Reading 3 consecutive lines with awk? (Or something else)-CodePudding

I have a huge file, and at some point it goes like this:

Bla bla bla
LAST ITERATION:        1780           6          12  0.689655172413793
  -8708.81862246834       -8698.33572943212       -2003.09638506407
 -9.912281246897692E-003
Bla bla bla

I would like to get all the numbers after "LAST ITERATION:" and put it in a line in a file. I have managed to get the two first lines with this:

awk 'a && NR==n{ print a,b,c,d,$1,$3 } /LAST ITERATION:/{ a=$3; b=$4; c=$5; d=$6; n=NR 1 }' ./$FOLDER/$NAMEDATA >> $NAMEOUTPUT

But I can't seem to find a way to get the last number which is on the 3rd line. Could anyone help me?

Thanks!

CodePudding user response：

If you are using gawk, where RS can be more than one character, an easier approach would be to use "LAST ITERATION:" as the record separator, and print the first 8 fields of the second record, with a space as the output record seperator:

gawk 'BEGIN{RS="LAST ITERATION:";ORS=" "}NR==2{for(n=1;n<=8;  n)print$n}'

Demo: https://ideone.com/UzwkdN

CodePudding user response：

You might use getline Variable to access future lines, I would use GNU AWK for this task following way, let file.txt content be

Bla bla bla
LAST ITERATION:        1780           6          12  0.689655172413793
  -8708.81862246834       -8698.33572943212       -2003.09638506407
 -9.912281246897692E-003
Bla bla bla

then

awk '/LAST ITERATION/{getline x1;getline x2;line=$0 " " x1 " " x2;sub(/LAST ITERATION:[[:space:]] /,"",line);gsub(/[[:space:]] /," ",line);print line}' file.txt

gives output

1780 6 12 0.689655172413793 -8708.81862246834 -8698.33572943212 -2003.09638506407 -9.912281246897692E-003

Explanation: when LAST ITERATION is encountered I save next line to variable x1 and next next line to variable x2, then construct line from space-sheared current line, next line and next next line, then remove LAST ITERATION: and following whitespaces characters using sub function and alter multiple whitespace characters to single spaces using gsub, after doing that I print said line. Disclaimer: this solution assumes there are always at least 2 lines after line with LAST ITERATION.

(tested in gawk 4.2.1)

CodePudding user response：

If you are using regular awk, you can concatenate the 3 lines starting with the line with LAST ITERATION: to a variable, and split it in the end:

awk 'BEGIN{ORS=" "}/LAST ITERATION:/{n=NR 3}NR<n{s=s$0}END{split(s,a);for(i=3;i<11;  i)print(a[i])}'

CodePudding user response：

If I read your question correctly, you say you want to see an entry within a file, followed by two following lines.
Next to that, you seem to want to work with awk.

My answer is: "Why awk? This can very easily be done using grep.", as follows:

grep has three switches to add more lines, next to the matching one:

-A n : add "n" lines after the match
-B n : add "n" lines before the match
-C n : add "n" lines, combined after and before

So, you can simply use grep -A 2 "match" filename.

Good luck

CodePudding user response：

For batch you can target lines using findstr:

The below can target single lines, consecutive lines or a list of lines.

@Echo off

REM :: CALL %0 <filepath> <integer|integer list>

    Set "File.Name="

    If exist "%~f1" (
        Set "File.Name=%~f1"
    )Else (
        1>&2 Echo(File Not Found
        Exit /B 1
    )
    
    If "%~2" == "" (
        1>&2 Echo(Missing Arg/s for target Line/s.
        Exit /B 2
    )

    Set Target.Lines=%*
    Set "Target.Lines=%Target.Lines:"=%"

    For /f "Delims=" %%G in ('CMD /V:ON /C "Echo(!Target.Lines:%~1 =!"')Do Set "Target.Lines=%%G"

    Set "Line.Count=0"
    Set "Highest.Target=0"
        Setlocal EnableExtensions EnableDelayedExpansion
            For %%G in (%Target.Lines%)Do if !Highest.Target! LSS %%G Set "Highest.Target=%%G"
        Endlocal & Set "Highest.Target=%Highest.Target%"

    For /f "Tokens=1* Delims=:" %%G in ('%SystemRoot%\System32\findstr.exe /VNLC:"{false.string[%~n0]}" "%File.Name%"')Do (
        Set "Line.Count=%%G"
        For %%i in (%Target.Lines%)Do if %%G EQU %%i Echo(%%H
    )

    If %Line.Count% LSS %Highest.Target% (
        1>&2 Echo(Lines missing. Lines in file: %Line.Count%. Highest Target: %Highest.Target%
    )

CodePudding user response：

  awk '
     gsub(/LAST ITERATION:/,""){
        gsub(/^ */,""); 
        o=sprintf("%s ",$0); 
        while(getline && $1 ~ /^[-]?[0-9] (\.[0-9]*)?/) {
           o=o sprintf("%s ", $0)
        } 
        print gensub(/  */," ", "g", o) > "output_file"
    }
  ' input_file 

$ cat output_file
1780 6 12 0.689655172413793 -8708.81862246834 -8698.33572943212 -2003.09638506407 -9.912281246897692E-003