Regex: Remove all match plus one char before all

[Alfalfa]

Hello,
I made this script to find all "|BS|" blocks in a string, and simulate a backspace, as if the string would be typed in a linetext input field. I came up with this:

#!/usr/bin/python3
import re

if __name__ == '__main__':
    string = "it |BS||BS||BS|this is one|BS||BS||BS|an example" #|BS| as in Backspace
    while re.search("\|BS\|", string):
        array = list(string)
        for m in re.finditer("\|BS\|", string):
            del array[m.start():m.end()]
            if m.start()-1 >= 0:
                del array[m.start()-1]
            string = ''.join(array)
            break
    print(string)

which output "this is an example"

It works, but feels very inefficient and I would like to improve it, avoiding the needs for an array or using a single pass regex. I looked at the documentation for re but yet I could not find a satisfying answer. Hopefully you can give me some tips 

Thanks!

[buran]

So actually you want to remove all |BS| as well as one or more optional spaces and one word before that?

---

#!/usr/bin/python3
import re

string = "it |BS||BS||BS|this is one|BS||BS||BS|an example" #|BS| as in Backspace
ptrn = re.compile(r'\w* ?\|BS\| ?') # (r'(\w* ?)(?=\|BS\|)(?<=[\w* ?])(\|BS\|)*') - this was the initail pattern
print(re.sub(ptrn, '', string))

pattern may not be the best, someone else may suggest something better

[Alfalfa]

Thanks for your reply! I want to remove all |BS|, as well as the optional character preceeding it.
Like so:

"it |BS||BS||BS||BS|this is one|BS||BS||BS|an example"
"it|BS||BS||BS|this is one|BS||BS||BS|an example"
"i|BS||BS|this is one|BS||BS||BS|an example"
"|BS|this is one|BS||BS||BS|an example"
"this is one|BS||BS||BS|an example"
"this is on|BS||BS|an example"
"this is o|BS|an example"
"this is an example"

[buran]

I don't get it. your code takes it |BS||BS||BS|this is one|BS||BS||BS|an example and returns this is an example. You said it works (i.e. that's the desired result) but inefficient. I give you more efficient code using single line re.sub. So now you display something totally different as desired output. Show us your code and ask specific question.

---

to make it more clear:

#!/usr/bin/python3
import re

string = "it |BS||BS||BS|this is one|BS||BS||BS|an example" #|BS| as in Backspace
ptrn = re.compile(r'\w* ?\|BS\| ?')
print(re.sub(ptrn, '', string))
print ('\n---- all matches follow----\n')
for match in re.finditer(ptrn, string):
    print(match.group())

and the output:

Output:
this is an example

---- all matches follow----

it |BS|
|BS|
|BS|
one|BS|
|BS|
|BS|

[Alfalfa]

Sorry for the misunderstanding, your solution work well. It is 5 times faster than the code in the original post. The quote I made was only about explaining what I want to do, in case someone would think of something better, as you suggested in your post.  However thank you again for your help, the code you provided is what I was looking for.

---

Oh I just noticed your pattern delete the whole word instead of a character.. Like I said I need to simulate a backspace event in a text field, so one |BS| should only remove a single (optionnal) char.

[buran]

Oh I just noticed your pattern delete the whole word instead of a character.. Like I said I need to simulate a backspace event in a text field, so one |BS| should only remove a single (optionnal) char.

well, you don't delete single character, you delete it - that's 3 characters and also one - that's 3 characters...

[Alfalfa]

string = "it |BS||BS||BS|this is one|BS||BS||BS|an example"

sorry I should have been more explicit. there are 3 |BS| there, and 3 char to remove.

[buran]

I think there is some misunderstanding. Show us your input string. Does it really have |BS| strings in it?

---

do you actually want to remove as much characters(incl. spaces) before the |BS| sequence as many |BS|? The best would be to show input string, your code and the result. and ask specific question.

[Alfalfa]

The input string really have |BS| in it, but it's content can correspond to anyning, so there is no typical string I can provide. Following the example I showed earlier, it should behave as follow:

input: "it |BS|this is an example"
outpt: "itthis is an example"

input: "it |BS||BS|this is an example"
outpt: "ithis is an example"

input: "it |BS||BS||BS|this is an example"
outpt: "this is an example"

input: "it |BS||BS||BS||BS|this is an example"
outpt: "this is an example"

[buran]

so actually the requested result is "For each "|BS| group remove as many chars before that group as many times |BS| is present in that group"

#!/usr/bin/python3
import re

strings = ['it |BS||BS||BS|this is one|BS||BS||BS|an example',
           'it |BS|this is an example',
           'it |BS||BS|this is an example',
           'it |BS||BS||BS|this is an example',
           'it |BS||BS||BS||BS|this is an example']
ptrn = re.compile(r'[\w ]?\|BS\|')
for string in strings:
   print(string)
   while True:
       after_sub = ptrn.sub('', string, count=1)
       if string == after_sub:
           break
       else:
           string = after_sub
   print(string)
   print('\n')

Output:
it |BS||BS||BS|this is one|BS||BS||BS|an example
this is an example


it |BS|this is an example
itthis is an example


it |BS||BS|this is an example
ithis is an example


it |BS||BS||BS|this is an example
this is an example


it |BS||BS||BS||BS|this is an example
this is an example