Unix & Linux Asked by user10345633 on January 19, 2021
I have checked all the previous posts related to this, but I cannot find the way that I wanted to do.
I have a file with some exponential numbers as below. I do not know which columns have exponential numbers.
file1.txt
1 499 5e-29 0.33 1.35 46.65
5 999 0.4444 3e-6 0.556 89.444
many more lines
I want to convert all the exponential numbers to decimal numbers.
If I want to convert only one number, I could do as below.
echo 12.34567E-3 | awk '{printf "%5.10fn", $1}'
But in this way, does anybody know how to do?
Thank you in advance!
You can loop for all fields of each line. Testing for if the field is a numerical value ($i+0==$i
) and (&&
) if it contains the character e
seems good. So we modify only these fields to decimals.
Here using GNU awk sprintf
function:
awk '{
for (i=1;i<=NF;i++) if ($i+0 == $i && $i ~ /e/) $i = sprintf("%.10f", $i)
} 1' file
.10f
into there.1
at the end means the default action, to print
the line.Correct answer by thanasisp on January 19, 2021
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