I was using python re
library and came across the following behavior.
>>> import re
>>> re.sub(pattern=".*", repl="r", string="hello")
'rr'
As you can see, for the pattern .*
and the replacement character( r
) re.sub
method returning rr
. But I was expecting the result as r
because .*
would match the entire string. Why is that?. I have also tested the same logic in Go but it was returning expected result.
package main
import (
"fmt"
"regexp"
)
func main() {
re := regexp.MustCompile(`.*`)
fmt.Println(re.ReplaceAllString("Hello", "r")) // Will print `r`
}
The following should start explaining what's going on:
>>> re.sub("x?", "_", "hello")
'_h_e_l_l_o_'
At every position in the string re.sub
tries to match x?
. It succeeds, because x?
can match the empty string, and replaces the empty string with _
.
In a similar fashion, in the following
>>> re.sub(".*", "r", "hello")
'rr'
we have that re.sub
tries to match .*
in position 0, succeeds, and consumes the whole string. Then it tries to match at the end position, succeeds (matching the empty string) and replaces it with r
again. The 'puzzling' behavior goes away if you disallow the empty match:
>>> re.sub(".+", "r", "hello")
'r'
In versions prior to Python 3.7 if re.sub
consumed the whole string it would then not try to match at the end again, whereas in Python 3.7+ it does. To be more specific, quoting the documentation of re.sub
:
Changed in version 3.7 : Empty matches for the pattern are replaced when adjacent to a previous non-empty match.
>>> matches = lambda r, s: [m.span() for m in re.finditer(r, s)]
>>> matches("x?", "x")
[(0, 1), (1, 1)]
>>> matches("x?", "y")
[(0, 0), (1, 1)]
>>> re.sub("x?", "r", "x")
'rr'
>>> re.sub("x?", "r", "y")
'ryr
>>> matches("x?", "x")
[(0, 1), (1, 1)]
>>> matches("x?", "y")
[(0, 0), (1, 1)]
>>> re.sub("x?", "r", "x")
'r'
>>> re.sub("x?", "r", "y")
'ryr'
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