Extracting part of lines with specific pattern using awk or sed

text-processing sed awk

105

Solution 1

With `grep`:

grep -oP 'sum=\K.*' inpufile > outputfile

grep with -P(perl-regexp) parameter supports \K, which use to ignoring the previously matched characters.

With `awk`:

awk -F"=" '{ print $NF; }' inputfile > outputfile

in awk the variable NF represent the total number of fields in a current record/line which is point to the last field number too and so $NF is its value accordingly.

With `sed`:

sed 's/^.*sum=//' inpufile > outputfile

^.*=sum replace all characters(.*) between starting of line(^) and last characters(sum=) with whitespace char.

Result:

-6.97168e-09
6.97168e-09
-5.12623e-12
5.12623e-12
-6.936e-09
6.97169e-09
-5.1e-12
5.12624e-12

With `cut`:

cut -d'=' -f2 inputfile > outputfile

if you want save same values into a same file and each separately, with awk you can do:

awk -F"=" '{print $NF >($NF); }' inputfile > outputfile

Solution 2

If I correctly understand the question you want to get only values after =, and store the these values in separate files, based on second field(?). If I'm right try something like this:

$ awk -F'[ =]' '{print $6>"file_"$2".txt"}' file

The result:

$ ls -1
  file_leftWallPhi.txt
  file_leftWallUSf.txt
  file_leftWallrhoPhi.txt
  file_leftWallrhoUSf.txt
  file_loweWallrhoPhi.txt
  file_loweWallrhoUSf.txt
  file_lowerWallPhi.txt
  file_lowerWallUSf.txt

$ cat  file_leftWallPhi.txt
  5.12623e-12

Solution 3

You can do it by sed

sed -E 's/^.* (\S+)\s*:.*=(\S+)/echo "\2" > "\1".txt/' file | bash

The script find out two pieces in line:

between spaces and : and should contain some(more then 0) non-space symbols ;
some(more then 0) non-space symbols after =;

and format from its in execution command which transfered through the pipe to bash

105

Author by

John Johnny

Updated on September 18, 2022

Comments

John Johnny over 1 year
I have a problem with following algorithm, What should be done to resolve this problem I type a year if the year is not a leap year I will need to type again untill the typed year is a leap year.
```
    int year = 0;        
    BOOL yearC;

    NSLog(@"Enter the year to be tested;");
    scanf ("%i", &year);
    yearC = ((year % 4 == 0 && year % 100 !=0)|| year % 400 == 0);
    if ( yearC  )
             NSLog (@"It's a leap year.");
    else 
    {
        NSLog (@"Nope, it's not a leap year.");
       scanf ("%i", &year);    
    }
}
```
- gasparuff almost 11 years
  
  Have you considered using a while-loop?
- Daij-Djan almost 11 years
  
  use a while loop to solve the assignment
- Git.Coach almost 11 years
  
  I would use a loop for this. In this case "while" should do the trick. ;-) Never give up, keep on improving! :)
- Desdenova almost 11 years
  
  stackoverflow.com/questions/2078372/…
Git.Coach almost 11 years

Aaaaand: Homework's done.
jimmij over 9 years

@KasiyA I cannot reproduce your problem with GNU awk 4.0.2. The command from my answer works also with -c option (compatibility mode with traditional UNIX awk where GNU extensions are disabled). Please be sure you have updated input file as the original question was edited and empty lines deleted.