I have this variable
x= "379_exp_mirror1.csv"
I need to extract the number ("379") at the beggining (which doesn't always have 3 characters), i.e. everything before the first "". And then I need to extract everything between the second "" and the ".", in this case "mirror1".
I have tried several combinations with sub and gsub with no success, can anyone give me some indications please?
Thank you
You can use regular expression. For your problem ^(?<Number>[0-9]*)_.* do the job
1/ Test your regular expression with this website : http://derekslager.com/blog/posts/2007/09/a-better-dotnet-regular-expression-tester.ashx
Or you can split string with underscore and then try parse (int.TryParse). I think the second is better but if you want to be a regular expression master try the first method
You can use sub to extract the substrings:
x <- "379_exp_mirror1.csv"
sub("_.*", "", x)
# [1] "379"
sub("^(?:.*_){2}(.*?)\\..*", "\\1", x)
# [1] "mirror1"
Another approach with gregexpr:
regmatches(x, gregexpr("^.*?(?=_)|(?<=_)[^_]*?(?=\\.)", x, perl = TRUE))[[1]]
# [1] "379" "mirror1"
May be you can try:
library(stringr)
x <- "379_exp_mirror1.csv"
str_extract_all(x, perl('^[0-9]+(?=_)|[[:alnum:]]+(?=\\.)'))[[1]]
#[1] "379" "mirror1"
Or
strsplit(x, "[._]")[[1]][c(T,F)]
#[1] "379" "mirror1"
Or
scan(text=gsub("[.]","_", x),what="",sep="_")[c(T,F)]
#Read 4 items
#[1] "379" "mirror1"
Related
I am in R and would like to extract a two digit number 38y from the following string:
"/Users/files/folder/file_number_23a_version_38y_Control.txt"
I know that _Control always comes after the 38y and that 38y is preceded by an underscore. How can I use strsplit or other R commands to extract the 38y?
You could use
regmatches(x, regexpr("[^_]+(?=_Control)", x, perl = TRUE))
# [1] "38y"
or equivalently
stringr::str_extract(x, "[^_]+(?=_Control)")
# [1] "38y"
Using gsub.
gsub('.*_(.*)_Control.*', '\\1', x)
# [1] "38y"
See demo with detailed explanation.
A possible solution:
library(stringr)
text <- "/Users/files/folder/file_number_23a_version_38y_Control.txt"
str_extract(text, "(?<=_)\\d+\\D(?=_Control)")
#> [1] "38y"
You can find an explanation of the regex part at:
https://regex101.com/r/PQSZHX/1
text <- c('d__Viruses|f__Closteroviridae|g__Closterovirus|s__Citrus_tristeza_virus',
'd__Viruses|o__Tymovirales|f__Alphaflexiviridae|g__Mandarivirus|s__Citrus_yellow_vein_clearing_virus',
'd__Viruses|o__Ortervirales|f__Retroviridae|s__Columba_palumbus_retrovirus')
I have tried but failed:
str_extract(text, pattern = 'f.*\\|')
How can I get
f__Closteroviridae
f__Alphaflexiviridae
f__Retroviridae
Any help will be high appreciated!
Make the regex non-greedy and since you don't want "|" in final output use positive lookahead.
stringr::str_extract(text, 'f.*?(?=\\|)')
#[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
In base R, we can use sub :
sub('.*(f_.*?)\\|.*', '\\1', text)
#[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
For a base R solution, I would use regmatches along with gregexpr:
m <- gregexpr("\\bf__[^|]+", text)
as.character(regmatches(text, m))
[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
The advantage of using gregexpr as above is that should an input contain more than one f__ matching term, we could also capture it. For example:
x <- 'd__Viruses|f__Closteroviridae|g__Closterovirus|f__some_virus'
m <- gregexpr("\\bf__[^|]+", x)
regmatches(x, m)[[1]]
[1] "f__Closteroviridae" "f__some_virus"
Data:
text <- c('d__Viruses|f__Closteroviridae|g__Closterovirus|s__Citrus_tristeza_virus',
'd__Viruses|o__Tymovirales|f__Alphaflexiviridae|g__Mandarivirus|s__Citrus_yellow_vein_clearing_virus',
'd__Viruses|o__Ortervirales|f__Retroviridae|s__Columba_palumbus_retrovirus')
I have a table with a string column formatted like this
abcdWorkstart.csv
abcdWorkcomplete.csv
And I would like to extract the last word in that filename. So I think the beginning pattern would be the word "Work" and ending pattern would be ".csv". I wrote something using grepl but not working.
grepl("Work{*}.csv", data$filename)
Basically I want to extract whatever between Work and .csv
desired outcome:
start
complete
I think you need sub or gsub (substitute/extract) instead of grepl (find if match exists). Note that when not found, it will return the entire string unmodified:
fn <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')
out <- sub(".*Work(.*)\\.csv$", "\\1", fn)
out
# [1] "start" "complete" "abcdNothing.csv"
You can work around this by filtering out the unchanged ones:
out[ out != fn ]
# [1] "start" "complete"
Or marking them invalid with NA (or something else):
out[ out == fn ] <- NA
out
# [1] "start" "complete" NA
With str_extract from stringr. This uses positive lookarounds to match any character one or more times (.+) between "Work" and ".csv":
x <- c("abcdWorkstart.csv", "abcdWorkcomplete.csv")
library(stringr)
str_extract(x, "(?<=Work).+(?=\\.csv)")
# [1] "start" "complete"
Just as an alternative way, remove everything you don't want.
x <- c("abcdWorkstart.csv", "abcdWorkcomplete.csv")
gsub("^.*Work|\\.csv$", "", x)
#[1] "start" "complete"
please note:
I have to use gsub. Because I first remove ^.*Work then \\.csv$.
For [\\s\\S] or \\d\\D ... (does not work with [g]?sub)
https://regex101.com/r/wFgkgG/1
Works with akruns approach:
regmatches(v1, regexpr("(?<=Work)[\\s\\S]+(?=[.]csv)", v1, perl = T))
str1<-
'12
.2
12'
gsub("[^.]","m",str1,perl=T)
gsub(".","m",str1,perl=T)
gsub(".","m",str1,perl=F)
. matches also \n when using the R engine.
Here is an option using regmatches/regexpr from base R. Using a regex lookaround to match all characters that are not a . after the string 'Work', extract with regmatches
regmatches(v1, regexpr("(?<=Work)[^.]+(?=[.]csv)", v1, perl = TRUE))
#[1] "start" "complete"
data
v1 <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')
I have a column as below.
9453, 55489, 4588, 18893, 4457, 2339, 45489HQ, 7833HQ
I would like to add leading zero if the number is less than 5 digits. However, some numbers have "HQ" in the end, some don't.(I did check other posts, they dont have similar problem in the "HQ" part)
so the finally desired output should be:
09453, 55489, 04588, 18893, 04457, 02339, 45489HQ, 07833HQ
any idea how to do this? Thank you so much for reading my post!
A one-liner using regular expressions:
my_strings <- c("9453", "55489", "4588",
"18893", "4457", "2339", "45489HQ", "7833HQ")
gsub("^([0-9]{1,4})(HQ|$)", "0\\1\\2",my_strings)
[1] "09453" "55489" "04588" "18893"
"04457" "02339" "45489HQ" "07833HQ"
Explanation:
^ start of string
[0-9]{1,4} one to four numbers in a row
(HQ|$) the string "HQ" or the end of the string
Parentheses represent capture groups in order. So 0\\1\\2 means 0 followed by the first capture group [0-9]{1,4} and the second capture group HQ|$.
Of course if there is 5 numbers, then the regex isn't matched, so it doesn't change.
I was going to use the sprintf approach, but found the the stringr package provides a very easy solution.
library(stringr)
x <- c("9453", "55489", "4588", "18893", "4457", "2339", "45489HQ", "7833HQ")
[1] "9453" "55489" "4588" "18893" "4457" "2339" "45489HQ" "7833HQ"
This can be converted with one simple stringr::str_pad() function:
stringr::str_pad(x, 5, side="left", pad="0")
[1] "09453" "55489" "04588" "18893" "04457" "02339" "45489HQ" "7833HQ"
If the number needs to be padded even if the total string width is >5, then the number and text need to be separated with regex.
The following will work. It combines regex matching with the very helpful sprintf() function:
sprintf("%05.0f%s", # this encodes the format and recombines the number with padding (%05.0f) with text(%s)
as.numeric(gsub("^(\\d+).*", "\\1", x)), #get the number
gsub("[[:digit:]]+([a-zA-Z]*)$", "\\1", x)) #get just the text at the end
[1] "09453" "55489" "04588" "18893" "04457" "02339" "45489HQ" "07833HQ"
Another attempt, which will also work in cases like "123" or "1HQR":
x <- c("18893","4457","45489HQ","7833HQ","123", "1HQR")
regmatches(x, regexpr("^\\d+", x)) <- sprintf("%05d", as.numeric(sub("\\D+$","",x)))
x
#[1] "18893" "04457" "45489HQ" "07833HQ" "00123" "00001HQR"
This basically finds any numbers at the start of the string (^\\d+) and replaces them with a zero-padded (via sprintf) string that was subset out by removing any non-numeric characters (\\D+$) from the end of the string.
We can use only sprintf() and gsub() by splitting up the parts then putting them back together.
sprintf("%05d%s", as.numeric(gsub("[^0-9]+", "", x)), gsub("[0-9]+", "", x))
# [1] "18893" "04457" "45489HQ" "07833HQ" "00123" "00001HQR"
Using #thelatemail's data:
x <- c("18893", "4457", "45489HQ", "7833HQ", "123", "1HQR")
I'm trying to use the stringr package in R to extract everything from a string up until the first occurrence of an underscore.
What I've tried
str_extract("L0_123_abc", ".+?(?<=_)")
> "L0_"
Close but no cigar. How do I get this one? Also, Ideally I'd like something that's easy to extend so that I can get the information in between the 1st and 2nd underscore and get the information after the 2nd underscore.
To get L0, you may use
> library(stringr)
> str_extract("L0_123_abc", "[^_]+")
[1] "L0"
The [^_]+ matches 1 or more chars other than _.
Also, you may split the string with _:
x <- str_split("L0_123_abc", fixed("_"))
> x
[[1]]
[1] "L0" "123" "abc"
This way, you will have all the substrings you need.
The same can be achieved with
> str_extract_all("L0_123_abc", "[^_]+")
[[1]]
[1] "L0" "123" "abc"
The regex lookaround should be
str_extract("L0_123_abc", ".+?(?=_)")
#[1] "L0"
Using gsub...
gsub("(.+?)(\\_.*)", "\\1", "L0_123_abc")
You can use sub from base using _.* taking everything starting from _.
sub("_.*", "", "L0_123_abc")
#[1] "L0"
Or using [^_] what is everything but not _.
sub("([^_]*).*", "\\1", "L0_123_abc")
#[1] "L0"
or using substr with regexpr.
substr("L0_123_abc", 1, regexpr("_", "L0_123_abc")-1)
#substr("L0_123_abc", 1, regexpr("_", "L0_123_abc", fixed=TRUE)-1) #More performant alternative
#[1] "L0"