*\\d+") - zero or more characters that are not a digit (\\D*) from the start (^) of the string, followed by one or more digits (\\d+) and this is captured as a group with parens ((...)).
Also, instead of gsub (global substitution) we need only sub as we need to match only a single instance (from the beginning)
This expression might be slightly safer,
^\s*(.+?)([0-9]+)
Demo
Another option instead of replace is to take your expression and use the match instead.
Your pattern will match till after the first digits by matching from the start of the string ^ 0+ times not a digit \D* followed by 1+ times a digit \d+:
^\\D*\\d+
Regex demo
If you use sub with perl=TRUE you could make use of \K to forget what was matched.
Then you might use:
^\\D*\\d+\\K.*
Regex demo
In the replacement use an empty string.
sub("^\\D*\\d+\\K.*", "", x, perl=TRUE)
You could also use your current regex pattern with stringr::str_extract:
x <- c("Hubert 208 apt 1", "Mass Av 300, block 3")
stringr::str_extract(x, "^\\D*\\d+")
[1] "Hubert 208" "Mass Av 300"
Related
Col
WBU-ARGU*06:03:04
WBU-ARDU*08:01:01
WBU-ARFU*11:03:05
WBU-ARFU*03:456
I have a column which has 75 rows of variables such as the col above. I am not quite sure how to use gsub or sub in order to get up until the integers after the first colon.
Expected output:
Col
WBU-ARGU*06:03
WBU-ARDU*08:01
WBU-ARFU*11:03
WBU-ARFU*03:456
I tried this but it doesn't seem to work:
gsub("*..:","", df$col)
Following may help you here too.
sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)
Output will be as follows.
> sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)
[1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456b"
Where Input for data frame is as follows.
dat <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456b")
df <- data.frame(dat)
Explanation: Following is only for explanation purposes.
sub(" ##using sub for global subtitution function of R here.
([^:]*) ##By mentioning () we are keeping the matched values from vector's element into 1st place of memory(which we could use later), which is till next colon comes it will match everything.
: ##Mentioning letter colon(:) here.
([^:]*) ##By mentioning () making 2nd place in memory for matched values in vector's values which is till next colon comes it will match everything.
.*" ##Mentioning .* to match everything else now after 2nd colon comes in value.
,"\\1:\\2" ##Now mentioning the values of memory holds with whom we want to substitute the element values \\1 means 1st memory place \\2 is second memory place's value.
,df$dat) ##Mentioning df$dat dataframe's dat value.
You may use
df$col <- sub("(\\d:\\d+):\\d+$", "\\1", df$col)
See the regex demo
Details
(\\d:\\d+) - Capturing group 1 (its value will be accessible via \1 in the replacement pattern): a digit, a colon and 1+ digits.
: - a colon
\\d+ - 1+ digits
$ - end of string.
R Demo:
col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("(\\d:\\d+):\\d+$", "\\1", col)
## => [1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
Alternative approach:
df$col <- sub("^(.*?:\\d+).*", "\\1", df$col)
See the regex demo
Here,
^ - start of string
(.*?:\\d+) - Group 1: any 0+ chars, as few as possible (due to the lazy *? quantifier), then : and 1+ digits
.* - the rest of the string.
However, it should be used with the PCRE regex engine, pass perl=TRUE:
col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("^(.*?:\\d+).*", "\\1", col, perl=TRUE)
## => [1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
See the R online demo.
sub("(\\d+:\\d+):\\d+$", "\\1", df$Col)
[1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
Alternatively match what you want (instead of subbing out what you don't want) with stringi:
stringi::stri_extract_first(df$Col, regex = "[A-Z-\\*]+\\d+:\\d+")
Slightly more concise stringr:
stringr::str_extract(df$Col, "[A-Z-\\*]+\\d+:\\d+")
# or
stringr::str_extract(df$Col, "[\\w-*]+\\d+:\\d+")
What's a good way to extract only the number 2007 from the following string:
some_string <- "1_2_start_2007_3_end"
The pattern to detect the year number in my case would be:
4 digits
surrounded by "_"
I am quite new to using regular expressions. I tried the following:
regexp <- "_+[0-9]+_"
names <- str_extract(files, regexp)
But this does not take into account that there are always 4 digits and outputs the underlines as well.
You may use a sub option, too:
some_string <- "1_2_start_2007_3_end"
sub(".*_(\\d{4})_.*", "\\1", some_string)
See the regex demo
Details
.* - any 0+ chars, as many as possible
_ - a _ char
(\\d{4}) - Group 1 (referred to via \1 from the replacement pattern): 4 digits
_.* - a _ and then any 0+ chars up to the end of string.
NOTE: akrun's str_extract(some_string, "(?<=_)\\d{4}") will extract the leftmost occurrence and my sub(".*_(\\d{4})_.*", "\\1", some_string) will extract the rightmost occurrence of a 4-digit substring enclosed with _. For my my solution to return the leftmost one use a lazy quantifier with the first .: sub(".*?_(\\d{4})_.*", "\\1", some_string).
R test:
some_string <- "1_2018_start_2007_3_end"
sub(".*?_(\\d{4})_.*", "\\1", some_string) # leftmost
## -> 2018
sub(".*_(\\d{4})_.*", "\\1", some_string) # rightmost
## -> 2007
We can use regex lookbehind to specify the _ and extract the 4 digits that follow
library(stringr)
str_extract(some_string, "(?<=_)\\d{4}")
#[1] "2007"
If the pattern also shows - both before and after the 4 digits, then use regex lookahead as well
str_extract(some_string, "(?<=_)\\d{4}(?=_)")
#[1] "2007"
Just to get a non-regex approach out there, in which we split on _ and convert to numeric. All non-numbers will be coerced to NA, so we use !is.na to eliminate them. We then use nchar to count the characters, and pull the one with 4.
i1 <- as.numeric(strsplit(some_string, '_')[[1]])
i1 <- i1[!is.na(i1)]
i1[nchar(i1) == 4]
#[1] 2007
This is the quickest regex I could come up with:
\S.*_(\d{4})_\S.*
It means,
any number of non-space characters,
then _
followed by four digits (d{4})
above four digits is your year captured using ()
another _
any other gibberish non space string
Since, you mentioned you're new, please test this and all other answers at https://regex101.com/, pretty good to learn regex, it explains in depth what your regex is actually doing.
If you just care about (year) then below regex is enough:
_(\d{4})_
I want to write a regex in R to remove all words of a string containing numbers.
For example:
first_text = "a2c if3 clean 001mn10 string asw21"
second_text = "clean string
Try with gsub
trimws(gsub("\\w*[0-9]+\\w*\\s*", "", first_text))
#[1] "clean string"
It is easier to select words with no numbers than to select and delete words with numbers:
> library(stringr)
> str1 <- "a2c if3 clean 001mn10 string asw21"
> paste(unlist(str_extract_all(str1, "(\\b[^\\s\\d]+\\b)")), collapse = " ")
[1] "clean string"
Note:
Backslashes have to be escaped in R to work properly, hence double backslashes
\b is word boundary
\s is white space
\d is digit character
a caret (^) inside square brackets is a negater: find characters that do not match ...
"+" after the character group inside [] means "1 or more" occurrences of those (non white space and non digit) characters
Just another alternative using gsub
trimws(gsub("[^\\s]*[0-9][^\\s]*", "", first_text, perl=T))
#[1] "clean string"
A bit longer than some of the answers but very tractable is to first convert the string to a vector of words, then check word by word if there are any numbers and use standard R subsetting.
first_text_vec <- strsplit(first_text, " ")[[1]]
first_text_vec
[1] "a2c" "if3" "clean" "001mn10" "string" "asw21"
paste(first_text_vec[!grepl("[0-9]", first_text_vec)], collapse = " ")
[1] "clean string"
I have a vector including certain strings, and I would like remove other parts in each string except the word including certain patter (here is mir).
s <- c("a mir-96 line (kk27)", "mir-133a cell",
"d mir-14-3p in", "m mir133 (sas)", "mir_23_5p r 27")
I want to obtain:
mir-96, mir-133a, mir-14-3p, mir133, mir_23_5p
I know the idea: use the gsub() and pattern is: a word beginning with (or including) mir.
But I have no idea how to construct such patter.
Or other idea?
Any help will be appreciated!
One way in base R would be splitting every string into words and then extracting only those with mir in it
unlist(lapply(strsplit(s, " "), function(x) grep("mir", x, value = TRUE)))
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p"
We can save the unlist step in lapply by using sapply as suggested by #Rich Scriven in comments
sapply(strsplit(s, " "), function(x) grep("mir", x, value = TRUE))
We can use sub to match zero or more characters (.*) followed by a word boundary (\\b) followed by the string (mir and one or more characters that are not a white space (\\S+), capture it as a group by placing inside the (...) followed by other characters, and in the replacement use the backreference of the captured group (\\1)
sub(".*\\b(mir\\S+).*", "\\1", s)
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p"
Update
If there are multiple 'mir.*' substring, then we want to extract strings having some numeric part
sub(".*\\b(mir[^0-9]*[0-9]+\\S*).*", "\\1", s1)
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p" "mir_23-5p"
data
s1 <- c("a mir-96 line (kk27)", "mir-133a cell", "d mir-14-3p in", "m mir133 (sas)",
"mir_23_5p r 27", "a mir_23-5p 1 mir-net")
I have two related questions regarding regular expressions in R:
[1]
I would like to convert sub-strings, containing punctuation followed by a letter, to an upper case letter.
Example:
Dr_dre to: DrDre
Captain.Spock to: CaptainSpock
spider-man to: spiderMan
[2]
I would like convert camel case strings to lower case strings with underscore delimiter.
Example:
EndOfFile to: End_of_file
CamelCase to: Camel_Case
ABC to: A_B_C
Thanks much,
Kamashay
We can use sub. We match one or more punctuation characters ([[:punct:]]+) followed by a single character which is captured as a group ((.)). In the replacement, the backreference for the capture group (\\1) is changed to upper case (\\U).
sub("[[:punct:]]+(.)", "\\U\\1", str1, perl = TRUE)
#[1] "DrDre" "CaptainSpock" "spiderMan"
For the second case, we use regex lookarounds i.e. match a letter ((?<=[A-Za-z])) followed by a capital letter and replace with _.
gsub("(?<=[A-Za-z])(?=[A-Z])", "_", str2, perl = TRUE)
#[1] "End_Of_File" "Camel_Case" "A_B_C"
data
str1 <- c("Dr_dre", "Captain.Spock", "spider-man")
str2 <- c("EndOfFile", "CamelCase", "ABC")