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.
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've scraped a large amount (10GB) of PDFs and converted them to text files, but due to the format of the original PDFs, there is an issue:
Many of the words which break across lines have a dash in them that artificially breaks up the word, like this:
You can see that this happened because the original PDFs files have breaks:
What would be the cleanest and fastest way to "join" every word instance that matches this pattern inside of a .txt file?
Perhaps some sort of Regex search, like for a [a-z]\-\s \w of some kind (word character followed by dash followed by space) would work?
Or would some sort of sed replacement work better?
Currently, I'm trying to get a sed regex to work, but I'm not sure how to translate this to use capture groups to replace the selected text:
sed -n '\%\w\- [a-z]%p' Filename.txt
My input text would look like this:
The dog rolled down the st- eep hill and pl- ayed outside.
And the output would be:
The dog rolled down the steep hill and played outside.
Ideally, the expression would also work for words split up by a newline, like this:
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
To this:
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
It's straightforward in sed:
sed -e ':a' -e '/-$/{N;s/-\n//;ba
}' -e 's/- //g' filename
This translates roughly as "if the line ends with a dash, read in the next line as well (so that you have a line with a carriage return in the middle) then excise the dash and carriage return, and loop back the beginning just in case this new line also ends with a dash. Then remove any instances of - ".
You may use this gnu-awk code:
cat file
The dog rolled down the st- eep hill and pl- ayed outside.
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
Then use awk like this:
awk 'p != "" {
w = $1
$1 = ""
sub(/^[[:blank:]]+/, ORS)
$0 = p w $0
p = ""
}
{
$0 = gensub(/([_[:alnum:]])-[[:blank:]]+([_[:alnum:]])/, "\\1\\2", "g")
}
/-$/ {
p = $0
sub(/-$/, "", p)
}
p == ""' file
The dog rolled down the steep hill and played outside.
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
If you can consider perl then this may also work for you:
Then use:
perl -0777 -pe 's/(\w)-\h+(\w)/$1$2/g; s/(\w)-\R(\w+)\s+/$1$2\n/g' file
You simply add backslash-parentheses (or use the -r or -E option if available to do away with the requirement to put backslashes before capturing parentheses) and recall the matched text with \1 for the first capturing parenthesis, \2 for the second, etc.
sed 's/\(\w\)\- \([a-z]\)/\1\2/g' Filename.txt
The \w escape is not standard sed but if it works for you, feel free to use it. Otherwise, it is easy to replace with [A-Za-z0-9_#] or whatever else you want to call "word characters".
I'm guessing not all of the matches will be hyphenated words so perhaps run the result through a spelling checker or something to verify whether the result is an English word. (I would probably switch to a more capable scripting language like Python for that, though.)
I have files where missing data is inserted as '+'. So lines look like this:
substring1+++++substring2++++++++++++++substring3+substring4
I wanna replace all repetitions of '+' >5 with 'MISSING'. This makes it more readable for my team and makes it easier to see the difference between missing data and data entered as '+' (up to 5 is allowed).
So far I have:
while read l; do
echo "${l//['([+])\1{5}']/'MISSING'}"
done < /path/file.txt
but this replaces every '+' with 'MISSING'. I need it to say 'MISSING' just once.
Thanks in advance.
You can't use regex in Bash variable expansion.
In your loop, you may use
sed 's/+\{1,\}/MISSING/g' <<< "$l"
Or, you may use sed directly on the file
sed 's/+\{1,\}/MISSING/g' /path/file.txt
The +\{1,\} POSIX BRE pattern matches a literal + (+) 1 or more times (\{1,\}).
See the sed demo online
sed 's/+\{1,\}/MISSING/g' <<< "substring1+++++substring2++++++++++++++substring3+substring4"
# => substring1MISSINGsubstring2MISSINGsubstring3MISSINGsubstring4
If you need to make changes to the same file use any technique described at sed edit file in place.
this is my first post on stackoverflow, please forgive me if I missed something important.
I am currently stuck with the follwing issue. The goal is, to replace port numbers dynamically based on a filelist I prepared with find. All of the ports in those files, start with the number "4" and have 5 digits.
Now the tricky part, I am replacing only digit #2 and #3, and keep positions 1, 4 and 5. Examples:
old port in file: 40380, 40381
new port in file: 41580, 40381
I am working on Sun Solaris 5.10 therefore I prefer perl for inline replacements
Finally the key question: how can I combine $1 (group 1) + $PIN_PINNO + $3 (group 3) so that the result would be: 41580
NEW_PINNO=15
LOGI=$HOME/filelist.txt
# port replacement
for file in `cat $LOGI`
do
perl -pe 's/[\:\>\=]\s*(4)(\d{2})(\d{2})\b/$1${NEW_PINNO}$3/g' $file
done
many thanks in advance
perl -pse 's/ [:>=]\s* \K (\d)\d\d(\d\d) \b/$1$pin$2/gx' -- -pin="$new_pinno" file
Your regex will match the [colon, greater than, equal sign] and the spaces, but you don't include them in the substitution. I'm using the \K directive to match those characters but then forget about them (ref: http://perldoc.perl.org/perlre.html#Lookaround-Assertions)
I'm using the -s option to enable "rudimentary switch parsing" to pass the shell variable into perl without playing quoting games. (ref: http://perldoc.perl.org/perlrun.html)
Testing
new_pinno=15
perl -pse 's/ [:>=]\s* \K (\d)\d\d(\d\d) \b/$1$pin$2/gx' -- -pin="$new_pinno" <<END
var1=40380
var2=40381
END
var1=41580
var2=41581
Notes
you should not use ALL_CAPS_VARNAMES in the shell, leave those to be reserved by the shell. One day, you'll use PATH=something and then wonder why your script is broken.
and #123's comment is valid. This is the safe way to read lines from a file:
while read -r file; do
perl ... "$file"
done < "$LOGI"
ref: http://mywiki.wooledge.org/DontReadLinesWithFor
perl -pe 's/[\:\>\=]\s*(4)(\d{2})(\d{2})\b/$1'${NEW_PINNO}'$3/g' $file
is ok.
The difference between single and double quotes is how bash treats its variables. In single quotes it won't expand them, but when enclosed in double quotes it will. You can open & close quotes as much as you want in a command line argument. It's only if there is a space that is outside of a pair of quotes (single or double) that determines the start of a new argument.
So you close the single quote, have the bash variable which will be expanded and then re-open the single quote. Enclosing the variable in double quotes ensures that if there are any spaces in the variable they'll not split the argument.
perl -pe 's/[\:\>\=]\s*(4)(\d{2})(\d{2})\b/$1'"${NEW_PINNO}"'$3/g' $file
I have an HTML file and would like to extract the text between <li> and </li> tags. There are of course a million ways to do this, but I figured it would be useful to get more into the habit of doing this in simple shell commands:
awk '/<li[^>]+><a[^>]+>([^>]+)<\/a>/m' cities.html
The problem is, this prints everything whereas I simply want to print the match in parenthesis -- ([^>]+) -- either awk doesn't support this, or I'm incompetent. The latter seems more likely. If you wanted to apply the supplied regex to a file and extract only the specified matches, how would you do it? I already know a half dozen other ways, but I don't feel like letting awk win this round ;)
Edit: The data is not well-structured, so using positional matches ($1, $2, etc.) is a no-go.
If you want to do this in the general case, where your list tags can contain any legal HTML markup, then awk is the wrong tool. The right tool for the job would be an HTML parser, which you can trust to get correct all of the little details of HTML parsing, including variants of HTML and malformed HTML.
If you are doing this for a special case, where you can control the HTML formatting, then you may be able to make awk work for you. For example, let's assume you can guarantee that each list element never occupies more than one line, is always terminated with </li> on the same line, never contains any markup (such as a list that contains a list), then you can use awk to do this, but you need to write a whole awk program that first finds lines that contain list elements, then uses other awk commands to find just the substring you are interested in.
But in general, awk is the wrong tool for this job.
gawk -F'<li>' -v RS='</li>' 'RT{print $NF}' file
Worked pretty well for me.
By your script, if you can get what you want (it means <li> and <a> tag is in one line.);
$ cat test.html | awk 'sub(/<li[^>]*><a[^>]*>/,"")&&sub(/<\/a>.*/,"")'
or
$ cat test.html | gawk '/<li[^>]*><a[^>]*>(.*?)<\/a>.*/&&$0=gensub(/<li[^>]*><a[^>]*>(.*?)<\/a>.*/,"\\1", 1)'
First one is for every awk, second one is for gnu awk.
There are several issues that I see:
The pattern has a trailing 'm' which is significant for multi-line matches in Perl, but Awk does not use Perl-compatible regular expressions. (At least, standard (non-GNU) awk does not.)
Ignoring that, the pattern seems to search for a 'start list item' followed by an anchor '<a>' to '</a>', not the end list item.
You search for anything that is not a '>' as the body of the anchor; that's not automatically wrong, but it might be more usual to search for anything that is not '<', or anything that is neither.
Awk does not do multi-line searches.
In Awk, '$1' denotes the first field, where the fields are separated by the field separator characters, which default to white space.
In classic nawk (as documented in the 'sed & awk' book vintage 1991) does not have a mechanism in place for pulling sub-fields out of matches, etc.
It is not clear that Awk is the right tool for this job. Indeed, it is not entirely clear that regular expressions are the right tool for this job.
Don't really know awk, how about Perl instead?
tr -d '\012' the.html | perl \
-e '$text = <>;' -e 'while ( length( $text) > 0)' \
-e '{ $text =~ /<li>(.*?)<\/li>(.*)/; $target = $1; $text = $2; print "$target\n" }'
1) remove newlines from file, pipe through perl
2) initialize a variable with the complete text, start a loop until text is gone
3) do a "non greedy" match for stuff bounded by list-item tags, save and print the target, set up for next pass
Make sense? (warning, did not try this code myself, need to go home soon...)
P.S. - "perl -n" is Awk (nawk?) mode. Perl is largely a superset of Awk, so I never bothered to learn Awk.