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.)
Related
I have a large binary file. I want to extract certain strings from it and copy them to a new text file.
For example, in:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^G
I want to take the number '7' (after the #^#^#E) and every character after it stopping at the Z ('ignoring the M-^G).
I want to copy this 7cacscKLrrok9bwC3Z64NTnZ to a new file.
There will be multiple such strings in one file. The end will always be denoted by the M- (which I don't want copied). The start will always be denoted by a 7 (which I do want copied).
Unfortunately, my knowledge of grep, sed, etc, does not extend to this level. Can someone please suggest a viable way to achieve this?
cat -v filename | grep [7][A-Z,a-z] will show all strings with a '7' followed by a letter but that's not much.
Thank you.
I've noticed that my requirements are rather more complicated.
(I've performed the correct - I hope - formatting this time). Thanks to 'tshiono' for his (?) answer to the earlier submission.
I want to check the ending of a string and, if it ends in M-, grep another string that follows it (with junk in between). If the string does not end in M-, then I don't want it copied (let alone any other strings).
So what I would like is:
grep -a -Po "7[[:alnum:]]+(?=M-)" file_name and if the ending is M- then grep -a -Po "5x[[:alnum:]]+(?=\^)" file_name to copy the string that starts with 5x and ends with a ^.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
However, if the ending is not M- (more precisely, if the ending is ^S), then do not try the second grep and do not record anything at all.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZ^SGwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be null (nothing copied) as the 7cacs... string ends in ^S.
Is grep the correct tool? Grep a file and if the condition in the grep command is 'yes' then issue a different grep command but if the condition is 'no' then do nothing.
Thanks again.
I have noticed one addition modification.
Can one add an OR command to the second part? Grep if the second string starts with 5x OR 6x?
In the example below, grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" filename | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)" will extract the strings starting with 7 and the strings starting with 5x.
How can one change the 5x to 5x or 6x?
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7AAAAAscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
In this example, the desired outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
7AAAAAscKLrrok9bwC3Z64NTnZ
6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
UPDATE MARCH 09:
I need to create a series of complex grep (or perl) commands to extract strings from a series of binary files.
I need two strings from the binary file.
The first string will always start with a 1.
The first string will end with a letter or number. The next letter will always be a lower case k. I do not want this k character.
The difficulty is that the ending k will not always be the first k in the string. It might be the first k but it might not.
After the k, there is a second string. The second string will always start with an A or a B.
The ending of the second string will be in one of two forms:
a) it will end with a space then display the first three characters from the first string in lower case followed by a )
b) it will end with a ^K then display the first three characters from the first string in lower case.
For example:
1pppsx9YPar8Rvs75tJYWZq3eo8PgwbckB4m4zT7Yg042KIDYUE82e893hY ppp)
Should be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc and B4m4zT7Yg042KIDYUE82e893hY - delete the k and the space then ppp.
For example:
1zzzsx9YPkr8Rvs75tJYWZq3eo8PgwbckA2m4zT7Yg042KIDYUE82e893hY^Kzzz
Should be:
1zzzsx9YPkar8Rvs75tJYWZq3eo8Pgwbc and A4m4zT7Yg042KIDYUE82e893hY - delete the second k and the ^Kzzz.
In the second example, we see that the first k is part of the first string. It is the k before the A that breaks up the first and second strings.
I hope there is a super grep expert who can help! Many thanks!
If your grep supports -P option, would you please try:
grep -a -Po "7[[:alnum:]]+(?=M-)" file
The -a option forces grep to read the input as a text file.
The -P option enables the perl-compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?=M-) is a zero-width lookahead assertion (introduced in
Perl) without including it in the result.
Alternatively you can also say with sed:
sed 's/M-/\n/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
The first sed command splits the input file into miltiple lines by
replacing the substring M- with a newline.
It has two benefits: it breaks the lines to allow multiple matches with
sed and excludes the unnecessary portion M- from the input.
The next sed command extracts the desired pattern from the input.
It assumes your sed accepts \n in the replacement, which is
a GNU extension (not POSIX compliant). Otherwise please try (in case you are working on bash):
sed 's/M-/\'$'\n''/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
[UPDATE]
(The requirement has been updated by the OP and the followings are solutions according to it.)
Let me assume the string which starts with 7 and ends with M- is always followed
by another (no more and no less than one) string which starts with 5x and ends
with ^ (ascii caret character) with junks in between.
Then would you please try the following:
grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)"
It executes the task in two steps (two cascaded greps).
The 1st grep narrows down the input data into the candidate substring
which will include the desired two sequences and junks in between.
The regex .*? in between matches any (ascii or binary) characters
except for a newline character.
The trailing ? enables the shortest match
which avoids the overrun due to the greedy nature of regex. The regex is intended to match junks in between.
The 2nd grep includes two regex's merged with a pipe | meaning logical OR.
Then it extracts two desired sequences.
A potential problem of grep solution is that grep is a line oriented command
and cannot include the newline character in the matched string.
If a newline character is included in the junks in between (I'm not sure about the possibility), the above solution will fail.
As a workaround, perl will provide flexible manipulations with binary data.
perl -0777 -ne '
while (/(7[[:alnum:]]+)M-.*?(5x[[:alnum:]]+)\^/sg) {
printf("%s\n%s\n", $1, $2);
}
' file
The regex is mostly same as that of grep because the -P option of grep means
perl-compatible.
It can capture multiple patterns at once in variables $1 and $2 hence just one regex is enough.
The -0777 option to the perl command tells perl to slurp all data
at once.
The s option at the end the regex makes a dot match a newline character.
The g option enables the global (multiple) match.
[UPDATE2]
In order to make the regex match either 5x or 6x, replace 5x with (5|6)x.
Namely:
grep -aPo "7[[:alnum:]]+M-.*?(5|6)x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|(5|6)x[[:alnum:]]+(?=\^)"
As mentioned before, the pipe | means OR. The OR operator has the lowest priority in the evaluation, hence you need to enclose them with parens in this case.
If there is a possibility any other number than 5 or 6 may appear, it will be safer to put [[:digit:]] instead, which matches any one digit betweeen 0 and 9:
grep -aPo "7[[:alnum:]]+M-.*?[[:digit:]]x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|[[:digit:]]x[[:alnum:]]+(?=\^)"
[UPDATE3]
(Answering the OP's requirement on March 9th)
Let me start with a perl code which regex will be relatively easier
to explain.
perl -0777 -ne 'while (/(1(.{3}).+)k([AB].*)[\013 ]\2/g){print "$1 $3\n"}' file
Output:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc B4m4zT7Yg042KIDYUE82e893hY
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc A2m4zT7Yg042KIDYUE82e893hY
[Explanation of regex]
(1(.{3}).+)k([AB].*)[\013 ]\2
( start of the 1st capture group referred by $1 later
1 literal "1"
( start of the 2nd capture group referred by \2 later
.{3} a sequence of the identical three characters such as ppp or zzz
) end of the 2nd capture group
.+ followed by any characters with "greedy" match which may include the 1st "k"
) end of the 1st capture group
k literal "k"
( start of the 3rd capture group referred by $3 later
[AB].* the character "A" or "B" followed by any characters
) end of the 3rd capture group
[\013 ] followed by ^K or a whitespace
\2 followed by the capture group 2 previously assigned
When implementing it with grep, we will encounter a limitation of grep.
Although we want to extract multiple patterns from the input file,
the -e option (which can specify multiple search patterns) does not
work with -P option. Then we need to split the regex into two patterns
such as:
grep -Po "(1(.{3}).+)(?=k([AB].*)[\013 ]\2)" file
grep -Po "(1(.{3}).+)k\K([AB].*)(?=[\013 ]\2)" file
And the result will be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc
B4m4zT7Yg042KIDYUE82e893hY
A2m4zT7Yg042KIDYUE82e893hY
Please be noted the order of output is not same as the order of appearance in the original file.
Another option will be to introduce ripgrep or rg which is a fast
and versatile version of grep. You may need to install ripgrep with
sudo apt install ripgrep or using other package handling tool.
An advantage of ripgrep is it supports -r (replace) option in which
you can make use of the backreferences:
rg -N -Po "(1(.{3}).+)k([AB].*)[\013 ]\2" -r '$1 $3' file
The -r '$1 $3' option prints the 1st and the 3rd capture groups and the result will be the same as perl.
In the general case, you can use the strings utility to pluck out ASCII from binary files; then of course you can try to grep that output for patterns that you find interesting.
Many traditional Unix utilities like grep have internal special markers which might get messed up by binary input. For example, the character \xFF was used for internal purposes by some versions of GNU grep so you can't grep for that character even if you can figure out a way to represent it in the shell (Bash supports $'\xff' for example).
A traditional approach would be to run hexdump or a similar utility, and then grep that for patterns. However, more modern scripting languages like Perl and Python make it easy to manipulate arbitrary binary data.
perl -ne 'print if m/\xff\xff/' </dev/urandom
This might work for you (GNU sed):
sed -En '/\n/!{s/M-\^G/\n/;s/7[^\n]*\n/\n&/};/^7[^\n]*/P;D' file
Split each line into zero or more lines that begin with 7 and end just before M-^G and only print such lines.
I've been looking around but couldn't find a way to do it with both AWK and SED.
I was wondering if there's a way to replace a string's start and end in a single command.
more specifically, there's a file with a lot of words in it, and I would like to add something before the word and after the word.
Thanks,
Roy
Since you said: more specifically, there's a file with a lot of words in it, and I would like to add something before the word and after the word.
The only thing you need is $& that is match itself. So you simply can write anything that you want just before and end of this whildcard. that's it.
For example say you have this file:
this is line 1.
this is line 2.
this is line 3.
And I tested with perl:
perl -lne 'print "beginning->", $&, "<-end" if /.+/g' file
which the output is:
beginning->this is line 1.<-end
beginning->this is line 2.<-end
beginning->this is line 3.<-end
May you would like to match only one word, so still this is a good solution such as:
perl -lne 'use English; print "$PREMATCH", "[$MATCH]","$POSTMATCH" if /line/g' file
Here I matched line and put around that: [ then $& then ]
the output
this is [line] 1.
this is [line] 2.
this is [line] 3.
NOTE
As you can see the only things you need just are prematch and match and postmatch. I tested it with perl for you, and if you are interesting in Perl you can use it or may you want to use Sed or Awk. Since you have no specific examples I tested with Perl.
If you want to wrap a particular word with markers you can use & in the replacement string to achieve what you want.
For example to put square brackets around every occurrence of the word bird:
$ echo "hello bird, are you really a bird?" | sed "s/\bbird\b/[&]/g"
hello [bird], are you really a [bird]?
to replace a string's start and end in a single command
Let's say we have a test file with line:
tag hello, world tag
To enclose each tag word with angle brackets < ... > we can apply:
awk approach with gsub() function:
awk '{ gsub(/\<tag\>/, "<&>"); print}' test_file
word boundaries \<, \> may differ depending on awk implementations
sed approach:
sed 's/\btag\b/<&>/g' test_file
The output(for both approaches):
<tag> hello, world <tag>
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
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input