The question is very clear in the title.
As added to base in 3.3.0, startsWith (and endsWith) are exactly this.
> startsWith("what", "wha")
[1] TRUE
> startsWith("what", "ha")
[1] FALSE
https://stat.ethz.ch/R-manual/R-devel/library/base/html/startsWith.html
Not inbuilt like that.
Options include grepl and substr.
x <- 'ABCDE'
grepl('^AB', x) # starts with AB?
grepl('DE$', x) # ends with DE?
substr(x, 1, 2) == 'AB'
substr('ABCDE', nchar(x)-1, nchar(x)) == 'DE'
The dplyr package's select statement supports starts_with and ends_with. For example, this selects the columns of the iris data frame that start with Petal
library(dplyr)
select(iris, starts_with("Petal"))
select supports other subcommands too. Try ?select .
The simplest way I can think of is to use the %like% operator:
library(data.table)
"foo" %like% "^f"
evaluates as TRUE - Starting with f
"foo" %like% "o$"
evaluates as TRUE - Ending with o
"bar" %like% "a"
evaluates as TRUE - Containing a
Borrowing some code from the dplyr package [see this] you could do something like this:
starts_with <- function(vars, match, ignore.case = TRUE) {
if (ignore.case) match <- tolower(match)
n <- nchar(match)
if (ignore.case) vars <- tolower(vars)
substr(vars, 1, n) == match
}
ends_with <- function(vars, match, ignore.case = TRUE) {
if (ignore.case) match <- tolower(match)
n <- nchar(match)
if (ignore.case) vars <- tolower(vars)
length <- nchar(vars)
substr(vars, pmax(1, length - n + 1), length) == match
}
This is relatively simple by using the substring function:
> strings = c("abc", "bcd", "def", "ghi", "xyzzd", "a")
> str_to_find = "de"
> substring(strings, 1, nchar(str_to_find)) == str_to_find
[1] FALSE FALSE TRUE FALSE FALSE FALSE
You cut each string to the desired length with substring. The length being the number of characters you are looking for at the beginning of each string.
来源:https://stackoverflow.com/questions/31467732/does-r-have-function-startswith-or-endswith-like-python