I would like to group all similar domains, not by alphabet.

guy038

This is the expected behaviour because these addresses contain 5 sections :

anaisdormoy.com.google.accounts.com
<----1----> <2>.<-3--> <--4---> <5>

And my regex works ONLY IF up to 4 dots only !

So, as I asked you in my previous post, just do the second part steps to determine how many sections contains your file !

See you later !

BR

guy038

Mohammad Al Thobiti

Could you tell us which is this number?

The number is 13 dots!

guy038

Hello @mohammad-al-thobiti and All,

Ah… OK !! So, give me some time to find out the correct regex S/R which could handle and revert up to 13 sections !!

See you later,

Best Regards

guy038

Coises

This post is deleted!

PeterJones

@Coises said in I would like to group all similar domains, not by alphabet.:

Commas are not allowed in domain names, so we can use a comma to distinguish between forward and reversed domain names

… which is why I used the historical standard of exclamation points for reversed domain names, rather than introducing ambiguity by re-using the period.

, and reverse them one part at a time:

I do like the way that simplifies the regex, to make it much more understandable and generic, at the expense of making the user click Replace All up to 13 times.

Ugh. You deleted your post while I was replying. It had good information. I am hoping you are going to re-post a slightly rephrased version eventually.

@guy038 said,

So, give me some time to find out the correct regex S/R which could handle and revert up to 13 sections !!

That’s why I just gave the generic format in my original regex, and explained how @Mohammad-Al-Thobiti could extend the idea to as many groups as was desired, because I had a feeling in my reply that the original three-section solution wasn’t going to be enough. Taking my generic formula and just appending copies of the two tokens that I supplied would have worked up to 9 capture groups – and I was hoping that the other regulars here would have let the OP try to learn from the example, rather than spoonfeeding.

And using the ${ℕ} substitution syntax and (?{ℕ}...) conditional replacement instead of $ℕ and (?ℕ...) in the replacement would have allowed for ten or more groups. Which is what I would have suggested when the OP came back with the inevitable “but what if I want more groups than 9? when I tried to 13 groups, it didn’t work”. Because my expression didn’t use nesting, it wouldn’t require any fancy thought on the part of the user, just literally copy/pasting more of the same sort of token, and the ability to interpret that they needed to count up with each ℕ in the replacement.

But @Coises’ currently-deleted suggestion of just doing a single pair replacement, run many times, would be even simpler to understand than my original suggestion, without the {ℕ} requirements. So I hope @Coises re-posts that solution once he’s comfortable with the wording, because it’s the best solution for easy extending to as many domain pieces as needed.

Coises

@PeterJones said in I would like to group all similar domains, not by alphabet.:

Ugh. You deleted your post while I was replying.

I am sorry about that. It contained a good idea, implemented incorrectly. My “solution” rotated the parts of the domain name rather than reversing them.

I didn’t know about exclamation points being a standard; I’ll use those instead of commas when I figure out how to do this correctly.

I also realized the original poster probably wants domains like xxx.com and xxx.org and xxx.co.uk to sort together, which adds an extra complexity.

PeterJones

@Coises said in I would like to group all similar domains, not by alphabet.:

I also realized the original poster probably wants domains like xxx.com and xxx.org and xxx.co.uk to sort together, which adds an extra complexity.

I think your solution, without that, is sufficient for any reasonable need. If the OP desires that complexity, they can take what we’ve already given them and read the documentation that I linked them to, and figure out the next level themselves. (But if you really want to spend your time on that, I’d recommend doing initial searches from blah.com, blah.co.uk, etc, and turn those into blah,com and blah,co,uk; then use ! as the machine separator. That way it will sort first by the blah, then by any more specific things above, which would keep blah.com and blah.co.uk near each other)

Coises

@Mohammad-Al-Thobiti

Since there are so many levels, and you’re working with a temporary file anyway, we could make this less tricky. At @PeterJones’ suggestion, here we use an exclamation point to distinguish between forward and reversed domain names.

Enter:

Find what : ([^.\s]+)\.([^!\s]+)
Replace with : \2!\1

Replace All repeatedly until it says 0 occurrences were replaced, then sort the file.

If you need to change back to normal domain names after the sort, use:

Find what : ([^!\s]+)!([^.\s]+)
Replace with : \2.\1

and Replace All repeatedly until it says 0 occurrences were replaced.

The steps above will sort first by the top-level domain (.com, .net, etc.). If you need to have, say, whatever.com and whatever.net and whatever.co.uk sort together, then after reversing the domain names, use something like:

Find what : ^([a-z]{2}![a-z]{2}|[^!]+)!(\S+)
Replace with : \2 \1

and Replace All once before sorting. (This is not guaranteed to be correct for every case of two-letter top level domains, but it should get the common ones right.)

To reverse, after sorting, use:

Find what : ^(\S+) (\S+)
Replace with : \2!\1

guy038

Hi, @mohammad-al-thobiti, @peterjones, @coises and All,

Ah, ah ah… I’m very happy to announce that I’ve found out a general regex which can handle any number of sections ;-))

So, let’s begin with this simple INPUT text containing from 2 to 13 sections ( one of each ), pasted in a new tab :

abc.def
abc.def.ghi
abc.def.ghi.jkl
abc.def.ghi.jkl.mno
abc.def.ghi.jkl.mno.pqr
abc.def.ghi.jkl.mno.pqr.stu
abc.def.ghi.jkl.mno.pqr.stu.vwx
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789.€±¶
abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789.€±¶.Ø÷ß

Move to the very beginning of the file ( Ctrl + Home )

First, we add the |. string at the beginning of every line, with the following regex S/R :

SEARCH (?x-s) ^ (?= . )
REPLACE |.

Thus, we get :

|.abc.def
|.abc.def.ghi
|.abc.def.ghi.jkl
|.abc.def.ghi.jkl.mno
|.abc.def.ghi.jkl.mno.pqr
|.abc.def.ghi.jkl.mno.pqr.stu
|.abc.def.ghi.jkl.mno.pqr.stu.vwx
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789.€±¶
|.abc.def.ghi.jkl.mno.pqr.stu.vwx.yz0.123.456.789.€±¶.Ø÷ß

Move to the very beginning of the file ( Ctrl + Home )

Now, this is the main regex S/R :

SEARCH (?x-s) ^ ( .* \| ) ( (?: \. (?: (?! \| ) \S )+ )+ ) ( \. (?: (?! \. ) \S )+ )
REPLACE \1\3|\2
Click 14th times on the Replace All button, till you get the message Replace All: 0 occurrences were replaced from caret to end-of-file

=> You should get the temporary text :

|.def|.abc
|.ghi|.def|.abc
|.jkl|.ghi|.def|.abc
|.mno|.jkl|.ghi|.def|.abc
|.pqr|.mno|.jkl|.ghi|.def|.abc
|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.123|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.456|.123|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.789|.456|.123|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.€±¶|.789|.456|.123|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc
|.Ø÷ß|.€±¶|.789|.456|.123|.yz0|.vwx|.stu|.pqr|.mno|.jkl|.ghi|.def|.abc

Move to the very beginning of the file ( Ctrl + Home )

Now, we just to get rid of all the | chars as well as the FIRST dot of each line, with the regex S/R :

SEARCH (?x) ^ \| \. | \|

REPLACE Leave EMPTY

And we get our expected OUTPUT text :

def.abc
ghi.def.abc
jkl.ghi.def.abc
mno.jkl.ghi.def.abc
pqr.mno.jkl.ghi.def.abc
stu.pqr.mno.jkl.ghi.def.abc
vwx.stu.pqr.mno.jkl.ghi.def.abc
yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc
123.yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc
456.123.yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc
789.456.123.yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc
€±¶.789.456.123.yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc
Ø÷ß.€±¶.789.456.123.yz0.vwx.stu.pqr.mno.jkl.ghi.def.abc

So, @mohammad-al-thobiti :

Apply all the above steps against your real file
Run the Edit > Line Operations > Sort Lines Lexicographically Ascending option ( No [rectangular] selection is needed as you just keep the addresses )
Possibly, add back the 127.0.0.1 IPV4 address, followed with two space chars with the regex S/R :
- SEARCH (?x-s) ^ (?= . )
- REPLACE (127.0.0.1 )

Best Regards,

guy038

P.S. :

I’ve just seen the @coises’s solution. I’m going to have a look at its solution which could be more simple than my regex solution !

guy038

Hi, @mohammad-al-thobiti, @peterjones, @coises and All,

@mohammad-al-thobiti, you should use the @coises’s approach, which works much better than mine !!

In addition, in my previous post, I omitted to add the reverse regex to use, once your sort would be done :-((

While the @coise’s method avoids any additional S/R and correctly mentions the reverse regex to run after the sort operation

Note that I slightly Modified the first two @coises’s regexes to get more rigorous ones ( See, at the end of this post the reason for these changes )

Thus, I would propose this road map :

First, from the @coises’s post, I would use this alternate regex formulation :
- SEARCH (?x) ( [^.\r\n]+ ) \. ( [^!\r\n]+ )
- REPLACE \2!\1
- Click 14th times on the Replace All button
Run the Edit > Line Operations > Sort Lines Lexicographically Ascending option ( No [rectangular] selection is needed as you just keep the addresses )
Thirdly, once the sort done, from the @coises’s post, use the alternate reverse regex S/R :
- SEARCH (?x) ( [^!\r\n]+ ) ! ( [^.\r\n]+ )
- REPLACE \2.\1
- Click 14th times on the Replace All button

=> You should get all your addresses back, in the right order

Best Regards,

guy038

P.S. :

@coises uses the \s class of characters, which is equivalent to any of the 25 characters, below, with the regex :

(?x) \t | \n | \x{000B} | \x{000C} | \r | \x{0020} | \x{0085} | \x{00A0} | x{1680} | [\x{2000}-\x{200B}] |\x{2028} | \x{2029} | \x{202F} | \x{3000}

In the highly unlikely event that one of these characters is included in some addresses, I preferred to use the [\r\n] regex, which ONLY avoids these 2 EOL chars in addresses, instead of using the \s regex !

Mohammad Al Thobiti

Thank you for your efforts, my friends.
I would like to tell you that the result is excellent.

But an idea came to me: why not just delete the subdomain and keep only the main domain and then delete the similar ones?
Is there a way to delete long link extensions? And keep the main domain?

Example:

Mohammad Al Thobiti

guy038

Hi, @mohammad-al-thobiti and All,

May I rephrase your question ? Let’s see if we mean the same goal !

So, for example, from the INPUT text, below :

abc.def.ghi.jkl.example.com
all.net
abc.def.example.com
abc.my_site.com
abc.def.ghi.all.net
my_site.com
abc.def.all.net
abc.def.ghi.jkl.mno.opq.my_site.com
example.com

With the following regex S/R :

SEARCH (?x) ^ (?: [\w-]+ \. )* ( [\w-]+ \. [\w-]+ ) $

REPLACE \1

We would get that text :

example.com
all.net
example.com
my_site.com
all.net
my_site.com
all.net
my_site.com
example.com

Then, using the Edit > Line Operations > Remove Duplicates Lines option, we would end up with this OUTPUT :

example.com
all.net
my_site.com

If this is exactly what you expect to, just go ahead !

BR

guy038

Mohammad Al Thobiti

I would like to thank you for your useful information and those who contributed to this topic.

This article has become my reference. It works nicely. Yes, we mean the same goal !

Right now. If you can sort or collect them from the most similar domains, please let me know.

There are too many lines, starting with the most similar domain.

Reason: To find the most “worried” domains because it takes many lines. I will delete them later, as you told us. However, the goal is to discover more domains with many characters or long URLs and I will block them in other programs.

How can this be done?

Coises

@Mohammad-Al-Thobiti said in I would like to group all similar domains, not by alphabet.:

If you can sort or collect them from the most similar domains, please let me know.

Here’s a way; enter:

Find what : ^(!*([^\r\n]+))\R(!*)\2$
Replace with : !\3\1

and Replace All repeatedly until there are no more changes.

Each line in the result will have an exclamation point at the beginning for each additional occurrence of the following text; so:

argh.com
argh.com
asdf.net
asdf.net
asdf.net
asdf.net
ef.org
ef.org
ef.org
ef.org
ef.org
fasde.com
fasde.com
fasde.com
fasde.com
gorch.net
gorch.net
gorch.net

would become:

!argh.com
!!!asdf.net
!!!!ef.org
!!!fasde.com
!!gorch.net

You can then sort that to put them in order by the number of exclamation points at the beginning.

Mohammad Al Thobiti

Thank you. I will clarify the issue.

Coises

@Mohammad-Al-Thobiti said in I would like to group all similar domains, not by alphabet.:

Thank you. I will clarify the issue.

Using the method I suggested, if you start with:

zindova.net
zindova.net
zinfandelreviews.net
zinfandelreviews.net
zingardi.net
zingcoach.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
ziobaweek.net
ziomik.net
zioninfosystems.net
zioninfosystems.net
zip-archive.net

and do the repeated Replace Alls, then sort, you’ll get:

!!!!!!!!!!!!!!!!!!!!!!!!zinoiosijek031.net
!zindova.net
!zinfandelreviews.net
!zioninfosystems.net
zingardi.net
zingcoach.net
ziobaweek.net
ziomik.net
zip-archive.net

The first line represents 25 occurrences — there are 24 leading exclamation points. Each of the next three lines represent two occurrences (one leading exclamation point). The remaining lines occurred only once.

Is that not what you needed to accomplish?

guy038

Hello, @mohammad-al-thobiti, @peterjones, @coises and All,

Oh, @coises, your method of finding out how many times each address occurs, is very elegant and really clever ! I’d never have thought of such sophistication on my own :-((

So, @mohammad-al-thobiti, starting with this INPUT file :

zindova.net
zindova.net
zinfandelreviews.net
zinfandelreviews.net
zingardi.net
zingcoach.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
zinoiosijek031.net
ziobaweek.net
ziomik.net
zioninfosystems.net
zioninfosystems.net
zip-archive.net
zip-archive.net
zip.net
zip.net
zip.net
zipaphoto.net
zipbah.net
zipexpose.net
ziph.next
ziph.next

I’m using the @coises’s regex, that I slightly modified :

SEARCH (?x-s) ^ ( !* ( .+ ) ) \R ( !* ) \2 $

REPLACE !\3\1

Run it several times, in order tp get the message Replace All: 0 occurrences were replaced

=> You should get this text :

!zindova.net
!zinfandelreviews.net
zingardi.net
zingcoach.net
!!!!!!!!!!!!!!!!!!!!!!!!zinoiosijek031.net
ziobaweek.net
ziomik.net
!zioninfosystems.net
!zip-archive.net
!!zip.net
zipaphoto.net
zipbah.net
zipexpose.net
!ziph.next

Now, we just add one ! character to get the exact number of occurrences of each address

SEARCH (?x-s) ^ (?= . )

REPLACE !

!!zindova.net
!!zinfandelreviews.net
!zingardi.net
!zingcoach.net
!!!!!!!!!!!!!!!!!!!!!!!!!zinoiosijek031.net
!ziobaweek.net
!ziomik.net
!!zioninfosystems.net
!!zip-archive.net
!!!zip.net
!zipaphoto.net
!zipbah.net
!zipexpose.net
!!ziph.next

Then, run the Edit > Line Operations > Sort Lines Lexicographically Ascending

=> We get this sorted text :

!!!!!!!!!!!!!!!!!!!!!!!!!zinoiosijek031.net
!!!zip.net
!!zindova.net
!!zinfandelreviews.net
!!zioninfosystems.net
!!zip-archive.net
!!ziph.next
!zingardi.net
!zingcoach.net
!ziobaweek.net
!ziomik.net
!zipaphoto.net
!zipbah.net
!zipexpose.net

Finally, runing the four regex S/R, below :

SEARCH  (?x-s) ^ !                         (?= [^!\r\n]+ )        REPLACE   1\t
SEARCH  (?x-s) ^ !!                        (?= [^!\r\n]+ )        REPLACE   2\t
SEARCH  (?x-s) ^ !!!                       (?= [^!\r\n]+ )        REPLACE   3\t
SEARCH  (?x-s) ^ !!!!!!!!!!!!!!!!!!!!!!!!! (?= [^!\r\n]+ )        REPLACE  25\t

=> You should end up with this OUTPUT text :

25	zinoiosijek031.net
3	zip.net
2	zindova.net
2	zinfandelreviews.net
2	zioninfosystems.net
2	zip-archive.net
2	ziph.next
1	zingardi.net
1	zingcoach.net
1	ziobaweek.net
1	ziomik.net
1	zipaphoto.net
1	zipbah.net
1	zipexpose.net

Which should be a practical document to exploit !

Best Regards,

guy038

Mohammad Al Thobiti

@ guy038 and All,

Thank you my friends for the amazing results.

We have received the correct information. The most commonly used domains cleverly.

In any case, as you can see in the picture, the outcome is wonderful.

How do I write the code like this that you mentioned:

SEARCH  (?x-s) ^ !                         (?= [^!\r\n]+ )        REPLACE   1\t
SEARCH  (?x-s) ^ !!                        (?= [^!\r\n]+ )        REPLACE   2\t
SEARCH  (?x-s) ^ !!!                       (?= [^!\r\n]+ )        REPLACE   3\t
SEARCH  (?x-s) ^ !!!!!!!!!!!!!!!!!!!!!!!!! (?= [^!\r\n]+ )        REPLACE  25\t

Did I write until I reached Col: 6,597? I think it is difficult and takes a lot of time. And I want to thank all of you for all your efforts on this fascinating topic.

Coises

@guy038 OK, I’ll take the bait… If you really want to count exclamation points:

Add the missing exclamation point, but also add a separator:

Find what : ^(!*)
Replace with : $1!/\t

Now, group by tens:

Find what : (!{10})+
Replace with : $0/

followed by:

Find what : !{10}
Replace with : !

Repeat the above two steps until the first step finds nothing.

Now, count the exclamation points in each digit and remove the forward slashes:

Find what : (?:(!{9})|(!{8})|(!{7})|(!{6})|(!{5})|(!{4})|(!{3})|(!{2})|(!{1})|())/
Replace with : (?{1}9)(?{2}8)(?{3}7)(?{4}6)(?{5}5)(?{6}4)(?{7}3)(?{8}2)(?{9}1)(?{10}0)

This assumes there are no exclamation points or forward slashes elsewhere in the text. Of course, the forward slash (/) can be replaced with any character that is not in use.