All values:
$ awk -F '[ -]*' '$0=$NF' /tmp/pwpower.log
114.10
130.09
Value on first line:
$ awk -F '[ -]*' 'NR==1{print $NF;exit}' /tmp/pwpower.log
114.10
Value on second line:
$ awk -F '[ -]*' 'NR==2{print $NF;exit}' /tmp/pwpower.log
130.09
Sum of all values:
$ awk -F '[ -]*' '{sum+=$NF} END{print sum}' /tmp/pwpower.log
244.19
don's might be better in most cases, but just in case the file is really big, and you can't get sed
to handle a script file that large (which can happen at around 5000+ lines of script), here it is with plain sed
:
sed -ne:t -e"/\n.*$match/D" \
-e'$!N;//D;/'"$match/{" \
-e"s/\n/&/$A;t" \
-e'$q;bt' -e\} \
-e's/\n/&/'"$B;tP" \
-e'$!bt' -e:P -e'P;D'
This is an example of what is called a sliding window on input. It works by building a look-ahead buffer of $B
-count lines before ever attempting to print anything.
And actually, probably I should clarify my previous point: the primary performance limiter for both this solution and don's will be directly related to interval. This solution will slow with larger interval sizes, whereas don's will slow with larger interval frequencies. In other words, even if the input file is very large, if the actual interval occurrence is still very infrequent then his solution is probably the way to go. However, if the interval size is relatively manageable, and is likely to occur often, then this is the solution you should choose.
So here's the workflow:
- If
$match
is found in pattern space preceded by a \n
ewline, sed
will recursively D
elete every \n
ewline that precedes it.
- I was clearing
$match
's pattern space out completely before - but to easily handle overlap, leaving a landmark seems to work far better.
- I also tried
s/.*\n.*\($match\)/\1/
to try to get it in one go and dodge the loop, but when $A/$B
are large, the D
elete loop proves considerably faster.
- Then we pull in the
N
ext line of input preceded by a \n
ewline delimiter and try once again to D
elete a /\n.*$match/
once again by referring to our most recently used regular expression w/ //
.
- If pattern space matches
$match
then it can only do so with $match
at the head of the line - all $B
efore lines have been cleared.
- So we start looping over
$A
fter.
- Each run of this loop we'll attempt to
s///
ubstitute for &
itself the $A
th \n
ewline character in pattern space, and, if successful, t
est will branch us - and our whole $A
fter buffer - out of the script entirely to start the script over from the top with the next input line if any.
- If the
t
est is not successful we'll b
ranch back to the :t
op label and recurse for another line of input - possibly starting the loop over if $match
occurs while gathering $A
fter.
- If we get past a
$match
function loop, then we'll try to p
rint the $
last line if this is it, and if !
not try to s///
ubstitute for &
itself the $B
th \n
ewline character in pattern space.
- We'll
t
est this, too, and if it is successful we'll branch to the :P
rint label.
- If not we'll branch back to
:t
op and get another input line appended to the buffer.
- If we make it to
:P
rint we'll P
rint then D
elete up to the first \n
ewline in pattern space and rerun the script from the top with what remains.
And so this time, if we were doing A=2 B=2 match=5; seq 5 | sed...
The pattern space for the first iteration at :P
rint would look like:
^1\n2\n3$
And that's how sed
gathers its $B
efore buffer. And so sed
prints to output $B
-count lines behind the input it has gathered. This means that, given our previous example, sed
would P
rint 1
to output, and then D
elete that and send back to the top of the script a pattern space which looks like:
^2\n3$
...and at the top of the script the N
ext input line is retrieved and so the next iteration looks like:
^2\n3\n4$
And so when we find the first occurrence of 5
in input, the pattern space actually looks like:
^3\n4\n5$
Then the D
elete loop kicks in and when it's through it looks like:
^5$
And when the N
ext input line is pulled sed
hits EOF and quits. By that time it has only ever P
rinted lines 1 and 2.
Here's an example run:
A=8 B=7 match='[24689]0'
seq 100 |
sed -ne:t -e"/\n.*$match/D" \
-e'$!N;//D;/'"$match/{" \
-e"s/\n/&/$A;t" \
-e'$q;bt' -e\} \
-e's/\n/&/'"$B;tP" \
-e'$!bt' -e:P -e'P;D'
That prints:
1
2
3
4
5
6
7
8
9
10
11
12
29
30
31
32
49
50
51
52
69
70
71
72
99
100
Best Answer
You can do:
This will look for lines ending in
&
, remove$
to match lines with&
anywhere in them.The
-B
switch tellsgrep
to output "context" lines that come before the lines that match. In this case, since you want one line of context, you need-B 1
.This switch is available in GNU
grep
but is not in the POSIX standard, though.Here's a simple
sed
solution that should help in case you don't have GNUgrep
:How it works
-n
switch suppresses the default "print" action ofsed
/&/!N
means: if the current line doesn't contain&
, append the next line to the pattern space./&/p
means: match&
and print the pattern space.