How to append values to an array in Bash by adding one element to the previous entry?

Question

I have a list of numbers 1 2 3 4 5 that I am trying to organize into an array where the values are in a sequence where the current value is the summation of the previous values in the array (like this): 1 3 6 10 15. My current code is as follows:

array=()
for n in `seq 1 5`
do
    if [ $n -eq 1 ]; then
        array+=($n)
    else
        value=$n
        index="$(($n-1))"
        array+=(`echo ${array[$index]}+$value`)
    fi

done

However, when I try checking the array echo "${array[@]}" I get 1 +2 +3 +4 +5. How can I best go about solving this problem?

This doesn't answer your question but your code boils down to: for n in {1..5}; do ((array[n] = n * (n+1) / 2)); done — xhienne
– xhienne, Commented Oct 6, 2018 at 14:21

Inian · Accepted Answer · 2018-10-06 11:26:30Z

2

It is quite simple if you know how to get the last element of the array in bash arrays!. You can just use a negative index ${myarray[-1]} to get the last element. You can do the same thing for the second-last, and so on; in Bash:

fbseries=()

for ((i=1; i<=5; i++)); do
    if [ "$i" -eq 1 ]; then
        fbseries+=("$i")
    else
        fbseries+=( $(( ${fbseries[-1]} + $i )) )
    fi
done

With the example and some modifications all you need is as above.

answered Oct 6, 2018 at 11:26

Inian

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Comments

Poshi · Accepted Answer · 2018-10-06 11:29:03Z

1

You are pretty close the a working code. Here I also added some improvements:

array=()
for n in {1..5}
do
    if [ "$n" -eq 1 ]; then
        array+=("$n")
    else
        value="$n"
        index="$((n-1))"
        array+=($((${array[$index]}+value)))
    fi
done

You can avoid using seq, and you don't need an echo but a calculus.

BTW, that's not a Fibonacci serie.

answered Oct 6, 2018 at 11:29

Poshi

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