Grep command to extract parameter values from file

Question

I am totally new to unix and need some help regarding grep command

I have an output file with 7 parameter values S1, S2..., S7. How can I use grep command to extract only the values and write them as an array to a new file, which I want to plot later.

<<<<< Function evaluation summary (APPROX_INTERFACE_1): 1808 total (1808 new, 0 duplicate)
<<<<< Best parameters          =
                      3.3500000000e+02 S1
                      1.0357218750e+00 S2
                      7.2046250000e+01 S3
                      7.2872921875e+01 S4
                      5.7405507812e+01 S5
                      5.0000054687e+01 S6
                      2.3999984375e+02 S7
<<<<< Best objective function  =
                      8.0674284316e-01

Answer 1

Grep is a line selection tool.

grep ' S.$' your.txt 

will give you the lines with the 7 values.

If you need to trim the values from each line you can use cut:

grep ' S.$' yout.txt |cut -d ' '  -f23

Use 23 if you have 22 spaces in the beggining of each line.

AWK will be a good alternative to both grep and cut.

Answer 2

If we can be sure that what we want is the first field of every line whose second field is a S followed by a number, we can do:

$ awk '$2~/^S[0-9]$/{print $1}' data
3.3500000000e+02
1.0357218750e+00
7.2046250000e+01
7.2872921875e+01
5.7405507812e+01
5.0000054687e+01
2.3999984375e+02

If you want those as values in an array, do:

array=( $(awk '$2~/^S[0-9]$/{print $1}' data) )

And if you want them in a new file, do:

awk '$2~/^S[0-9]$/{print $1}' data > new.data