Marked text manipulation

Scott Sumner

How do I copy all them in one click?

In one click? Not possible [to the best of my knowledge…always have to add that disclaimer :-) ].

The closest thing I can think of would be “one keypress”, which would involve tying the Pythonscript I provided before to a shortcut-mapper key-combination, so that it gets executed when the key-combo is pressed.

Suncatcher

@guy038 bookmark feature copies the whole line, and I need only the certain pieces of line, as you can see from the screenshot.
If I used bookmarking it’ll copy all lines (because I have matches on each line) and that’s not what I want. I want just highlighted pieces.

@Scott-Sumner could you provide the sample of such script?

Scott Sumner

@Suncatcher

In my immediately previous post, I said “the Pythonscript I provided before”. What I meant by that, if it was unclear, was to scroll up in this thread to a previous posting by me. It starts with “At one time I had a need to do” and contains the code I was talking about.

guy038

Hi, Suncatcher,

OK, I see what you mean ! But, how would you like that the extracted matches were displayed and which separator do you need between these different matches ?

For instance, from your picture :

In line 2, do you want : “134.170.110.” OR “134. 170. 110.” OR, even, the string “134 170 110”, without dot OR else ?
In line 4, do you want : “157.240.” OR “157. 240.” OR some spaces, to indicate the missing second group as “157. 240.” OR else ?
In line 13 do you want: “185.222.185.223.” OR with a dash to show it’s from a range of IPV4 addresses, as “185.222. - 185.223.” OR else ?

See you soon,

Cheers,

guy038

Suncatcher

@Scott-Sumner I beg my pardon! I am so inattentive. The script makes exactly what I want. Thank you a lot!

Suncatcher

@guy038 Ideally I want the highlighted pieces to be copied exactly like they are placed on initial page, with the exact amount of spaces as before as a separator.
For example, from line 2 should be copied:

 134.170.110.

from line 12:

 185.   220.

from line 13:

 185.   222.    185.   223.

guy038

Suncatcher,

From what you said, I used the FOUR following rules :

If a line does NOT contain a string of THREE digits, followed by a DOT, this line is completely deleted
Any range of characters, ENDING a line, which does NOT contain a string of THREE digits, followed by a DOT, is deleted, too
Any string of THREE digits, followed by a DOT, is UNCHANGED
Any other single character is REPLACED by a single SPACE character

This leads to the following S/R, with the Regular expression search mode CHECKED :

SEARCH (?-s)^\R|(?!.*\d{3}\.).+|(\d{3}\.)|(.)

REPLACE (?1\1)(?2 )

For instance, from this example text, below :

134.170.110.48
85.33.98.0 - 85.33.99.255
185.33.220.38
200.25.6.78
65.55.52.23
5.155.52.23
12.3.8.145
185.33.220.0 - 185.33.223.255
1.23.137.2
1.2.3.4
25.155.52.153
67.42.95.0 - 67.42.95.99
31.53.61.99 - 31.53.61.100
58.33.99.0 - 58.33.101.1

We would get the replaced text, below :

134.170.110.
185.   220.
200.
  155.
185.   220.    185.   223.
     137.
   155.
                   101.

As you may notice, as expected, some lines have been deleted :

The lines without any three consecutive digits, at all
The lines with, ONLY, one block of three digits, at the END of a line, without the final DOT character

REMARK :

If you prefer to keep a blank line, in case NO block of three consecutive digits exists, in a line, just change the search regex to :

SEARCH (?-s)(?!.*\d{3}\.).+|(\d{3}\.)|(.)

This time, you would obtain :

134.170.110.

185.   220.
200.

  155.

185.   220.    185.   223.
     137.

   155.


                   101.

Best Regards,

guy038

Suncatcher

@guy038 well, RegExp approach is not bad but seems to be disposable. I tried to adapt it to another pattern and that’s what I’ve got.
I set the pattern which consists of dot and two digits.

And rewrote your expressions accordingly - (?-s)(?!.*\.\d{2}\.).+|(\.\d{2}\.)|(.)
And that’s what it output to me

And that is obviously not the thing supposed to be there. To achieve our aim regexp should be rewritten every time. Pythonscript seems to be more universal and consistent approach.

guy038

Hi, Suncatcher and Scott,

Ah… Of course, my previous regex was much to closed to, your specific regex \d{3}\. . In addition, I tried, hard, from what you said, to keep the exact position where the different matches were !

So, I decided to run the Scott script to see how this script re-writes the different matches :-) BTW, Scott, very nice script for people, who does not worry about regex problems or details ;-))

So, starting from my original example text, below :

134.170.110.48
85.33.98.0 - 85.33.99.255
185.33.220.38
200.25.6.78
65.55.52.23
5.155.52.23
12.3.8.145
185.33.220.0 - 185.33.223.255
1.23.137.2
1.2.3.4
25.155.52.153
67.42.95.0 - 67.42.95.99
31.53.61.99 - 31.53.61.100
58.33.99.0 - 58.33.101.1

The Scott’s script, put the following text, in the clipboard, in that form :

.17
.11
.48
.33.98
.33.99.25
.33.22
.38
.25
.78
.55.52.23
.15
.52.23
.14
.33.22
.33.22
.25
.23.13
.15
.52.15
.42.95
.42.95.99
.53.61.99
.53.61.10
.33.99
.33.10

From this modified text, we are able to deduce that the script follows TWO main rules :

If two or more matches are adjacent, there are rewritten as a single unit, in a same line
As soon as two consecutive matches, are NON adjacent, there are displayed in two consecutive lines

Quite different than before, isn’t it ?!

I, then, realized that you can start, with my general regex S/R , exposed in my THIRD post, on that topic, which is :

SEARCH (?s)^.*?(Your regex to match)|(?s).*\z

REPLACE (?1\1\r\n)

With a minor modification, this new general S/R, below, will adopt the same output displaying, than the Scott’s script :-))

SEARCH (?s)^.*?((?:Your regex to match)+)|(?s).*\z ( GENERAL regex syntax )

REPLACE (?1\1\r\n)

So, if we use your second regex \.\d{2}, that we insert in the general regex syntax, above, we obtain the practical S/R:

SEARCH (?s)^.*?((?:\.\d{2})+)|(?s).*\z

REPLACE (?1\1\r\n)

which gives, after replacement, the expected output text, identical to Scott’s script one !

NOTES :

Compared to my THIRD post, the second part, of this new general regex, has changed into : ((?:Your regex to match)+) :
- It represents any consecutive and adjacent matches of your regex, which is stored as group 1 and output on a single line, followed by a line break
- The inner parentheses , (?:Your regex to match) , stands for a non-capturing group, containing a single match of your regex
The IMPORTANT and P.S. sections, of my THIRD post, are still pertinent

Best Regards,

guy038

P.S. :

I did an other tests, with your first regex \d{3}\. and, also, with the simple regex \d\.\d, leading to the appropriate following S/R :

SEARCH (?s)^.*?((?:\d\.\d)+)|(?s).*\z

REPLACE (?1\1\r\n)

After replacement, the resulting text is, as expected :

4.1
0.1
0.4
5.33.98.0
5.33.99.2
5.33.2
0.3
0.25.6
5.55.52.2
5.1
5.52.2
2.3
8.1
5.33.2
0.0
5.33.2
3.2
1.23.1
7.2
1.2
3.4
5.1
5.52.1
7.42.95.0
7.42.95.9
1.53.61.9
1.53.61.1
8.33.99.0
8.33.1
1.1

and corresponds, exactly, to the text, generated by the Scott’s script :-))

Scott Sumner

So I’m currently trying to copy some multi-line (red)marked text out of a large (~70MB) file, and my Pythonscript technique for doing so (see earlier posting in this thread) works but is super-slow on a large file; it iterates through the file one position at a time (pos += 1). Is there a faster way to code it, given the functions we have at our disposal for doing this? @Claudia-Frank , ideas? :-)

dail

@Scott-Sumner

See next post, this is wrong!

AFAIK using indicatorStart() and indicatorEnd() is quite efficient finding marked locations. The code you posted above doesn’t seem to be utilizing this as efficiently as it could. I have no way of testing this following code but you should be able to do something like this:

start = 0
end = 0
while True:
	start = editor.indicatorStart(SCE_UNIVERSAL_FOUND_STYLE, end)
	if start == 0:
		break
	end = editor.indicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, start)
	accum_text += editor.getTextRange(start, end) + '\r\n'

~~Again this hasn’t been tested so there may be corner cases you need to check for…such as using start + 1 when calling indicatorEnd() but this is the gist of it.~~

dail

@Scott-Sumner

Woops sorry about the above, it was way off. Here is a small LuaScript which works (I’m sure you can easily translate it into Python)

SCE_UNIVERSAL_FOUND_STYLE = 31
start = editor:IndicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, -1)
while start ~= 0 and start ~= editor.Length do
	endd = editor:IndicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, start)
	print(editor:textrange(start, endd))
	start = editor:IndicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, endd)
end

Note: The one major initial bug I know if is that it is incorrect if the very first character of the file is marked.

Claudia Frank

@dail, @Scott-Sumner

This is strange, isn’t it? You have to use IndicatorEnd to find the start position but it is like it is…

Cheers
Claudia

dail

@Claudia-Frank

Yeah I ran into this as well when modifying my DoxyIt plugin…the way I came to think of it now is that it finds the end of the range you specify by pos. And technically a range that is not marked has an end…which is the start of the range you want…oh well :)

Claudia Frank

@dail

yeah, :-) sounds … logical … some how . … still confusing :-)
And what makes it confusing even more, what you already said, is, that if you do
editor.indicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, -1) you will get the end position.
Aahhhh :-D

What I meant is about

Note: The one major initial bug I know if is that it is incorrect if the very first character of the file is marked.

Cheers
Claudia

Scott Sumner

@dail , @Claudia-Frank :

Thanks for your inputs, I used the basic ideas but came up with my own Pythonscript version that is much faster than my original PS version on large files, and seems to correctly handle the oddities of the editor.indicatorEnd() function previously mentioned.

So here is RedmarkedTextToClipboard2.py:

def RTTC2__main():

    SCE_UNIVERSAL_FOUND_STYLE = 31  # N++ red-"mark" feature highlighting style indicator number
    ind_end_ret_vals_list = []
    ierv = 0
    while True:
        ierv = editor.indicatorEnd(SCE_UNIVERSAL_FOUND_STYLE, ierv)
        # editor.indicatorEnd() returns 0 if no redmarked text exists
        # editor.indicatorEnd() returns last pos in file if no more redmarked text beyond the 'ierv' argument value
        if ierv == 0 or len(ind_end_ret_vals_list) > 0 and ierv == ind_end_ret_vals_list[-1]: break
        ind_end_ret_vals_list.append(ierv)
    if len(ind_end_ret_vals_list) > 0:
        if editor.indicatorValueAt(SCE_UNIVERSAL_FOUND_STYLE, 0) == 1:
            # compensate for weirdness with editor.indicatorEnd() when a match starts at the zero position
            zero = 0; ind_end_ret_vals_list.insert(0, zero)  # insert at BEGINNING of list
        if editor.indicatorValueAt(SCE_UNIVERSAL_FOUND_STYLE, ind_end_ret_vals_list[-1]) == 0:
            # remove end-of-file position unless it is part of the match
            ind_end_ret_vals_list.pop()
    start_end_pos_tup_list = zip(*[iter(ind_end_ret_vals_list)]*2)  # see https://stackoverflow.com/questions/14902686/turn-flat-list-into-two-tuples
    accum_text = ''
    for (start_pos, end_pos) in start_end_pos_tup_list:
        accum_text += editor.getTextRange(start_pos, end_pos) + '\r\n'
    if len(accum_text) > 0: editor.copyText(accum_text)  # put results in clipboard

RTTC2__main()

Claudia Frank

@Scott-Sumner

Hi Scott,
a nice one - good performance improvement. :-)

If you are still looking for performance increase,
a general suggestion would be to use as less global objects as possible within a loop as
the cost of loading global is expensive.
Cache global objects at the beginning of the script.
Creating the tuple list from the beginning should be faster than creating from a flat list.
Meaning do your two indicatorEnd calls and create a tuple from the results which than
gets added to a list.
Maybe a list comprehension to create the accum_text is faster as well - but not really sure
as it would need to call the global object.
All in all I assume this might make it faster up to 3-5%, not sure if it is worth thinking about it.

Cheers
Claudia

Scott Sumner

@Claudia-Frank

not sure if it is worth thinking about it.

I should have posted my before-and-after timing, but really, the “before” was “forever” on my 70MB data file! The “after” was extremely quick, certainly on par with how long it took Notepad++'s Mark feature to redmark my desired text. Therefore performance was rated “very acceptable” for the new version. And that’s really all the performance I care about, so further optimizations aren’t worth it to me. Probably some of those optimizations you suggest would make the code less readable, too, so I’m definitely not wanting to go there (although now I leave myself open to comments on how readable/unreadable the existing code is). :-D

I probably would have written this better the first time around if how these “indicator” functions worked was better documented!

Until Notepad++ natively allows a non-destructive (@guy038’s regex method is destructive…but there is UNDO…hmmm) copy of all regex-matched text, this little script will serve me nicely, now on all files big and small.

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Kashif Rana

@Scott-Sumner

I am in the same situation but regular expression method is not working for me to copy match text.

I want to grab all occurrences in configuration file where first line starts from ‘object’ and immediately second line starts with ‘nat’

object network obj_any
nat (inside,outside) dynamic interface
object network obj-test
nat (DMZ1,outside) static 10.206.49.180
object network obj-192.168.236.200
nat (DMZ1,outside) static 10.206.74.60
object network obj-192.168.236.8
nat (DMZ1,outside) static 10.206.49.183 tcp 8080 80
object network obj-192.168.236.9
nat (DMZ1,outside) static 10.206.49.178 tcp 1002 22
object network obj-192.168.236.10
nat (DMZ1,outside) static 10.206.49.178 tcp 8080 80
object network obj-192.168.236.13
nat (DMZ1,outside) static 10.206.74.58 dns

I wrote regular expression ^object.\R\snat.* to grab both lines
starting with ‘object’ and with ‘nat’ but when I am replacing it with
(?1\1), it is deleting the matched lines. Any dea what could be the correct replace string to keep only matced two lines