Bash- parse file into arrays

Question

i have this file

i want to have arrays for each Seqs (Seq1, Seq2) like this:

2ndColumn=(1,2,3) 
3rdColumn=(5,6,9)

i wrote this but it does not break the while loop..

#!/bin/bash
2ndColumn=()
3rdColumn=()
while read line 
do
if [[ $line == S* ]] 
 echo "$line"
else
 i=0
 while [[ $line != S* ]]
 do
  2ndColumn[i]="$(echo $line | cut -d\  -f2)"
  3rdColumn[i]="$(echo $line | cut -d\  -f3)"
  i=$((i+1))
  read line
 done
 echo "${2ndColumn[@]} and ${3rdColumn[@]}"
fi
done < file
exit 0

this script will iterate forever, it does not get out of while loop. please give a hand to this stupid humanbeing :(

The reason you're looping forever is because your inner loop never checks whether read succeeded or failed. — Barmar
– Barmar, Commented Dec 1, 2012 at 11:07
When the read fails, line is set to the empty string, and [[ $line != S* ]] is true, so the while-loop keeps looping. The outer loop uses the success of read as its while condition. — Barmar
– Barmar, Commented Dec 1, 2012 at 17:49

reece · Accepted Answer · 2012-12-01 10:54:41Z

1

I would restructure it to use a single loop instead of a nested loop with nested read calls on stdin:

2ndColumn=()
3rdColumn=()
i=0
while read line 
do
    if [[ $line == S* ]] 
        echo "$line ==> ${2ndColumn[@]} and ${3rdColumn[@]}"
        # reset the lists...
        2ndColumn=()
        3rdColumn=()
        i=0
    else
        2ndColumn[i]="$(echo $line | cut -d\  -f2)"
        3rdColumn[i]="$(echo $line | cut -d\  -f3)"
        i=$((i+1))
    fi
done

This will avoid issues with the inner read call, which is probably being blocked when fed a stdin file handle.

answered Dec 1, 2012 at 10:54

reece

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Comments

Gilles Quénot · Accepted Answer · 2012-12-01 12:25:27Z

you can't change the cursor line of a while read

I wrote this for you with a little hack to prefix the array names with seq N+1 :

#!/bin/bash

file=file.txt

while read line; do
    if [[ $line == S* ]]; then
        echo "$line"
        i=0
        ((Seq++))
    else
        declare seq${Seq}_2ndColumn[i]="$(echo $line | cut -d\  -f2)"
        declare seq${Seq}_3rdColumn[i]="$(echo $line | cut -d\  -f3)"
        ((i++))
    fi
done < "$file"

echo "${!seq*} arrays are declared"

OUTPUT

seq1_2ndColumn seq1_3rdColumn seq2_2ndColumn seq2_3rdColumn arrays are declared

EXPLANATIONS

${!pattern*} is a nice feature of bash to display variables beginning with pattern
(( )) is an arithmetic command, which returns an exit status of 0 if the expression is nonzero, or 1 if the expression is zero. Also used as a synonym for "let", if side effects (assignments) are needed. See http://mywiki.wooledge.org/ArithmeticExpression

BONUS

If you have the control on what's in your array, try doing this at the end ::

for s in ${!seq*}; do
    printf '\t%s\n' $(eval echo \${$s[@]})
done

See See http://mywiki.wooledge.org/BashFAQ/048

new OUTPUT

seq1_2ndColumn
        1
        2
        3
seq1_3rdColumn
        5
        6
        9
seq2_2ndColumn
        2
        4
        8
seq2_3rdColumn
        7
        9
        12

Gilles Quénot · Accepted Answer · 2012-12-01 10:47:31Z

0

file=/PATH/TO/file.txt

arr1=( $(awk '/^Seq/{l++} l==1{print $2}' "$file") )
arr2=( $(awk '/^Seq/{l++} l==1{print $3}' "$file") )

echo "arr1:"
printf '\t%s\n' ${arr1[@]}

echo "arr2:"
printf '\t%s\n' ${arr2[@]}

answered Dec 1, 2012 at 10:47

Gilles Quénot

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1 Comment

teutara Over a year ago

hey sputnick. as i understand your solution will get all values into array at once, but i have to get the two arrays with respect to header (Seq1,Seq2..) so that i will introduce another calculations..

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