It looks like the 'N' command works on every other line:
$ cat in.txt
a
b
c
d
$ sed '=;N' in.txt
1
a
b
3
c
d
Maybe that would be natural because command 'N' joins the next line and changes the current line number. But (I saw this here):
$ sed 'N;$!P;$!D;$d' thegeekstuff.txt
The above example deletes the last two lines of a file. This works not only for even-line-numbered files but also for odd-line-numbered files. In this example 'N' command runs on every line. What's the difference?
And could you tell me why I cannot see the last line when I run sed like this:
# sed N odd-lined-file.txt
Excerpt from info sed:
`sed' operates by performing the following cycle on each lines of
input: first, `sed' reads one line from the input stream, removes any
trailing newline, and places it in the pattern space. Then commands
are executed; each command can have an address associated to it:
addresses are a kind of condition code, and a command is only executed
if the condition is verified before the command is to be executed.
...
When the end of the script is reached, unless the `-n' option is in
use, the contents of pattern space are printed out to the output
stream,
...
Unless special commands (like 'D') are used, the pattern space is
deleted between two cycles
...
`N'
Add a newline to the pattern space, then append the next line of
input to the pattern space. If there is no more input then `sed'
exits without processing any more commands.
...
`D'
Delete text in the pattern space up to the first newline. If any
text is left, restart cycle with the resultant pattern space
(without reading a new line of input), otherwise start a normal
new cycle.
This should pretty much resolve your query. But still I will try to explain your three different cases:
CASE 1:
sed reads a line from input. [Now there is 1 line in pattern space.]
= Prints the current line no.
N reads the next line into pattern space.[Now there are 2 lines in pattern space.]
If there is no next line to read then sed exits here. [ie: In case of odd lines, sed exits here - and hence the last line is swallowed without printing.]
sed prints the pattern space and cleans it. [Pattern space is empty.]
If EOF reached sed exits here. Else Restart the complete cycle from step 1. [ie: In case of even lines, sed exits here.]
Summary: In this case sed reads 2 lines and prints 2 lines at a time. Last line is swallowed it there are odd lines (see step 3).
CASE 2:
sed reads a line from input. [Now there is 1 line in pattern space.]
N reads the next line into pattern space. [Now there are 2 lines in pattern space.]
If it fails exit here. This occurs only if there is 1 line.
If its not last line($!) print the first line(P) from pattern space. [The first line from pattern space is printed. But still there are 2 lines in pattern space.]
If its not last line($!) delete the first line(D) from pattern space [Now there is only 1 line (the second one) in the pattern space.] and restart the command cycle from step 2. And its because of the command D (see the excerpt above).
If its last line($) then delete(d) the complete pattern space. [ie. reached EOF ] [Before beginning this step there were 2 lines in the pattern space which are now cleaned up by d - at the end of this step, the pattern space is empty.]
sed automatically stops at EOF.
Summary: In this case :
sed reads 2 lines first.
if there is next line available to read, print the first line and read the next line.
else delete both lines from cache. This way it always deletes the last 2 line.
CASE 3:
Its the same case as CASE:1, just remove the Step 2 from it.
Related
I saw a sed command to delete the last 10 rows of data:
sed -e :a -e '$d;N;2,10ba' -e 'P;D'
But I don't understand how it works. Can someone explain it for me?
UPDATE:
Here is my understanding of this command:
The first script indicates that a label “a” is defined.
The second script indicates that it first determines whether the
line currently reading pattern space is the last line. If it is,
execute the "d" command to delete it and restart the next cycle; if
not, skip the "d" command; then execute "N" command: append a new
line from the input file to the pattern space, and then execute
"2,10ba": if the line currently reading the pattern space is a line
in the 2nd to 10th lines, jump to label "a".
The third script indicates that if the line currently read into
pattern space is not a line from line 2 to line 10, first execute "P" command: the first line
in pattern space is printed, and then execute "D" command: the first line in pattern
space is deleted.
My understanding of "$d" is that "d" will be executed when sed reads the last line into the pattern space. But it seems that every time "ba" is executed, "d" will be executed, regardless of Whether the current line read into pattern space is the last line. why?
:a is a label. $ in the address means the last line, d means delete. N stands for append the next line into the pattern space. 2,10 means lines 2 to 10, b means branch (i.e. goto), P prints the first line from the pattern space, D is like d but operates on the pattern space if possible.
In other words, you create a sliding window of the size 10. Each line is stored into it, and once it has 10 lines, lines start to get printed from the top of it. Every time a line is printed, the current line is stored in the sliding window at the bottom. When the last line gets printed, the sliding window is deleted, which removes the last 10 lines.
You can modify the commands to see what's getting deleted (()), stored (<>), and printed by the P ([]):
$ printf '%s\n' {1..20} | \
sed -e ':a ${s/^/(/;s/$/)/;p;d};s/^/</;s/$/>/;N;2,10ba;s/^/[/;s/$/]/;P;D'
[<<<<<<<<<<1>
[<2>
[<3>
[<4>
[<5>
[<6>
[<7>
[<8>
[<9>
[<10>
(11]>
12]>
13]>
14]>
15]>
16]>
17]>
18]>
19]>
20])
a simpler resort, if your data in 'd' file by gnu sed,
sed -Ez 's/(.*\n)(.*\n){10}$/\1/' d
^
pointed 10 is number of last line to remove
just move the brace group to invert, ie. to get only the last 10 lines
sed -Ez 's/.*\n((.*\n){10})$/\1/' d
I have the following (gnu) sed script, which is intended to parse another sed script, and output distinct commands on a separate line.
In words, this script should put a newline after each semicolon ;, except semicolons that are inside a matching or substitution command.
Sed script:
#!/bin/sed -rf
# IDEA:
# replace ';' by ';\n' except when it's inside a match expression or subst. expression.
# Ignored patterns:
/^#/b # commented lines
/^$/b # empty lines
# anything in a single line, without semicolon except at the end
/^[^\n;]*;?$/b
# Processed patterns (put on separate lines):
# Any match preceding a semicolon, or the end of the line, or a substitution
s_/^[^/]+/[^;s]*;?_&\n_; t printtopline
s/^\\(.)[^\1]+\1[^;s]*;?/&\n/;t printtopline
# Any substitution (TODO)
# Any other command, separated by semicolon
s/\;/\;\n/; t printtopline;
:printtopline
P;D; # print top line, delete it, start new cycle
For example, I tested it with the following file (actually adapted from an answer of #ctac_ to one of my previous sed questions):
Input file:
#!/bin/sed -f
#/^>/N;
:A;
/\n>/!{s/\n/ /;N;bA}; # join next line if not a sequence label
#h;
#s/\(.*\)\n.*/\1/p;
s/^>//g;P
#x;
#s/.*\n//;
D
bA;
Output
The above script produces the right output, for example, the line /\n>/!{s/\n/ /;N;bA}; # join next line if not a sequence label becomes:
/\n>/!{s/\n/ /;
N;
bA};
# join next line if not a sequence label
Question
However, could you help me understand why this part of the script works:
s/\;/\;\n/; t printtopline;
:printtopline
?
I seems to me that the branching command t printtopline is useless here. I thought whatever the success of the substitution, the next thing to be executed would be :printtopline.
However, if I comment out the t command, or if I replace it with b, the script produces the following output lines:
/\n>/!{s/\n/ /;
N;bA}; # join next line if not a sequence label
From info sed, here is the explanation of t:
't LABEL'
Branch to LABEL only if there has been a successful 's'ubstitution
since the last input line was read or conditional branch was taken.
The LABEL may be omitted, in which case the next cycle is started.
Why isn't the t command immediately followed by its label not behaving like no command at all or the b command?
The crucial part is this:
Branch to label only if there has been a successful substitution since the last input line was read or conditional branch was taken.
I.e. t looks into the past and takes into account the success of all recent substitutions up to the most recent
input, or
conditional branch.
Consider the input line you're asking about. After all the substitutions we have
/\n>/!{s/\n/ /;
N;bA}; # join next line if not a sequence label
in our pattern space when we reach P;D;. The P commands outputs the first line, then D deletes the first line and restarts the main loop. Now we just have
N;bA}; # join next line if not a sequence label
Note that this didn't involve reading any additional lines. No input occurred; D just removed parts of the pattern space.
We process the remaining text (which does nothing because none of the other patterns match) until we reach this part of the code:
s_/^[^/]+/[^;s]*;?_&\n_; t printtopline
The substitution fails (the pattern space doesn't contain /^). But the t command doesn't check the status of just this one s command; it looks at the history of all substitutions since the most recent input or conditional branch taken.
The most recent input occurred when /\n>/!{s/\n/ /;N;bA}; was read.
The most recent conditional branch taken was
s/\;/\;\n/; t printtopline;
:printtopline
in the original version of your code. Since then no other substitution succeeded, so the t command does nothing. The rest of the program continues as expected.
But in the modified version of your code there was no conditional branch at this point (b is an unconditional branch):
s/\;/\;\n/; b printtopline;
:printtopline
That means the t from s_/^[^/]+/[^;s]*;?_&\n_; t printtopline "sees" the s/\;/\;\n/; as having succeeded, so it immediately jumps to the P;D; part. This is what outputs
N;bA}; # join next line if not a sequence label
unmodified.
In summary: t makes a difference here not because of its immediate effect of jumping to a label, but because it serves as a dynamic delimiter for the next t that gets executed. Without t here, the previously executed s command is taken into account for the next t.
Part 1 - how the P;D; sequence works.
Compare this two command's outputs: sed 's/;/;\n/' and sed 's/;/;\n/; P;D;'.
First:
$ sed 's/;/;\n/' <<< 'one;two;three;four'
one;
two;three;four
Second:
$ sed 's/;/;\n/; P;D;' <<< 'one;two;three;four'
one;
two;
three;
four
Why the difference? I will to explain.
The first command substitutes only the first occurrence of the ; character. To substitute all occurrences, the g modifier should be added to the s command: sed 's/;/;\n/g'.
The second command works this way:
sed 's/;/;\n/; - the same as the first command - no difference. Before this command the pattern space is one;two;three;four, after - one\ntwo;three;four.
P; -
from man: "Print up to the first embedded newline of the current pattern space."
That is, it prints up to first newline - one. The pattern space stay unchanged: one\ntwo;three;four
D; -
from man: "If pattern space contains no newline, start a normal new cycle as if the d command was
issued. Otherwise, delete text in the pattern space up to the first newline, and restart
cycle with the resultant pattern space, without reading a new line of input."
In the our case, pattern space has newline - one\ntwo;three;four. The D; removes the one\n part and repeat all commands cycle from the beginning. Now, the pattern space is: two;three;four.
That is, again sed 's/;/;\n/; - pattern space: two\nthree;four, then P; - print two, pattern space unchanged: two\nthree;four, D; - removes two\n, pattern space becomes: three;four. Etc.
Part 2 - what happening with branching.
I looked at the sed source code and found next information:
When the s command is executing and having match, the replaced flag is setting to the true:
/* We found a match, set the 'replaced' flag. */
replaced = true;
The t command is executing, if the replaced flag is true. And it is changing this flag to the false:
case 't':
if (replaced)
{
replaced = false;
So, in the first, s/\;/\;\n/; t printtopline; case, the substitution is successful - therefore, replaced flag is setting to the true. Then, the following t command is running and changing replaced flag back to the false.
In the second case, without t command - s/\;/\;\n/;, substitution is successful, too - therefore, replaced flag is setting to the true.
But now, this flag is stored to the next cycle, initiated by the D command. So, then the first t command appears in the new cycle - s_/^[^/]+/[^;s]*;?_&\n_; t printtopline, it checks the replaced flag, sees, that the flag is true and jumps to the label :printtopline, omitting all other commands before the label.
The pattern space doesn't have newlines, so P;D; sequence just prints pattern space and starts the next cycle with the new line of input.
I found this magical command on the unix forum to move the last line of a file to the beginning of the file. I use sed quite a bit but not to this extent. Can someone explain each part to me?
sed '1h;1d;$!H;$!d;G' infile
Yes, it uses exotic commands.
1h: put first line in the "hold" space (sed has 2 spaces: 1 hold space to keep data and the pattern space: actual processed line)
1d: delete first line
$!H: append all lines BUT the last one (and the first one since d command skips to the next line) into the "hold" space
$!d: delete (do not print) all lines except the last one
G: Append a newline to the contents of the pattern space (this is the last line, the only one able to reach that part of the script), and then append the contents of the hold space to that of the pattern space, pattern space which is printed right away. Swap done.
Opinion based comment: I must admit I would never have thought of doing that using sed, and I would have had to make a test to convince me of what this command was doing... in awk, it is much much easier to do that.
But sed has a special place in my heart with it's cryptic commands. I wonder if there are some sed candidates to CodeGolf :)
reference manual: https://www.gnu.org/software/sed/manual/sed.html
some exotic things you can do with sed (my best 1999 read): http://sed.sourceforge.net/grabbag/tutorials/do_it_with_sed.txt
Here is the same command in a more procedural-looking pseudocode:
for line in infile:
# Always do this: Copy the current line to the pattern
pattern = line
# Process the script
if first line:
hold = pattern # 1h
pattern = ""; continue # 1d
elif not last line:
hold = hold + "\n" + pattern # $!H
pattern = ""; continue # $!d
pattern = pattern + "\n" + hold # G
# Always do this after the script is completed.
# Due to the continue statements above, this
# isn't always reached, and in this case
# is only reached for the last line.
print pattern
d clears the pattern space and continues to the next input line without executing the rest of the script.
h copies the pattern space to the hold space.
H appends a newline to the hold space, then appends the pattern space to the hold space.
G is like H, but in the other direction; it copies the hold space to the pattern space.
The overall affect on a file with N lines is to build up a copy of lines 1 through N-1 in the hold space. When the pattern holds line N, append the hold space to the pattern space and print the pattern space to standard output.
I have a file that, occasionally, has split lines. The split is signaled by the fact that the line starts with '+' (possibly preceeded by spaces).
line 1
line 2
+ continue 2
line 3
...
I'd like join the split line back:
line 1
line 2 continue 2
line 3
...
using sed. I'm not clear how to join a line with the preceeding one.
Any suggestion?
This might work for you:
sed 'N;s/\n\s*+//;P;D' file
These are actually four commands:
N
Append line from the input file to the pattern space
s/\n\s*+//
Remove newline, following whitespace and the plus
P
print line from the pattern space until the first newline
D
delete line from the pattern space until the first newline, e.g. the part which was just printed
The relevant manual page parts are
Selecting lines by numbers
Addresses overview
Multiline techniques - using D,G,H,N,P to process multiple lines
Doing this in sed is certainly a good exercise, but it's pretty trivial in perl:
perl -0777 -pe 's/\n\s*\+//g' input
I'm not partial to sed so this was a nice challenge for me.
sed -n '1{h;n};/^ *+ */{s// /;H;n};{x;s/\n//g;p};${x;p}'
In awk this is approximately:
awk '
NR == 1 {hold = $0; next}
/^ *\+/ {$1 = ""; hold=hold $0; next}
{print hold; hold = $0}
END {if (hold) print hold}
'
If the last line is a "+" line, the sed version will print a trailing blank line. Couldn't figure out how to suppress it.
You can use Vim in Ex mode:
ex -sc g/+/-j -cx file
g global search
- select previous line
j join with next line
x save and close
Different use of hold space with POSIX sed... to load the entire file into the hold space before merging lines.
sed -n '1x;1!H;${g;s/\n\s*+//g;p}'
1x on the first line, swap the line into the empty hold space
1!H on non-first lines, append to the hold space
$ on the last line:
g get the hold space (the entire file)
s/\n\s*+//g replace newlines preceeding +
p print everything
Input:
line 1
line 2
+ continue 2
+ continue 2 even more
line 3
+ continued
becomes
line 1
line 2 continue 2 continue 2 even more
line 3 continued
This (or potong's answer) might be more interesting than a sed -z implementation if other commands were desired for other manipulations of the data you can simply stick them in before 1!H, while sed -z is immediately loading the entire file into the pattern space. That means you aren't manipulating single lines at any point. Same for perl -0777.
In other words, if you want to also eliminate comment lines starting with *, add in /^\s*\*/d to delete the line
sed -n '1x;/^\s*\*/d;1!H;${g;s/\n\s*+//g;p}'
versus:
sed -z 's/\n\s*+//g;s/\n\s*\*[^\n]*\n/\n/g'
The former's accumulation in the hold space line by line keeps you in classic sed line processing land, while the latter's sed -z dumps you into what could be some painful substring regexes.
But that's sort of an edge case, and you could always just pipe sed -z back into sed. So +1 for that.
Footnote for internet searches: This is SPICE netlist syntax.
A solution for versions of sed that can read NUL separated data, like here GNU Sed's -z:
sed -z 's/\n\s*+//g'
Compared to potong's solution this has the advantage of being able to join multiple lines that start with +. For example:
line 1
line 2
+ continue 2
+ continue 2 even more
line 3
becomes
line 1
line 2 continue 2 continue 2 even more
line 3
I would like to operate on the last 7 lines of a file with sed regardless of the filelength.
According to a related question this type of range won't work: $-6,$ {..commands..}
What is the equivalent that will?
Pipe the output of tail -7 into sed.
tail -7 test.txt | sed -e "s/e/WWW/"
More info on Pipes here.
You could just switch from sed(1) to ed(1), the commands are about the same. In this case, the command is the same, except with no limitations on address range.
$ cat > fl7.ed
ed - $1 << \eof
1,7s/$/ (one of the first seven lines)/
$-6,$s/$/ (one of the last seven lines)/
w
q
eof
$ sh fl7.ed yourfile
perl -lne 'END{print join$\,#a,"-",#b}push#a,$_ if#a<6;push#b,$_;shift#b if#b>7'
In the END{} block you can do whatever is required; #a contains the first 6, #b the last 7 lines as requested.
This should work for you:
sed '1{N;N;N;N;N};N;$s/foo/bar/g;P;D' inputfile
Explanation:
1{N;N;N;N;N} - when the first line is read, load the pattern space with five more lines (total: 6 at this point)
N - append another line
$s/foo/bar/g - when the last line is read, perform some operation on the entire contents of pattern space (the last seven lines of the file). Operations can be more complex than shown here
P - print the test before the first newline in pattern space
D - delete the text just printed and loop to the beginning of the script (the "append another line" step - the first instruction is skipped since it only applies to the first line in the file)
This might work for you:
sed ':a;1,6{$!N;ba};${s/foo/bar/g;q};N;D' file
Explanation:
Create a loop label. :a
Gather lines 1 to 6 in the pattern space (PS). 1,6{$!N;ba}
If it's the last line, process the PS and quit, therefore printing out the last seven lines. ${s/foo/bar/g;q}
If it's not the last line, append the next line to the PS. N
Delete upto the first newline and begin a new cycle without reading a new line. D