awk, multiply column value, compare to a constant and find first match before exiting

Question

I have a data2process.tempdata file with data in six columns as follows.

# Angle(deg)    D1      D2          D3          D4          D5
0.50    0.00E+00    0.00E+00    1.97E-10    3.94E-09    1.97E-10
1.00    0.00E+00    0.00E+00    1.70E-10    3.41E-09    1.70E-10
1.50    0.00E+00    0.00E+00    1.30E-10    2.59E-09    1.30E-10
2.00    0.00E+00    0.00E+00    9.49E-11    1.90E-09    9.49E-11
2.50    0.00E+00    0.00E+00    6.99E-11    1.40E-09    6.99E-11
3.00    0.00E+00    0.00E+00    5.27E-11    1.05E-09    5.27E-11
3.50    0.00E+00    0.00E+00    4.12E-11    8.23E-10    4.12E-11
4.00    0.00E+00    0.00E+00    3.33E-11    6.66E-10    3.33E-11
4.50    0.00E+00    0.00E+00    2.78E-11    5.56E-10    2.78E-11
5.00    0.00E+00    0.00E+00    2.37E-11    4.74E-10    2.37E-11

I want to multiply sixth column(D5) with 3.6E8 and find the first instance when the product becomes equal or less than 0.05. At that point, I want to save the corresponding value in the first column, print it and exit. In the (edited) data shown above, the first instance when the condition is met at 1.5 deg (1.3e-10*3.6e8=0.0468) but it holds true after that too. So echo $ maxangle should be 1.5, not 5.0 which here is the last entry which also satisfies the condition.

awk version is GNU Awk 5.0.1, API: 2.0 (GNU MPFR 4.0.2, GNU MP 6.2.0)

I was trying with the following awk commands.

maxangle=$(awk -v maxt=0.05; $6 *=3.6e8 '{if($6<=maxt){maxt=$6; maxa=$1}}END{print maxa ; exit} ' data2process.tempdata)

This gives an error *=3.6e8: command not found

maxangle=$(awk -v maxt=0.05 v6="$arg6"*3.6e8 '{if($v6<maxt){maxt=$1; maxa=$1}}END{print maxa ; exit} ' data2process.tempdata)

gives v6=*3.6e8 syntax error

maxangle=$(awk -v maxt=0.05 '{$6 = $6 * 3.6e8 ; if($6<=maxt){maxt=$1; maxa=$1}}  END{print maxa ;}' data2process.tempdata)
        echo $maxangle

gives me 5.0.

Thanks a lot for all your help. I am sure I can use all of it.

Answer 1

If you divide your 0.05 constant by 3.6E8 you'll have a value you can directly compare with field 6. This avoids any possibility of numeric underflow or loss of precision when comparing very large and very small values:

awk -v maxt=0.05 '
    BEGIN { tdash = maxt / 3.6E8 }
    NR>1 && $6 < tdash { print $1; exit }
' data2process.tempdata

Answer 2

OK I understand.

You want something like

awk -v maxt=0.05 '{t6 = $6 * 3.6e8 ;
        if(t6<=maxt){maxt=$1; maxa=$1}}
     END{print maxa ;}' data

You must be inside awk to multiply value fetched from file, so thing like v6="$arg6"*3.6e8 outside awk's code won't work.

which give 0

• (with initials data) condition is met on first line, maxt and maxa are set 0, condition is never met again.

---

There is no need to store value (unless doing other computation within awk)

awk -v maxt=0.05 '$6 * 3.6e8 <= maxt {print $1 ; exit ;}' data

where

• -v maxt=0.05 assign value to maxt
• $6 * 3.6e8 <= maxt before { test for sucessfull condition (this is called a "pattern" in awk)
• {print $1 ; exit ;} print searched value and exit. (this is called an "action" in awk)

and with new data give 1.5