This keeps everything before a period:
gsub("\\..*","", data$column )
how to keep everything after the period?
To remove all the characters before a period in a string(including period).
gsub("^.*\\.","", data$column )
Example:
> data <- 'foobar.barfoo'
> gsub("^.*\\.","", data)
[1] "barfoo"
To remove all the characters before the first period(including period).
> data <- 'foo.bar.barfoo'
> gsub("^.*?\\.","", data)
[1] "bar.barfoo"
You could use stringi with lookbehind regex
library(stringi)
stri_extract_first_regex(data1, "(?<=\\.).*")
#[1] "bar.barfoo"
stri_extract_first_regex(data, "(?<=\\.).*")
#[1] "barfoo"
If the string doesn't have ., this retuns NA (it is not clear about how to deal with this in the question)
stri_extract_first_regex(data2, "(?<=\\.).*")
#[1] NA
###data
data <- 'foobar.barfoo'
data1 <- 'foo.bar.barfoo'
data2 <- "foobar"
If you don't want to think about the regex for this the qdap package has the char2end function that grabs from a particular character until the end of the string.
data <- c("foo.bar", "foo.bar.barfoo")
library(qdap)
char2end(data, ".")
## [1] "bar" "bar.barfoo"
use this :
gsub(".*\\.","", data$column )
this will keep everything after period
require(stringr)
I run a course on Data Analysis and the students came up with this solution :
get_after_period <- function(my_vector) {
# Return a string vector without the characters
# before a period (excluding the period)
# my_vector, a string vector
str_sub(my_vector, str_locate(my_vector, "\\.")[,1]+1)
}
Now, just call the function :
my_vector <- c('foobar.barfoo', 'amazing.point')
get_after_period(my_vector)
[1] "barfoo" "point"
Related
I'd like to match everything between the first and last underscore. I use R.
What I have until now is this:
p.subject <- c('bla_bla', 'bla', 'bla_bla_bla', 'bla_bla_bla_bla')
sub('[^_]*_(.*)_[^_]*', x = p.subject, replacement = '\\1', perl = T)
Where 'bla' is any character except an underscore...
The result I'd like would be something like this:
c(NA, NA, bla, bla_bla)
I can't figure it out! Why does the first pattern match? It shouldn't because the pattern must have 2 underscores! Or do I have to use some kind of lookahead expression?
Your help is very welcome!
You can use gsub:
vec <- gsub("(^[^_]+)_?|_?([^_]+$)", "", p.subject)
vec <- ifelse(nchar(vec) == 0 , NA, vec)
vec
[1] NA NA "bla" "bla_bla"
Data:
dput(p.subject)
c("bla_bla", "bla", "bla_bla_bla", "bla_bla_bla_bla")
Here is another option using str_extract. We use regex lookarounds to extract the pattern between the first and the last occurrence of a specified character i.e. _.
library(stringr)
str_extract(p.subject, "(?<=[^_]{1,30}_).*(?=_[^_]+)")
#[1] NA NA "bla" "bla_bla"
NOTE: We didn't use any ifelse.
data
p.subject <- c('bla_bla', 'bla', 'bla_bla_bla', 'bla_bla_bla_bla')
I have the following dataset
> head(names$SAMPLE_ID)
[1] "Bacteria|Proteobacteria|Gammaproteobacteria|Pseudomonadales|Moraxellaceae|Acinetobacter|"
[2] "Bacteria|Firmicutes|Bacilli|Bacillales|Bacillaceae|Bacillus|"
[3] "Bacteria|Proteobacteria|Gammaproteobacteria|Pasteurellales|Pasteurellaceae|Haemophilus|"
[4] "Bacteria|Firmicutes|Bacilli|Lactobacillales|Streptococcaceae|Streptococcus|"
[5] "Bacteria|Firmicutes|Bacilli|Lactobacillales|Streptococcaceae|Streptococcus|"
[6] "Bacteria|Firmicutes|Bacilli|Lactobacillales|Streptococcaceae|Streptococcus|"
I want to extract the last word between || as a new variable i.e.
Acinetobacter
Bacillus
Haemophilus
I have tried using
library(stringr)
names$sample2 <- str_match(names$SAMPLE_ID, "|.*?|")
We can use
library(stringi)
stri_extract_last_regex(v1, '\\w+')
#[1] "Acinetobacter"
data
v1 <- "Bacteria|Proteobacteria|Gammaproteobacteria|Pseudomonadales|Moraxellaceae|Acinetobacter|"
Using just base R:
myvar <- gsub("^..*\\|(\\w+)\\|$", "\\1", names$SAMPLE_ID)
^.*\\|\\K.*?(?=\\|)
Use \K to remove rest from the final matche.See demo.Also use perl=T
https://regex101.com/r/fM9lY3/45
x <- c("Bacteria|Firmicutes|Bacilli|Lactobacillales|Streptococcaceae|Streptococcus|",
"Bacteria|Firmicutes|Bacilli|Lactobacillales|Streptococcaceae|Streptococcus|" )
unlist(regmatches(x, gregexpr('^.*\\|\\K.*?(?=\\|)', x, perl = TRUE)))
# [1] "Streptococcus" "Streptococcus"
The ending is all you need [^|]+(?=\|$)
Per #RichardScriven :
Which in R would be regmatches(x, regexpr("[^|]+(?=\\|$)", x, perl = TRUE)
You can use package "stringr" as well in this case. Here is the code:
v<- "Bacteria|
Proteobacteria|Gammaproteobacteria|Pseudomonadales|Moraxellaceae|Acinetobacter|"
v1<- str_replace_all(v, "\\|", " ")
word(v1,-2)
Here I used v as the string. The basic theory is to replace all the | with spaces, and then get the last word in the string by using function word().
I have some string
string <- "abbccc"
I want to replace the chains of the same letter to just one letter and number of occurance of this letter. So I want to have something like this:
"ab2c3"
I use stringi package to do this, but it doesn't work exactly like I want. Let's say I already have vector with parts for replacement:
vector <- c("b2", "c3")
stri_replace_all_regex(string, "([a-z])\\1{1,8}", vector)
The output:
[1] "ab2b2" "ac3c3"
The output I want: [1] "ab2c3"
I also tried this way
stri_replace_all_regex(string, "([a-z])\\1{1,8}", vector, vectorize_all=FALSE)
but i get error
Error in stri_replace_all_regex(string, "([a-z])\\1{1,8}", vector, vectorize_all = FALSE) :
vector length not consistent with other arguments
Not regex but astrsplit and rle with some paste magic:
string <- c("abbccc", "bbaccc", "uffff", "aaabccccddd")
sapply(lapply(strsplit(string, ""), rle), function(x) {
paste(x[[2]], ifelse(x[[1]] == 1, "", x[[1]]), sep="", collapse="")
})
## [1] "ab2c3" "b2ac3" "uf4" "a3bc4d3"
Not a stringi solution and not a regex either, but you can do it by splitting the string and using rle:
string <- "abbccc"
res<-paste(collapse="",do.call(paste0,rle(strsplit(string,"",fixed=TRUE)[[1]])[2:1]))
gsub("1","",res)
#[1] "ab2c3"
I have some data frame, df with a column with dates that are in the following format:
pv$day
01/01/13 00:00:00
03/01/13 00:02:03
04/03/13 00:10:15
....
I would like to eliminate the timestamp, just leaving the date (e.g. 01/01/13 for the first row). I have tried both using sapply() to apply the strsplit() function, and tried to filter the content using a regex, but don't seem to have quite gotten it right in either case. This:
sapply(pv$day, function(x) strsplit(toString(x), ' '))
gives me the column with the correct split, but indexing with either [1] or [[1]] does not return the first element of the split.
What is the best way to go about this?
You can use sub:
vec <- c("01/01/13 00:00:00", "03/01/13 00:02:03", "04/03/13 00:10:15")
sub(" .+", "", vec)
# [1] "01/01/13" "03/01/13" "04/03/13"
A simple, flexible solution is to use strptime and strftime. Here is an example that uses your dates from the example above:
# Your dates
t <- c("01/01/13 00:00:00","03/01/13 00:02:03", "04/03/13 00:10:15")
# Convert character strings to dates
z <- strptime(t, "%d/%m/%y %H:%M:%OS")
# Convert dates to string, omitting the time
z.date <- strftime(z,"%d/%m/%y")
# Print the first date
z.date[1]
Here's a nice way to use sapply, it uses strsplit to split at the space
> d <- c("01/01/13 00:00:00", "03/01/13 00:02:03", "04/03/13 00:10:15")
> sapply(strsplit(d, " "), `[`, 1)
# [1] "01/01/13" "03/01/13" "04/03/13"
And also, you could use stringr::word if you just want a character vector.
> library(stringr)
> word(d)
# [1] "01/01/13" "03/01/13" "04/03/13"
Here is an approach using a look around assertion:
vec <- c("01/01/13 00:00:00", "03/01/13 00:02:03", "04/03/13 00:10:15")
gsub(pattern = "(?=00).*$", replacement = "", vec, perl = TRUE)
[1] "01/01/13 " "03/01/13 " "04/03/13 "
The pattern looks for anything at the end of a string that begins with double 00, and removes it.
I have to extract values between a very peculiar feature in R. For eg.
a <- "{1:0987617820}{2:q312132498s7yd09f8sydf987s6df8797yds9f87098}
{3:{112:123123214321}}{4:20:asdasd3214213}"
This is my example string and I wish to extract text between {[0-9]: and } such that my output for the above string looks like
## output should be
"0987617820" "q312132498s7yd09f8sydf987s6df8797yds9f87098", "{112:123123214321}" "20:asdasd3214213"
This is a horrible hack and probably breaks on your real data. Ideally you could just use a parser but if you're stuck with regex... well... it's not pretty
a <- "{1:0987617820}{2:q312132498s7yd09f8sydf987s6df8797yds9f87098}
{3:{112:123123214321}}{4:20:asdasd3214213}"
# split based on }{ allowing for newlines and spaces
out <- strsplit(a, "\\}[[:space:]]*\\{")
# Make a single vector
out <- unlist(out)
# Have an excess open bracket in first
out[1] <- substring(out[1], 2)
# Have an excess closing bracket in last
n <- length(out)
out[length(out)] <- substring(out[n], 1, nchar(out[n])-1)
# Remove the number colon at the beginning of the string
answer <- gsub("^[0-9]*\\:", "", out)
which gives
> answer
[1] "0987617820"
[2] "q312132498s7yd09f8sydf987s6df8797yds9f87098"
[3] "{112:123123214321}"
[4] "20:asdasd3214213"
You could wrap something like that in a function if you need to do this for multiple strings.
Using PERL. This way is a bit more robust.
a = "{1:0987617820}{2:q312132498s7yd09f8sydf987s6df8797yds9f87098}{3:{112:123123214321}}{4:20:asdasd3214213}"
foohacky = function(str){
#remove opening bracket
pt1 = gsub('\\{+[0-9]:', '##',str)
#remove a closing bracket that is preceded by any alphanumeric character
pt2 = gsub('([0-9a-zA-Z])(\\})', '\\1',pt1, perl=TRUE)
#split up and hack together the result
pt3 = strsplit(pt2, "##")[[1]][-1]
pt3
}
For example
> foohacky(a)
[1] "0987617820"
[2] "q312132498s7yd09f8sydf987s6df8797yds9f87098"
[3] "{112:123123214321}"
[4] "20:asdasd3214213"
It also works with nesting
> a = "{1:0987617820}{{3:{112:123123214321}}{4:{20:asdasd3214213}}"
> foohacky(a)
[1] "0987617820" "{112:123123214321}" "{20:asdasd3214213}"
Here's a more general way, which returns any pattern between {[0-9]: and } allowing for a single nest of {} inside the match.
regPattern <- gregexpr("(?<=\\{[0-9]\\:)(\\{.*\\}|.*?)(?=\\})", a, perl=TRUE)
a_parse <- regmatches(a, regPattern)
a <- unlist(a_parse)