I have a vector:
c("BAAAVAST", "BAACEZ", "BAAGECBA", "LOL")
And I would like to remove "BAA" from the words that contain it. And to those words I would like to append ".PR".
Desired outcome:
c("AVAST.PR", "CEZ.PR", "GECBA.PR", "LOL")
Any ideas? Ideally using stringr. Thank you a lot.
You could use the following solution:
gsub("BAA(.*)", "\\1\\.PR", vec)
[1] "AVAST.PR" "CEZ.PR" "GECBA.PR" "LOL"
You could use
library(stringr)
# optimized thanks to Anoushiravan
str_replace(c("BAAAVAST", "BAACEZ", "BAAGECBA", "LOL"), "BAA(\\w*)", "\\1.PR")
#> [1] "AVAST.PR" "CEZ.PR" "GECBA.PR" "LOL"
use \\w* if you want to match word characters only or .* if there are no limitations to the characters.
This is verbose than the other answers. It finds strings with 'BAA' and appends 'PR.' to it.
inds <- grepl('BAA', vec, fixed = TRUE)
vec[inds] <- paste(sub('BAA', '', vec[inds]), 'PR', sep = '.')
vec
#[1] "AVAST.PR" "CEZ.PR" "GECBA.PR" "LOL"
Related
Is there an R function to get only the part of a string before the 2nd capital character appears?
For example:
Example <- "MonkeysDogsCats"
Expected output should be:
"Monkeys"
Maybe something like
stringr::str_extract("MonkeysDogsCats", "[A-Z][a-z]*")
#[1] "Monkeys"
Here is an alternative approach:
Here we first put a space before all uppercase and then extract the first word:
library(stringr)
word(gsub("([a-z])([A-Z])","\\1 \\2", Example), 1)
[1] "Monkeys"
A base solution with sub():
x <- "MonkeysDogsCats"
sub("(?<=[a-z])[A-Z].*", "", x, perl = TRUE)
# [1] "Monkeys"
Another way using stringr::word():
stringr::word(x, 1, sep = "(?=[A-Z])\\B")
# [1] "Monkeys"
If the goal is strictly to capture any string before the 2nd capital character, one might want pick a solution it'll also work with all types of strings including numbers and special characters.
strings <- c("MonkeysDogsCats",
"M4DogsCats",
"M?DogsCats")
stringr::str_remove(strings, "(?<=.)[A-Z].*")
Output:
[1] "Monkeys" "M4" "M?"
It depends on what you want to allow to match. You can for example match an uppercase char [A-Z] optionally followed by any character that is not an uppercase character [^A-Z]*
If you don't want to allow whitespace chars, you can exclude them [^A-Z\\s]*
library(stringr)
str_extract("MonkeysDogsCats", "[A-Z][^A-Z]*")
Output
[1] "Monkeys"
R demo
If there should be an uppercase character following, and there are only lowercase characters allowed:
str <- "MonkeysDogsCats"
regmatches(str, regexpr("[A-Z][a-z]*(?=[A-Z])", str, perl = TRUE))
Output
[1] "Monkeys"
R demo
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')
The string is as shown below:
s <- "12N10-3A 12N10-3A-1 12N10-3A-2 YB10L-A2"
I can get the strings except from second one.
gsub("\\s.*","",s) #12N10-3A
gsub(".*\\s","",s) #YB10L-A2
gsub(".*\\s.*\\s(.*).*\\s(.*)","\\1",s) #12N10-3A-2
How to get the second string from s and what's short approach for each code line? I tried what I learnt on regex101.com
We can use stri_extract_last from stringi
library(stringi)
stri_extract_last(s, regex = '\\S+')
#[1] "YB10L-A2"
Or use word from stringr
library(stringr)
word(s, 4)
#[1] "YB10L-A2"
Just use strsplit:
items <- strsplit(s, "\\s+")[[1]]
If you want to access the last item, then just use:
items[4]
[1] "YB10L-A2"
If you really wanted to isolate the last term using sub, then here is one way:
sub(".*\\s+", "", s)
Hi have data which contains two or more dots. My requirement is to get string from first to second dot.
E.g string <- "abcd.vdgd.dhdsg"
Result expected =vdgd
I have used
pt <-strapply(string, "\\.(.*)\\.", simplify = TRUE)
which is giving correct data but for string having more than two dots its not working as expected.
e.g string <- "abcd.vdgd.dhdsg.jsgs"
its giving dhdsg.jsgs but expected is vdgd
Could anyone help me.
Thanks & Regards,
In base R we can use strsplit
ss <- "abcd.vdgd.dhdsg"
unlist(strsplit(ss, "\\."))[2]
#[1] "vdgd"
Or using gregexpr with regmatches
unlist(regmatches(ss, gregexpr("[^\\.]+", ss)))[2]
#[1] "vdgd"
Or using gsub (thanks #TCZhang)
gsub("^.+?\\.(.+?)\\..*$", "\\1", ss)
#[1] "vdgd"
Another option:
string <- "abcd.vdgd.dhdsg.jsgs"
library(stringr)
str_extract(string = string, pattern = "(?<=\\.).*?(?=\\.)")
[1] "vdgd"
I like this one because the str_extract function will return the first instance of the correct pattern, but you could also use str_extract_all to get all instances.
str_extract_all(string = string, pattern = "(?<=\\.).*?(?=\\.)")
[[1]]
[1] "vdgd" "dhdsg"
From here, you could index to get any position between two dots you want.
Another solution with the qdapRegex package:
library(qdapRegex)
ex_between("abcd.vdgd.dhdsg.jsgs", ".", ".")[[1]][1]
# "vdgd"
You can use read.table as well if you wish.Here providing the string as given in your problem and selecting the separator as dot("."), Once the column is converted into a data.frame, you may choose to select whatever column you want to pick(In this case it is column number 2).
read.table(text=string, sep=".",stringsAsFactors = FALSE)[,2]
Output:
> read.table(text=string, sep=".",stringsAsFactors = FALSE)[,2]
[1] "vdgd"
Here is a fun easy way via stringr
stringr::word(string, 2, sep = '\\.')
Here are two options that are vectorized over the input string vector:
You can try tstrsplit from data.table, which is vectorized over string:
> string <- c("abcd.vdgd.dhdsg", "abcd.vdgd.dhdsg.jsgs")
> tstrsplit(string, '.', fixed = TRUE)[[2]]
[1] "vdgd" "vdgd"
or regex:
> sub('.*?\\.(.*?)\\..*', '\\1', string)
[1] "vdgd" "vdgd"`
How do I remove part of a string? For example in ATGAS_1121 I want to remove everything before _.
Use regular expressions. In this case, you can use gsub:
gsub("^.*?_","_","ATGAS_1121")
[1] "_1121"
This regular expression matches the beginning of the string (^), any character (.) repeated zero or more times (*), and underscore (_). The ? makes the match "lazy" so that it only matches are far as the first underscore. That match is replaced with just an underscore. See ?regex for more details and references
You can use a built-in for this, strsplit:
> s = "TGAS_1121"
> s1 = unlist(strsplit(s, split='_', fixed=TRUE))[2]
> s1
[1] "1121"
strsplit returns both pieces of the string parsed on the split parameter as a list. That's probably not what you want, so wrap the call in unlist, then index that array so that only the second of the two elements in the vector are returned.
Finally, the fixed parameter should be set to TRUE to indicate that the split parameter is not a regular expression, but a literal matching character.
If you're a Tidyverse kind of person, here's the stringr solution:
R> library(stringr)
R> strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
R> strings %>% str_replace(".*_", "_")
[1] "_1121" "_1432" "_1121"
# Or:
R> strings %>% str_replace("^[A-Z]*", "")
[1] "_1121" "_1432" "_1121"
Here's the strsplit solution if s is a vector:
> s <- c("TGAS_1121", "MGAS_1432")
> s1 <- sapply(strsplit(s, split='_', fixed=TRUE), function(x) (x[2]))
> s1
[1] "1121" "1432"
Maybe the most intuitive solution is probably to use the stringr function str_remove which is even easier than str_replace as it has only 1 argument instead of 2.
The only tricky part in your example is that you want to keep the underscore but its possible: You must match the regular expression until it finds the specified string pattern (?=pattern).
See example:
strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
strings %>% stringr::str_remove(".+?(?=_)")
[1] "_1121" "_1432" "_1121"
Here the strsplit solution for a dataframe using dplyr package
col1 = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
col2 = c("T", "M", "A")
df = data.frame(col1, col2)
df
col1 col2
1 TGAS_1121 T
2 MGAS_1432 M
3 ATGAS_1121 A
df<-mutate(df,col1=as.character(col1))
df2<-mutate(df,col1=sapply(strsplit(df$col1, split='_', fixed=TRUE),function(x) (x[2])))
df2
col1 col2
1 1121 T
2 1432 M
3 1121 A