I am wondering, if one can print the matched strings as it is in each line... using grep or sed?
TestCase1: File1 contains below text
The Sun
Thunder The Rain They say
They say The dance
If I use this command:
egrep -o 'The|They' File1
The output I get is:
The
The
They
They
The
But, my expected output should be as below:
The
The They
They The
I am aware that, In grep the option -o, --only-matching prints only the matched non-empty) parts of a matching line, with each such part on a separate output line.
Edit: Please also suggest, if one wants to have a filter with exact word match with multiple match strings
i.e. <The> and <They> exact word match? Space separated words simply.
TestCase2: File2 contains below text
The Sun
Thunder The Rain They say
They say The dance
They're dancing with them in the dorm
The sun is shining the east and they scream.
Output is:
The
The They
They the
the
The the they
How to approach this?
With GNU awk for FPAT:
$ awk -v FPAT='\\<[Tt]hey?\\>' '{$1=$1}1' file
The
The They
They The
They the
The the they
Note that that can't NOT identify They when it appears in They're. If that's really an issue and you want to look for space-separated complete strings then this might be what you want:
$ awk '{c=0; for (i=1;i<=NF;i++) if ($i ~ /^[Tt]hey?$/) printf "%s%s", (c++?OFS:""), $i; print ""}' file
The
The They
They The
the
The the they
If not, let us know.
The above was run against this iteration of the OPs posted sample input:
$ cat file
The Sun
Thunder The Rain They say
They say The dance
They're dancing with them in the dorm
The sun is shining the east and they scream.
Best do it with Perl:
~$ perl -nE 'say /They? /g' File1
The
The They
They The
EDIT : Add new conditions. The regex still matches all but the lowercase the. Adding the i flag makes the match case-insensitive and matches all your test strings.
$ perl -nE 'say /They? /ig' File1
The
The They
They The
the
The the they
There is a little bit of a trick here: the match also picks up the space after the ? and prints it in the output. E.g. the first line of output is realy: "The_\n" - where "_" = space character. This may or may not be acceptable. One way to remove the spaces and reassemble the string would be:
$ perl -nE 'say join " ", map {substr $_,0,-1} /They? /ig' File1
As to your question about matching full words <The> and <They>, as you put it, the ? in They? indicates that the 'y' is optional. I.e. matches 0 or 1 times. Therefore the pattern is considering 'The' and 'They' as full words, one or the other, followed by a space. You could rewrite the pattern as:
$ perl -nE 'say /(?:They|The) /ig' File1
And effect the same output.
Now that you are considering lowercase the you may run into more edge case "gotchas" like words that end in "the". "loathe" and "tythe" come to mind.
$ echo "I'm loathe to cringe and tythe socks" >> File1
$ perl -nE 'say /They? /ig' File1
The
The They
They The
the
The the they
the the <--- not wanted!
You can then add the \b test in to match on word boundaries (as in zdim's answer):
$ perl -nE 'say /\bThey? /ig' File1
The
The They
They The
the
The the they
<-- But you get this empty line where no match occurs
So to refine further, you could only print if the line matches. Like this:
$ perl -nE 'say /\bThey? /ig if /\bThey? /i' File1
The
The They
They The
the
The the they
Then, I'm sure, you can find more edge cases that will blow it all up and force further refinement.
Things are not fully specified so here are a couple of possibilities
To catch all words starting with The, and print them with a space in between
perl -wnE'say join " ", /\bThe\w*/g' file
where \b is a word-boundary, a zero-width anchor, and \w is a word character. Using \S (a non-space character) is yet more permissive.
For only The or They can instead use
perl -wnE'say join " ", /\bThey?\b/g' file
where y? makes y optional.
To allow the as well use [tT] instead of T in the pattern, or /i for either case for all chars.
It's been clarified in coments that punctuation after The|They isn't allowed, and that low case t is. Then we need to constrain the match by space, not word boundary, and use [tT] as mentioned
perl -wnE'say join " ", /\b([Tt]hey?)\s/g' file
Now the capturing parenthesis () are needed since \s does consume, unlike \b before.
This prints the desired output with the provided input.
awk to the rescue!
$ awk -v p="They?" '$0~p{for(i=1;i<=NF;i++) if($i~p) printf "%s",$i OFS; print ""}' file
The
The They
They The
try one more awk:
awk '{while(match($0,/The|They/)){string=substr($0,RSTART,RLENGTH);VAL=VAL?VAL OFS string:string;$0=substr($0,RSTART+RLENGTH+1);};print VAL;VAL=""}' Input_file
NON-ONE line form of solution as follows too.
awk '{
while(match($0,/The|They/)){
string=substr($0,RSTART,RLENGTH);
VAL=VAL?VAL OFS string:string;
$0=substr($0,RSTART+RLENGTH+1);
};
print VAL;
VAL=""
}
' Input_file
Will add the explanation shortly for same.
Have a file eg. Inventory.conf with lines like:
Int/domain—home.dir=/etc/int
I need to replace / and — before the = but not after.
Result should be:
Int_domain_home_dir=/etc/int
I have tried several sed commands but none seem to fit my need.
Sed with a t loop (BRE):
$ sed ':a;s/[-/—.]\(.*=\)/_\1/;ta;' <<< "Int/domain—home.dir=/etc/int"
Int_domain_home_dir=/etc/int
When one of the -/—. character is found, it's replaced with a _. Following text up to = is captured and output using backreference. If the previous substitution succeeds, the t command loops to label :a to check for further replacements.
Edit:
If you're under BSD/Mac OSX (thanks #mklement0):
sed -e ':a' -e 's/[-/—.]\(.*=\)/_\1/;ta'
You're asking for a sed solution, but an awk solution is simpler and performs better in this case, because you can easily split the line into 2 fields by = and then selectively apply gsub() to only the 1st field in order to replace the characters of interest:
$ awk -F= '{ gsub("[./-]", "_", $1); print $1 FS $2 }' <<< 'Int/domain-home.dir=/etc/int'
Int_domain_home_dir=/etc/int
-F= tells awk to split the input into fields by =, which with the input at hand results in $1 (1st field) containing the first half of the line, before the =, and $2 (2nd field) the 2nd half, after the =; using the -F option sets variable FS, the input field separator.
gsub("[./-]", "_", $1) globally replaces all characters in set [./-] with _ in $1 - i.e., all occurrences of either ., / or - in the 1st field are replaced with a _ each.
print $1 FS $2 prints the result: the modified 1st field ($1), followed by FS (which is =), followed by the (unmodified) 2nd field ($2).
Note that I've used ASCII char. - (HYPHEN-MINUS, codepoint 0x2d) in the awk script, even though your sample input contains the Unicode char. — (EM DASH, U+2014, UTF-8 encoding 0xe2 0x80 0x94).
If you really want to match that, simply substitute it in the command above, but note that the awk version on macOS won't handle that properly.
Another option is to use iconv with ASCII transliteration, which tranlates the em dash into a regular ASCII -:
iconv -f utf-8 -t ascii//translit <<< 'Int/domain—home.dir=/etc/int' |
awk -F= '{ gsub("[./-]", "_", $1); print $1 FS $2 }'
perl allows for an elegant solution too:
$ perl -F= -ane '$F[0] =~ tr|-/.|_|; print join("=", #F)' <<<'Int/domain-home.dir=/etc/int'
Int_domain_home_dir=/etc/int
-F=, just like with Awk, tells Perl to use = as the separator when splitting lines into fields
-ane activates field splitting (a), turns off implicit output (n), and e tells Perl that the next argument is an expression (command string) to execute.
The fields that each line is split into is stored in array #F, where $F[0] refers to the 1st field.
$F[0] =~ tr|-/.|-| translates (replaces) all occurrences of -, /, and . to _.
print join("=", #F) rebuilds the input line from the fields - with the 1st field now modified - and prints the result.
Depending on the Awk implementation used, this may actually be faster (see below).
That sed isn't the best tool for this job is also reflected in the relative performance of the solutions:
Sample timings from my macOS 10.12 machine (GNU sed 4.2.2, Mawk awk 1.3.4, perl v5.18.2, using input file file, which contains 1 million copies of the sample input line) - take them with a grain of salt, but the ratios of the numbers are of interest; fastest solutions first:
# This answer's awk answer.
# Note: Mawk is much faster here than GNU Awk and BSD Awk.
$ time awk -F= '{ gsub("[./-]", "_", $1); print $1 FS $2 }' file >/dev/null
real 0m0.657s
# This answer's perl solution:
# Note: On macOS, this outperforms the Awk solution when using either
# GNU Awk or BSD Awk.
$ time perl -F= -ane '$F[0] =~ tr|-/.|_|; print join("=", #F)' file >/dev/null
real 0m1.656s
# Sundeep's perl solution with tr///
$ time perl -pe 's#^[^=]+#$&=~tr|/.-|_|r#e' file >/dev/null
real 0m2.370s
# Sundeep's perl solution with s///
$ time perl -pe 's#^[^=]+#$&=~s|[/.-]|_|gr#e' file >/dev/null
real 0m3.540s
# Cyrus' solution.
$ time sed 'h;s/[^=]*//;x;s/=.*//;s/[/.-]/_/g;G;s/\n//' file >/dev/null
real 0m4.090s
# Kenavoz' solution.
# Note: The 3-byte UTF-8 em dash is NOT included in the char. set,
# for consistency of comparison with the other solutions.
# Interestingly, adding the em dash adds another 2 seconds or so.
$ time sed ':a;s/[-/.]\(.*=\)/_\1/;ta' file >/dev/null
real 0m9.036s
As you can see, the awk solution is fastest by far, with the line-internal-loop sed solution predictably performing worst, by a factor of about 12.
With GNU sed:
echo 'Int/domain—home.dir=/etc/int' | sed 'h;s/[^=]*//;x;s/=.*//;s/[/—.]/_/g;G;s/\n//'
Output:
Int_domain_home_dir=/etc/int
See: man sed. I assume you want to replace dots too.
If perl solution is okay:
$ echo 'Int/domain-home.dir=/etc/int' | perl -pe 's#^[^=]+#$&=~s|[/.-]|_|gr#e'
Int_domain_home_dir=/etc/int
^[^=]+ string matching from start of line up to but not including the first occurrence of =
$&=~s|[/.-]|_|gr perform another substitution on matched string
replace all / or . or - characters with _
the r modifier would return the modified string
the e modifier allows to use expression instead of string in replacement section
# is used as delimiter to avoid having to escape / inside the character class [/.-]
Also, as suggested by #mklement0, we can use translate instead of inner substitute
$ echo 'Int/domain-home.dir=/etc/int' | perl -pe 's#^[^=]+#$&=~tr|/.-|_|r#e'
Int_domain_home_dir=/etc/int
Note that I've changed sample input, - is used instead of — which is what OP seems to want based on comments
I have a 10,000,000 digit string of numbers. Numbers are not separated by anything, they are all crammed together like this (its a long string of the first 10,000,000 digits in pi):
1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
I'm trying to use sed to replace each number with a hex color value. Here's my code:
sed -e 's/0/#F5F5F5/' -e 's/1/#FFE4B5/' -e 's/2/#98FB98/' -e 's/3/#ADFF2F/' -e 's/4/#FF69B4/' -e 's/5/#BA55D3/' -e 's/6/#FF6347/' -e 's/7/#2E8B57/' -e 's/8/#8B4513/' -e 's/9/#000000/' < pi > pi2
trouble is, sed starts converting numbers from my hexcode substitutions. I want those conversions to stay untouched. How do I prevent that? Hope this is clear enough.
ending up with results like this:
#FFE#FF#FF634#2E#8B4513B57#000000B4B#BA55D34159#98FB986535897932384626433832795#F5F5F528841971693993751
You can stick with this approach if you add one sed command to the beginning and make a small change to each of your existing commands:
sed -e 's/\(.\)/\1,/g' -e 's/0,/#F5F5F5/g' -e 's/1,/#FFE4B5/g' ... < pi > pi2
1) Mark each integer in the input by following it with a comma.
2) When you do the substitution, only replace integers that are followed by a comma, and remove the comma when you're done with each one.
Also, as a bug fix, you should also add the g option on each substitution.
try:
LC_LANG=C sed -e 's/[0-9]/,_\1_,/g' <pi >pi2| sed -e 's/,_0_,/#F5F5F5/g;s/,_1_,/#FFE4B5/g .....'
and change the "delimiters" ,_ and _, accordingly depending on the content of the file. (so that it never interfere with the content).
And maybe you need to add a space or ; after each color codes?... and maybe also a newline once in a while ^^
Probably easier to use perl:
perl -pe 'BEGIN{ $/=\1 }; s/0/#F5F5F5/ or s/1/#FFE4B5/ or s/2/#98FB98/ ...'
$ cat tst.awk
BEGIN{
codes="F5F5F5 FFE4B5 98FB98 ADFF2F FF69B4 BA55D3 FF6347 2E8B57 8B4513 000000"
split(codes,hex)
FS=""
}
{ for (i=1;i<=NF;i++)
printf "#%s",hex[($i)+1]
print ""
}
$ echo 1415 | awk -f tst.awk
#FFE4B5#FF69B4#FFE4B5#BA55D3
$ echo 1042 | awk -f tst.awk
#FFE4B5#F5F5F5#FF69B4#98FB98
The above just creates an array mapping numbers to the codes in the string and then for each digit on the input prints the string from the array. Awk arrays start at 1, not zero, hence adding 1 to $i.
How can I replace a newline ("\n") with a space ("") using the sed command?
I unsuccessfully tried:
sed 's#\n# #g' file
sed 's#^$# #g' file
How do I fix it?
sed is intended to be used on line-based input. Although it can do what you need.
A better option here is to use the tr command as follows:
tr '\n' ' ' < input_filename
or remove the newline characters entirely:
tr -d '\n' < input.txt > output.txt
or if you have the GNU version (with its long options)
tr --delete '\n' < input.txt > output.txt
Use this solution with GNU sed:
sed ':a;N;$!ba;s/\n/ /g' file
This will read the whole file in a loop (':a;N;$!ba), then replaces the newline(s) with a space (s/\n/ /g). Additional substitutions can be simply appended if needed.
Explanation:
sed starts by reading the first line excluding the newline into the pattern space.
Create a label via :a.
Append a newline and next line to the pattern space via N.
If we are before the last line, branch to the created label $!ba ($! means not to do it on the last line. This is necessary to avoid executing N again, which would terminate the script if there is no more input!).
Finally the substitution replaces every newline with a space on the pattern space (which is the whole file).
Here is cross-platform compatible syntax which works with BSD and OS X's sed (as per #Benjie comment):
sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/ /g' file
As you can see, using sed for this otherwise simple problem is problematic. For a simpler and adequate solution see this answer.
Fast answer
sed ':a;N;$!ba;s/\n/ /g' file
:a create a label 'a'
N append the next line to the pattern space
$! if not the last line, ba branch (go to) label 'a'
s substitute, /\n/ regex for new line, / / by a space, /g global match (as many times as it can)
sed will loop through step 1 to 3 until it reach the last line, getting all lines fit in the pattern space where sed will substitute all \n characters
Alternatives
All alternatives, unlike sed will not need to reach the last line to begin the process
with bash, slow
while read line; do printf "%s" "$line "; done < file
with perl, sed-like speed
perl -p -e 's/\n/ /' file
with tr, faster than sed, can replace by one character only
tr '\n' ' ' < file
with paste, tr-like speed, can replace by one character only
paste -s -d ' ' file
with awk, tr-like speed
awk 1 ORS=' ' file
Other alternative like "echo $(< file)" is slow, works only on small files and needs to process the whole file to begin the process.
Long answer from the sed FAQ 5.10
5.10. Why can't I match or delete a newline using the \n escape
sequence? Why can't I match 2 or more lines using \n?
The \n will never match the newline at the end-of-line because the
newline is always stripped off before the line is placed into the
pattern space. To get 2 or more lines into the pattern space, use
the 'N' command or something similar (such as 'H;...;g;').
Sed works like this: sed reads one line at a time, chops off the
terminating newline, puts what is left into the pattern space where
the sed script can address or change it, and when the pattern space
is printed, appends a newline to stdout (or to a file). If the
pattern space is entirely or partially deleted with 'd' or 'D', the
newline is not added in such cases. Thus, scripts like
sed 's/\n//' file # to delete newlines from each line
sed 's/\n/foo\n/' file # to add a word to the end of each line
will NEVER work, because the trailing newline is removed before
the line is put into the pattern space. To perform the above tasks,
use one of these scripts instead:
tr -d '\n' < file # use tr to delete newlines
sed ':a;N;$!ba;s/\n//g' file # GNU sed to delete newlines
sed 's/$/ foo/' file # add "foo" to end of each line
Since versions of sed other than GNU sed have limits to the size of
the pattern buffer, the Unix 'tr' utility is to be preferred here.
If the last line of the file contains a newline, GNU sed will add
that newline to the output but delete all others, whereas tr will
delete all newlines.
To match a block of two or more lines, there are 3 basic choices:
(1) use the 'N' command to add the Next line to the pattern space;
(2) use the 'H' command at least twice to append the current line
to the Hold space, and then retrieve the lines from the hold space
with x, g, or G; or (3) use address ranges (see section 3.3, above)
to match lines between two specified addresses.
Choices (1) and (2) will put an \n into the pattern space, where it
can be addressed as desired ('s/ABC\nXYZ/alphabet/g'). One example
of using 'N' to delete a block of lines appears in section 4.13
("How do I delete a block of specific consecutive lines?"). This
example can be modified by changing the delete command to something
else, like 'p' (print), 'i' (insert), 'c' (change), 'a' (append),
or 's' (substitute).
Choice (3) will not put an \n into the pattern space, but it does
match a block of consecutive lines, so it may be that you don't
even need the \n to find what you're looking for. Since GNU sed
version 3.02.80 now supports this syntax:
sed '/start/,+4d' # to delete "start" plus the next 4 lines,
in addition to the traditional '/from here/,/to there/{...}' range
addresses, it may be possible to avoid the use of \n entirely.
A shorter awk alternative:
awk 1 ORS=' '
Explanation
An awk program is built up of rules which consist of conditional code-blocks, i.e.:
condition { code-block }
If the code-block is omitted, the default is used: { print $0 }. Thus, the 1 is interpreted as a true condition and print $0 is executed for each line.
When awk reads the input it splits it into records based on the value of RS (Record Separator), which by default is a newline, thus awk will by default parse the input line-wise. The splitting also involves stripping off RS from the input record.
Now, when printing a record, ORS (Output Record Separator) is appended to it, default is again a newline. So by changing ORS to a space all newlines are changed to spaces.
GNU sed has an option, -z, for null-separated records (lines). You can just call:
sed -z 's/\n/ /g'
The Perl version works the way you expected.
perl -i -p -e 's/\n//' file
As pointed out in the comments, it's worth noting that this edits in place. -i.bak will give you a backup of the original file before the replacement in case your regular expression isn't as smart as you thought.
Who needs sed? Here is the bash way:
cat test.txt | while read line; do echo -n "$line "; done
In order to replace all newlines with spaces using awk, without reading the whole file into memory:
awk '{printf "%s ", $0}' inputfile
If you want a final newline:
awk '{printf "%s ", $0} END {printf "\n"}' inputfile
You can use a character other than space:
awk '{printf "%s|", $0} END {printf "\n"}' inputfile
tr '\n' ' '
is the command.
Simple and easy to use.
Three things.
tr (or cat, etc.) is absolutely not needed. (GNU) sed and (GNU) awk, when combined, can do 99.9% of any text processing you need.
stream != line based. ed is a line-based editor. sed is not. See sed lecture for more information on the difference. Most people confuse sed to be line-based because it is, by default, not very greedy in its pattern matching for SIMPLE matches - for instance, when doing pattern searching and replacing by one or two characters, it by default only replaces on the first match it finds (unless specified otherwise by the global command). There would not even be a global command if it were line-based rather than STREAM-based, because it would evaluate only lines at a time. Try running ed; you'll notice the difference. ed is pretty useful if you want to iterate over specific lines (such as in a for-loop), but most of the times you'll just want sed.
That being said,
sed -e '{:q;N;s/\n/ /g;t q}' file
works just fine in GNU sed version 4.2.1. The above command will replace all newlines with spaces. It's ugly and a bit cumbersome to type in, but it works just fine. The {}'s can be left out, as they're only included for sanity reasons.
Why didn't I find a simple solution with awk?
awk '{printf $0}' file
printf will print the every line without newlines, if you want to separate the original lines with a space or other:
awk '{printf $0 " "}' file
The answer with the :a label ...
How can I replace a newline (\n) using sed?
... does not work in freebsd 7.2 on the command line:
( echo foo ; echo bar ) | sed ':a;N;$!ba;s/\n/ /g'
sed: 1: ":a;N;$!ba;s/\n/ /g": unused label 'a;N;$!ba;s/\n/ /g'
foo
bar
But does if you put the sed script in a file or use -e to "build" the sed script...
> (echo foo; echo bar) | sed -e :a -e N -e '$!ba' -e 's/\n/ /g'
foo bar
or ...
> cat > x.sed << eof
:a
N
$!ba
s/\n/ /g
eof
> (echo foo; echo bar) | sed -f x.sed
foo bar
Maybe the sed in OS X is similar.
Easy-to-understand Solution
I had this problem. The kicker was that I needed the solution to work on BSD's (Mac OS X) and GNU's (Linux and Cygwin) sed and tr:
$ echo 'foo
bar
baz
foo2
bar2
baz2' \
| tr '\n' '\000' \
| sed 's:\x00\x00.*:\n:g' \
| tr '\000' '\n'
Output:
foo
bar
baz
(has trailing newline)
It works on Linux, OS X, and BSD - even without UTF-8 support or with a crappy terminal.
Use tr to swap the newline with another character.
NULL (\000 or \x00) is nice because it doesn't need UTF-8 support and it's not likely to be used.
Use sed to match the NULL
Use tr to swap back extra newlines if you need them
You can use xargs:
seq 10 | xargs
or
seq 10 | xargs echo -n
cat file | xargs
for the sake of completeness
If you are unfortunate enough to have to deal with Windows line endings, you need to remove the \r and the \n:
tr '\r\n' ' ' < $input > $output
I'm not an expert, but I guess in sed you'd first need to append the next line into the pattern space, bij using "N". From the section "Multiline Pattern Space" in "Advanced sed Commands" of the book sed & awk (Dale Dougherty and Arnold Robbins; O'Reilly 1997; page 107 in the preview):
The multiline Next (N) command creates a multiline pattern space by reading a new line of input and appending it to the contents of the pattern space. The original contents of pattern space and the new input line are separated by a newline. The embedded newline character can be matched in patterns by the escape sequence "\n". In a multiline pattern space, the metacharacter "^" matches the very first character of the pattern space, and not the character(s) following any embedded newline(s). Similarly, "$" matches only the final newline in the pattern space, and not any embedded newline(s). After the Next command is executed, control is then passed to subsequent commands in the script.
From man sed:
[2addr]N
Append the next line of input to the pattern space, using an embedded newline character to separate the appended material from the original contents. Note that the current line number changes.
I've used this to search (multiple) badly formatted log files, in which the search string may be found on an "orphaned" next line.
In response to the "tr" solution above, on Windows (probably using the Gnuwin32 version of tr), the proposed solution:
tr '\n' ' ' < input
was not working for me, it'd either error or actually replace the \n w/ '' for some reason.
Using another feature of tr, the "delete" option -d did work though:
tr -d '\n' < input
or '\r\n' instead of '\n'
I used a hybrid approach to get around the newline thing by using tr to replace newlines with tabs, then replacing tabs with whatever I want. In this case, " " since I'm trying to generate HTML breaks.
echo -e "a\nb\nc\n" |tr '\n' '\t' | sed 's/\t/ <br> /g'`
You can also use this method:
sed 'x;G;1!h;s/\n/ /g;$!d'
Explanation
x - which is used to exchange the data from both space (pattern and hold).
G - which is used to append the data from hold space to pattern space.
h - which is used to copy the pattern space to hold space.
1!h - During first line won't copy pattern space to hold space due to \n is
available in pattern space.
$!d - Clear the pattern space every time before getting the next line until the
the last line.
Flow
When the first line get from the input, an exchange is made, so 1 goes to hold space and \n comes to pattern space, appending the hold space to pattern space, and a substitution is performed and deletes the pattern space.
During the second line, an exchange is made, 2 goes to hold space and 1 comes to the pattern space, G append the hold space into the pattern space, h copy the pattern to it, the substitution is made and deleted. This operation is continued until EOF is reached and prints the exact result.
Bullet-proof solution. Binary-data-safe and POSIX-compliant, but slow.
POSIX sed
requires input according to the
POSIX text file
and
POSIX line
definitions, so NULL-bytes and too long lines are not allowed and each line must end with a newline (including the last line). This makes it hard to use sed for processing arbitrary input data.
The following solution avoids sed and instead converts the input bytes to octal codes and then to bytes again, but intercepts octal code 012 (newline) and outputs the replacement string in place of it. As far as I can tell the solution is POSIX-compliant, so it should work on a wide variety of platforms.
od -A n -t o1 -v | tr ' \t' '\n\n' | grep . |
while read x; do [ "0$x" -eq 012 ] && printf '<br>\n' || printf "\\$x"; done
POSIX reference documentation:
sh,
shell command language,
od,
tr,
grep,
read,
[,
printf.
Both read, [, and printf are built-ins in at least bash, but that is probably not guaranteed by POSIX, so on some platforms it could be that each input byte will start one or more new processes, which will slow things down. Even in bash this solution only reaches about 50 kB/s, so it's not suited for large files.
Tested on Ubuntu (bash, dash, and busybox), FreeBSD, and OpenBSD.
In some situations maybe you can change RS to some other string or character. This way, \n is available for sub/gsub:
$ gawk 'BEGIN {RS="dn" } {gsub("\n"," ") ;print $0 }' file
The power of shell scripting is that if you do not know how to do it in one way you can do it in another way. And many times you have more things to take into account than make a complex solution on a simple problem.
Regarding the thing that gawk is slow... and reads the file into memory, I do not know this, but to me gawk seems to work with one line at the time and is very very fast (not that fast as some of the others, but the time to write and test also counts).
I process MB and even GB of data, and the only limit I found is line size.
Finds and replaces using allowing \n
sed -ie -z 's/Marker\n/# Marker Comment\nMarker\n/g' myfile.txt
Marker
Becomes
# Marker Comment
Marker
You could use xargs — it will replace \n with a space by default.
However, it would have problems if your input has any case of an unterminated quote, e.g. if the quote signs on a given line don't match.
On Mac OS X (using FreeBSD sed):
# replace each newline with a space
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g; ta'
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g' -e ta
To remove empty lines:
sed -n "s/^$//;t;p;"
Using Awk:
awk "BEGIN { o=\"\" } { o=o \" \" \$0 } END { print o; }"
A solution I particularly like is to append all the file in the hold space and replace all newlines at the end of file:
$ (echo foo; echo bar) | sed -n 'H;${x;s/\n//g;p;}'
foobar
However, someone said me the hold space can be finite in some sed implementations.
Replace newlines with any string, and replace the last newline too
The pure tr solutions can only replace with a single character, and the pure sed solutions don't replace the last newline of the input. The following solution fixes these problems, and seems to be safe for binary data (even with a UTF-8 locale):
printf '1\n2\n3\n' |
sed 's/%/%p/g;s/#/%a/g' | tr '\n' # | sed 's/#/<br>/g;s/%a/#/g;s/%p/%/g'
Result:
1<br>2<br>3<br>
It is sed that introduces the new-lines after "normal" substitution. First, it trims the new-line char, then it processes according to your instructions, then it introduces a new-line.
Using sed you can replace "the end" of a line (not the new-line char) after being trimmed, with a string of your choice, for each input line; but, sed will output different lines. For example, suppose you wanted to replace the "end of line" with "===" (more general than a replacing with a single space):
PROMPT~$ cat <<EOF |sed 's/$/===/g'
first line
second line
3rd line
EOF
first line===
second line===
3rd line===
PROMPT~$
To replace the new-line char with the string, you can, inefficiently though, use tr , as pointed before, to replace the newline-chars with a "special char" and then use sed to replace that special char with the string you want.
For example:
PROMPT~$ cat <<EOF | tr '\n' $'\x01'|sed -e 's/\x01/===/g'
first line
second line
3rd line
EOF
first line===second line===3rd line===PROMPT~$