Bash- create an array and increment each index- then the greatest index

Question

know question is not clear at a glance, i have this table:

   ID Start End
   1  1     4
   2  2     5
   3  4     9
   4  8     10

I want to set these in an order (illustration below). I need an array that its indices will increment by one with respect to start and the end positions, and get the greatest index of all. For example:

1. ####
2.  ####
3.    ######
4.        ### 

so array will be;
    array =(1,2,2,3,2,1,1,2,2,1)

i did not start to write anything because i could not figure whether that is possible with bash. please advice..

I am having trouble discerning the how the sequence 1,2,2,3,2,1,1,2,2,1 relates to your table of ranges. — ddoxey
– ddoxey, Commented Nov 30, 2012 at 14:23
@ddoxey those are just for display, i mean i want to fill them to have those values, so i that i can the greatest one.. — teutara
– teutara, Commented Nov 30, 2012 at 14:26

choroba · Accepted Answer · 2012-11-30 14:28:56Z

2

Just loop over all the elements of each interval:

#! /bin/bash

array=()
while read id start end ; do
    for (( i=start ; i<=end ; i++ )) ; do
        let array[i]++
    done
done << EOF
1  1     4
2  2     5
3  4     9
4  8     10
EOF

echo "${array[@]}"

answered Nov 30, 2012 at 14:28

choroba

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