Bash script to convert column data into rows

Question

I have data in a comma separate file. This is the structure of the input file.

code, day_1, day_2, day_3

1235, 0, 21, 6
1236, 9, 3, 4
1237, 1, 7, 3

I would like transform it, and display the column data as row entries

code, day, count

1235, day_1, 0
1235, day_2, 21
1235, day_3, 6
1236, day_1, 9
1236, day_2, 3
1236, day_3, 4
1237, day_1, 1
1237, day_2, 7
1237, day_3, 3

Here's my code and output.

Code

#!/bin/bash
file='test_file.in'
IFS=$','
while read -r code day_1 day_2 day_3
do
        #echo "$code, $branch, $description"
        count=1
        while [ $count -le 3 ]
        do

                day_var=""'$day_'"${count}"
                #echo $day_var
                echo "$code, day_${count}, $day_var"
                count=$[$count+1]
        done
done < $file

Output:

120006, day_1, $day_1 120006, day_2, $day_2 120006, day_3, $day_3 120007, day_1, $day_1 120007, day_2, $day_2 120007, day_3, $day_3 140197, day_1, $day_1 140197, day_2, $day_2 140197, day_3, $day_3

Kindly suggest a bash script solution for me.

M. Nejat Aydin · Accepted Answer · 2020-07-22 09:40:01Z

1

Here is the corrected version of your code:

#!/bin/bash

file='test_file.in'
while IFS=, read -r code day_1 day_2 day_3
do
        #echo "$code, $branch, $description"
        count=1
        while [ $count -le 3 ]
        do
                day_var=day_$count
                echo "$code, $day_var, ${!day_var}"
                count=$((count+1))
        done
done < "$file"

Compare it with yours.

But my approach would be:

#!/bin/bash

while IFS=, read -r code day1 day2 day3; do
    printf '%s, day_1,%s\n%s, day_2,%s\n%s, day_3,%s\n' \
        "$code" "$day1" "$code" "$day2" "$code" "$day3"
done < file

answered Jul 22, 2020 at 9:40

M. Nejat Aydin

10.3k1 gold badge10 silver badges22 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Grooth Over a year ago

Thank you so much. Really appreciated.

kvantour · Accepted Answer · 2020-07-22 09:43:45Z

0

A quick awk solution:

awk 'BEGIN{FS=OFS=","}{for(i=2;i<=NF;++i) print $1,"day_"(i-1),$i}' file

Why is your code not working:

There are many reasons why this is not the best code, but the main are in the following two lines:

day_var=""'$day_'"${count}"
echo "$code, day_${count}, $day_var"

You would like to use a variable name which you actually created from variables. This is called indirection and there is a nice BashFAQ/006 about it.

When we would fix your code, it would be something like this:

#!/usr/bin/evn bash
file='test_file.in'
while IFS=, read -r code day_1 day_2 day_3
do
        for (( count=1;count<=3; count++)); do
                day_var="day_${count}"
                echo "$code, day_${count}, ${!day_var}"
        done
done < $file

edited Jul 22, 2020 at 9:43

answered Jul 22, 2020 at 9:32

kvantour

26.9k4 gold badges57 silver badges81 bronze badges

Collectives™ on Stack Overflow

Bash script to convert column data into rows

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related