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