Create an array with a sequence of numbers in bash

Question

I would like to write a script that will create me an array with the following values:

{0.1 0.2 0.3 ... 2.5} 

Until now I was using a script as follows:

plist=(0.1 0.2 0.3 0.4) for i in ${plist[@]}; do     echo "submit a simulation with this parameter:"     echo "$i" done 

But now I need the list to be much longer ( but still with constant intervals).

Is there a way to create such an array in a single command? what is the most efficient way to create such a list?

Answer 1

Using seq you can say seq FIRST STEP LAST. In your case:

seq 0 0.1 2.5 

Then it is a matter of storing these values in an array:

vals=($(seq 0 0.1 2.5)) 

You can then check the values with:

$ printf "%s\n" "${vals[@]}" 0,0 0,1 0,2 ... 2,3 2,4 2,5 

Yes, my locale is set to have commas instead of dots for decimals. This can be changed setting LC_NUMERIC="en_US.UTF-8".

By the way, brace expansion also allows to set an increment. The problem is that it has to be an integer:

$ echo {0..15..3} 0 3 6 9 12 15