I have a file with anywhere between a dozen and hundreds of matches on the search
/playOrder="(\d+)"/
These are in the index file of an ePub ebook, in case anyone is wondering.
Is it possible to have a perl regex replace what finds all these, and "magically" renumber them all to a sequence, starting from 1?
Posting comment as answer, as requested by OP:
perl -pe 's/playOrder="\K\d+"/++$i . q(")/ge' infile > outfile
This one-liner is using a replacement field which is created by evaluation, creating a sequence like 1", 2"...
Further optimization can be made if using a lookahead assertion instead of inserting a new double quote ":
perl -pe 's/playOrder="\K\d+(?=")/++$i/ge' infile > outfile
Related
I have the following string:
load Add 20 percent
to accommodate
I want to get to:
load Add 20 percent to accommodate
With, e.g., regex in sublime, this is easily done by:
Regex:
([a-z])\n\s([a-z])
Replace:
$1 $2
However, in Perl, if I input this command, (adapted to test if I can match the pattern in any case):
perl -pi.orig -e 's/[a-z]\n.+to/TEST/g' file
It doesn't match anything.
Does anyone know why Perl would be different in this case, and what the correct formulation of the Perl command should be?
By default, Perl -p flag read input lines one by one. You can't thus expect your regex to match anything after \n.
Instead, you want to read the whole input at once. You can do this by using the flag -0777 (this is documented in perlrun):
perl -0777 -pi.orig -e 's/([a-z])\n\s(to)/$1 $2/' file
Just trying to help and reminding below your initial proposal for perl regex:
perl -pi.orig -e 's/[a-z]\n.+to/TEST/g' file
Note that in perl regex, [a-z] will match only one character, NOT including any whitespace. Then as a start please include a repetition specifier and include capability to also 'eat' whitespaces. Also to keep the recognized (but 'eaten') 'to' in the replacement, you must put it again in the replacement string, like finally in the below example perl program:
$str = "load Add 20 percent
to accommodate";
print "before:\n$str\n";
$str =~ s/([ a-z]+)\n\s*to/\1 to/;
print "after:\n$str\n";
This program produces the below input:
before:
load Add 20 percent
to accommodate
after:
load Add 20 percent to accommodate
Then it looks like that if I understood well what you want to do, your regexp should better look like:
s/([ a-z]+)\n\s*to/\1 to/ (please note the leading whitespace before 'a-z').
I have some 'fastq' format DNA sequence files (basically just text files) like this:
#Sample_1
ACTGACTGACTGACTGACTGACTGACTG
ACTGACTGACTGACTGACTGACTGACTG
+
BBBBBBBBBBBBEEEEEEEEEEEEEEEE
EHHHHKKKKKKKKKKKKKKNQQTTTTTT
#
+
#
+
#Sample_4
ACTGACTGACTGACTGACTGACTGACTG
ACTGACTGACTGACTGACTGACTGACTG
+
BBBBBBBBBBBBEEEEEEEEEEEEEEEE
EHHHHKKKKKKKKKKKKKKNQQTTTTTT
My ultimate goal is to turn these into 'fasta' format files, but to do that I need to get rid of the two empty sequences in the middle.
EDIT
The desired output would look like this:
#Sample_1
ACTGACTGACTGACTGACTGACTGACTG
ACTGACTGACTGACTGACTGACTGACTG
+
BBBBBBBBBBBBEEEEEEEEEEEEEEEE
EHHHHKKKKKKKKKKKKKKNQQTTTTTT
#Sample_4
ACTGACTGACTGACTGACTGACTGACTG
ACTGACTGACTGACTGACTGACTGACTG
+
BBBBBBBBBBBBEEEEEEEEEEEEEEEE
EHHHHKKKKKKKKKKKKKKNQQTTTTTT
All of the dedicated software I tried (Biopython, stand alone programs, perl scripts posted by others) crash at the empty sequences. This is really just a problem of searching for the string #\n+ and replacing it with nothing. I googled this and read several posts and tried about a million options with sed and couldn't figure it out. Here are some things that didn't work:
sed s/'#'/,/'+'// test.fastq > test.fasta
sed s/'#,+'// test.fastq > test.fasta
Any insights would be greatly appreciated.
PS. I've got a Mac.
Try:
sed "/^[#+]*$/d" test.fastq > test.fasta
The /d option tells sed to "delete" the matching line (i.e. not print it).
^ and $ mean "start of string" and "end of string" respectively, i.e. the line must be an exact match.
So, the above command basically says:
Print all lines that do not only contain # or +, and write the result to test.fasta.
Edit: I misunderstood the question slightly, sorry. If you want to only remove pairs of consecutive lines like
#
+
then you need to perform a multi-line search and replace.
Although this can be done with sed, it's perhaps easier to use something like a perl script instead:
perl -0pe 's/^#\n\+\n//gm' test.fastq > test.fasta
The -0 option turns Perl into "file slurp" mode, where Perl reads the entire input file in one shot (instead of line by line). This enables multi-line search and replace.
The -pe option allows you to run Perl code (pattern matching and replacement in this case) and display output from the command line.
^#\n\+\n is the pattern to match, which we are replacing with nothing (i.e. deleting).
/gm makes the substitution multiline and global.
You could also instead pass -i as the first parameter to perl, to edit the file inline.
This may not be the most elegant solution in the world, but you can use tr to replace the \n with a null character and back.
cat test.fastq | tr '\n' '\0' | sed 's/#\x0+\x0//g' | tr '\0' '\n' > test.fasta
Try this:
sed '/^#$/{N;/\n+$/d}' file
When # is found, next line is appended to the pattern space with N.
If $ is found in next line, the d command deletes both lines.
I'm trying to write a bash script that would be able to grep table names from across files (within a directory) that partially match a string.
For my case, I'd like to return all table references following a certain convention (case insensitive):
tblpl
tbljoin
tbldim
This would ideally return a list like this:
product.dbo.tblplColors
product..tblplMonograms
solr.dbo.tbljoinSkuCategory
Matching one table name format at a time would also be alright if that helped reduce some of the complexity. To clarify, this would return just the table names- not the file name/all of the file contents. It's safe to say the end of the table name would be delimited by a space since it's SQL.
Where I've started:
grep -rio 'tblpl*[^ ]' d:/sqldirectoryhere > c:/Users/foo/Desktop/tables.txt
Any help/pointers are appreciated here- thanks!
Edit: Both of these answers nailed my use case. I ended up adding the extended regex (so huge thanks for that recommendation) but I have to give credit to the person who wrote the bulk of it. Thanks all!
My extended use case ended up being a way to return this list of tables and then script it to a query-friendly format so I could throw these into a WHERE IN clause. In case anyone ever needs it:
grep -rioE --no-filename '[a-zA-Z_.]+\.tbl(pl|join|dim)[a-zA-Z_]+' {DIRECTORY_HERE} | sed -n 's/.*/\x27&\x27/; $! s/$/,/; 1 h; 1 ! H; $ { x; s/\n/ /g; p; }'
Returns formatted as: 'db.tblplColorSwatches', 'db.tbljoinCustomerSegment'...
It finds any mixed sequence of letters and periods followed by .tblpl or .tbljoin or .tbldim followed by one or more letters (see regex101 link)
try this regular expression:
[a-zA-Z.]+\.tbl(pl|join|dim)[a-zA-Z]+
I would use the -E flag to use extended regular expression:
grep -rioE '[a-z]*\.[a-z]*\.tbl(pl|join|dim)[a-z]*' d:/sqldirectoryhere
I have a very large file, containing the following blocks of lines throughout:
start :234
modify 123 directory1/directory2/file.txt
delete directory3/file2.txt
modify 899 directory4/file3.txt
Each block starts with the pattern "start : #" and ends with a blank line. Within the block, every line starts with "modify # " or "delete ".
I need to modify the path in each line, specifically appending a directory to the front. I would just use a general regex to cover the entire file for "modify #" or "delete ", but due to the enormous amount of other data in that file, there will likely be other matches to this somewhat vague pattern. So I need to use multi-line matching to find the entire block, and then perform edits within that block. This will likely result in >10,000 modifications in a single pass, so I'm also trying to keep the execution down to less than 30 minutes.
My current attempt is a sed one-liner:
sed '/^start :[0-9]\+$/ { :a /^[modify|delete] .*$/ { N; ba }; s/modify [0-9]\+ /&Appended_DIR\//g; s/delete /&Appended_DIR\//g }' file_to_edit
Which is intended to find the "start" line, loop while the lines either start with a "modify" or a "delete," and then apply the sed replacements.
However, when I execute this command, no changes are made, and the output is the same as the original file.
Is there an issue with the command I have formed? Would this be easier/more efficient to do in perl? Any help would be greatly appreciated, and I will clarify where I can.
I think you would be better off with perl
Specifically because you can work 'per record' by setting $/ - if you're records are delimited by blank lines, setting it to \n\n.
Something like this:
#!/usr/bin/env perl
use strict;
use warnings;
local $/ = "\n\n";
while (<>) {
#multi-lines of text one at a time here.
if (m/^start :\d+/) {
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
Each iteration of the loop will pick out a blank line delimited chunk, check if it starts with a pattern, and if it does, apply some transforms.
It'll take data from STDIN via a pipe, or myscript.pl somefile.
Output is to STDOUT and you can redirect that in the normal way.
Your limiting factor on processing files in this way are typically:
Data transfer from disk
pattern complexity
The more complex a pattern, and especially if it has variable matching going on, the more backtracking the regex engine has to do, which can get expensive. Your transforms are simple, so packaging them doesn't make very much difference, and your limiting factor will be likely disk IO.
(If you want to do an in place edit, you can with this approach)
If - as noted - you can't rely on a record separator, then what you can use instead is perls range operator (other answers already do this, I'm just expanding it out a bit:
#!/usr/bin/env perl
use strict;
use warnings;
while (<>) {
if ( /^start :/ .. /^$/)
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
We don't change $/ any more, and so it remains on it's default of 'each line'. What we add though is a range operator that tests "am I currently within these two regular expressions" that's toggled true when you hit a "start" and false when you hit a blank line (assuming that's where you would want to stop?).
It applies the pattern transformation if this condition is true, and it ... ignores and carries on printing if it is not.
sed's pattern ranges are your friend here:
sed -r '/^start :[0-9]+$/,/^$/ s/^(delete |modify [0-9]+ )/&prepended_dir\//' filename
The core of this trick is /^start :[0-9]+$/,/^$/, which is to be read as a condition under which the s command that follows it is executed. The condition is true if sed currently finds itself in a range of lines of which the first matches the opening pattern ^start:[0-9]+$ and the last matches the closing pattern ^$ (an empty line). -r is for extended regex syntax (-E for old BSD seds), which makes the regex more pleasant to write.
I would also suggest using perl. Although I would try to keep it in one-liner form:
perl -i -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit
Or you can use redirection of stdout:
perl -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit > new_file
with gnu sed (with BRE syntax):
sed '/^start :[0-9][0-9]*$/{:a;n;/./{s/^\(modify [0-9][0-9]* \|delete \)/\1NewDir\//;ba}}' file.txt
The approach here is not to store the whole block and to proceed to the replacements. Here, when the start of the block is found the next line is loaded in pattern space, if the line is not empty, replacements are performed and the next line is loaded, etc. until the end of the block.
Note: gnu sed has the alternation feature | available, it may not be the case for some other sed versions.
a way with awk:
awk '/^start :[0-9]+$/,/^$/{if ($1=="modify"){$3="newdirMod/"$3;} else if ($1=="delete"){$2="newdirDel/"$2};}{print}' file.txt
This is very simple in Perl, and probably much faster than the sed equivalent
This one-line program inserts Appended_DIR/ after any occurrence of modify 999 or delete at the start of a line. It uses the range operator to restrict those changes to blocks of text starting with start :999 and ending with a line containing no printable characters
perl -pe"s<^(?:modify\s+\d+|delete)\s+\K><Appended_DIR/> if /^start\s+:\d+$/ .. not /\S/" file_to_edit
Good grief. sed is for simple substitutions on individual lines, that is all. Once you start using constructs other than s, g, and p (with -n) you are using the wrong tool. Just use awk:
awk '
/^start :[0-9]+$/ { inBlock=1 }
inBlock { sub(/^(modify [0-9]+|delete) /,"&Appended_DIR/") }
/^$/ { inBlock=0 }
{ print }
' file
start :234
modify 123 Appended_DIR/directory1/directory2/file.txt
delete Appended_DIR/directory3/file2.txt
modify 899 Appended_DIR/directory4/file3.txt
There's various ways you can do the above in awk but I wrote it in the above style for clarity over brevity since I assume you aren't familiar with awk but should have no trouble following that since it reuses your own sed scripts regexps and replacement text.
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
basic
I want unix sed command such that only basic that is not in quotes should be changed.[change basic to ring]
Expected output:
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If we disallow escaping quotes, then any basic that is not within " is preceded by an even number of ". So this should do the trick:
sed -r 's/^([^"]*("[^"]*){2}*)basic/\1ring/' file
And as ДМИТРИЙ МАЛИКОВ mentioned, adding the --in-place option will immediately edit the file, instead of returning the new contents.
How does this work?
We anchor the regular expression to the beginning of each line with ". Then we allow an arbitrary number of non-" characters (with [^"]*). Then we start a new subpattern "[^"]* that consists of one " and arbitrarily many non-" characters. We repeat that an even number of times (with {2}*). And then we match basic. Because we matched all of that stuff in the line before basic we would replace that as well. That's why this part is wrapped in another pair of parentheses, thus capturing the line and writing it back in the replacement with \1 followed by ring.
One caveat: if you have multiple basic occurrences in one line, this will only replace the last one that is not enclosed in double quotes, because regex matches cannot overlap. A solution would be a lookbehind, but since this would be a variable-length lookbehind, which is only supported by the .NET regex engine. So if that is the case in your actual input, run the command multiple times until all occurrences are replaced.
$> sed -r 's/^([^\"]*)(basic)([^\"]*)$/\1ring\3/' file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If you wanna edit file in place use --in-place option.
This might work for you (GNU sed):
sed -r 's/^/\n/;ta;:a;s/\n$//;t;s/\n("[^"]*")/\1\n/;ta;s/\nbasic/ring\n/;ta;s/\n([^"]*)/\1\n/;ta' file
Not a sed solution, but it substitutes words not in quotes
Assuming that there is no escaped quotes in strings, i.e. "This is a trap \" hehe", awk might be able to solve this problem
awk -F\" 'BEGIN {OFS=FS}
{
for(i=1; i<=NF; i++){
if(i%2)
gsub(/basic/,"ring",$i)
}
print
}' inputFile
Basically the words that are not in quotes are in odd-numbered fields, and the word "basic" is replaced by "ring" in these fields.
This can be written as a one-liner, but for clarity's sake I've written it in multiple lines.
If basic is at the beginning of line:
sed -e 's/^basic/ring/' file0