CODEWITHSUNDEEP
bashshellscriptingfor-loop
snowmonkey
snowmonkey

Reputation: 287

bash shell script two variables in for loop

I am new to shell scripting. so kindly bear with me if my doubt is too silly.

I have png images in 2 different directories and an executable which takes an images from each directory and processes them to generate a new image.

I am looking for a for loop construct which can take two variables simultaneously..this is possible in C, C++ etc but how do I accomplish something of the following. The code is obviously wrong. 

#!/bin/sh

im1_dir=~/prev1/*.png  
im2_dir=~/prev3/*.png
index=0

for i,j in $im1_dir $im2_dir  # i iterates in im1_dir and j iterates in im2_dir 
do
  run_black.sh $i $j  
done

thanks!

Upvotes: 11

Views: 112397

Answers (8)

Dennis Williamson
Dennis Williamson

Reputation: 359925

Here are a few additional ways to do what you're looking for with notes about the pros and cons.

The following only works with filenames that do not include newlines. It pairs the files in lockstep. It uses an extra file descriptor to read from the first list. If im1_dir contains more files, the loop will stop when im2_dir runs out. If im2_dir contains more files, file1 will be empty for all unmatched file2. Of course if they contain the same number of files, there's no problem.

#!/bin/bash
im1_dir=(~/prev1/*.png) 
im2_dir=(~/prev3/*.png)

exec 3< <(printf '%s\n' "${im1_dir[@]}")

while IFS=$'\n' read -r -u 3 file1; read -r file2
do
    run_black "$file1" "$file2"
done < <(printf '%s\n' "${im2_dir[@]}")

exec 3<&-

You can make the behavior consistent so that the loop stops with only non-empty matched files no matter which list is longer by replacing the semicolon with a double ampersand like so:

while IFS=$'\n' read -r -u 3 file1 && read -r file2

This version uses a for loop instead of a while loop. This one stops when the shorter of the two lists run out.

#!/bin/bash
im1_dir=(~/prev1/*.png) 
im2_dir=(~/prev3/*.png)

for ((i = 0; i < ${#im1_dir[@]} && i < ${#im2_dir[@]}; i++))
do
    run_black "${im1_dir[i]}" "${im2_dir[i]}"
done

This version is similar to the one immediately above, but if one of the lists runs out it wraps around to reuse the items until the other one runs out. It's very ugly and you could do the same thing another way more simply.

#!/bin/bash
im1_dir=(~/prev1/*.png) 
im2_dir=(~/prev3/*.png)

for ((i = 0, j = 0,
          n1 = ${#im1_dir[@]}, 
          n2 = ${#im2_dir[@]}, 
          s = n1 >= n2 ? n1 : n2, 
          is = 0, js = 0; 

      is < s && js < s; 

      i++, is = i, i %= n1, 
          j++, js = j, j %= n2))
do
    run_black "${im1_dir[i]}" "${im2_dir[i]}"
done

This version only uses an array for the inner loop (second directory). It will only execute as many times as there are files in the first directory.

#!/bin/bash
im1_dir=~/prev1/*.png
im2_dir=(~/prev3/*.png)

for file1 in $im1_dir
do
    run_black "$file1" "${im2_dir[i++]}"
done

Upvotes: 7

Ben Usman
Ben Usman

Reputation: 8387

The accepted answer can be further simplified using the ${!array[@]} syntax to iterate over array's indexes:

a=(x y z); b=(q w e); for i in ${!a[@]}; do echo ${a[i]}-${b[i]}; done

Upvotes: 1

Thanh
Thanh

Reputation: 17

It is very simple you can use two for loop functions in this problem.

#bin bash
index=0
for i in ~/prev1/*.png  
do
    for j ~/prev3/*.png
    do 
        run_black.sh $i $j
    done
done

Upvotes: 0

John Haberstroh
John Haberstroh

Reputation: 485

I have this problem for a similar situation where I want a top and bottom range simultaneously. Here was my solution; it's not particularly efficient but it's easy and clean and not at all complicated with icky BASH arrays and all that nonsense.

SEQBOT=$(seq 0  5  $((PEAKTIME-5)))
SEQTOP=$(seq 5  5  $((PEAKTIME-0)))

IDXBOT=0
IDXTOP=0

for bot in $SEQBOT; do
    IDXTOP=0
    for top in $SEQTOP; do
        if [ "$IDXBOT" -eq "$IDXTOP" ]; then
            echo $bot $top
        fi
        IDXTOP=$((IDXTOP + 1))
    done
    IDXBOT=$((IDXBOT + 1))
done

Upvotes: 0

Rashedul Islam
Rashedul Islam

Reputation: 939

This might be another way to use two variables in the same loop. But you need to know the total number of files (or, the number of times you want to run the loop) in the directory to use it as the value of iteration i.

Get the number of files in the directory:

ls /path/*.png | wc -l

Now run the loop:

im1_dir=(~/prev1/*.png) 
im2_dir=(~/prev3/*.png)

for ((i = 0; i < 4; i++)); do run_black.sh ${im1_dir[i]} ${im2_dir[i]}; done

For more help please see this discussion.

Upvotes: 0

Fritz G. Mehner
Fritz G. Mehner

Reputation: 17188

Another solution. The two lists with filenames are pasted into one.

paste <(ls --quote-name ~/prev1/*.png) <(ls --quote-name ~/prev3/*.png) | \
while read args ; do
  run_black $args
done

Upvotes: 0

jordanm
jordanm

Reputation: 34924

If you are depending on the two directories to match up based on a locale sorted order (like your attempt), then an array should work.

im1_files=(~/prev1/*.png)
im2_files=(~/prev3/*.png)

for ((i=0;i<=${#im1_files[@]};i++)); do
   run_black.sh "${im1_files[i]}" "${im2_files[i]}"
done

Upvotes: 18

Hai Vu
Hai Vu

Reputation: 40688

If you don't mind going off the beaten path (bash), the Tool Command Language (TCL) has such a loop construct:

#!/usr/bin/env tclsh

set list1 [glob dir1/*]
set list2 [glob dir2/*]

foreach item1 $list1 item2 $list2 {
    exec command_name $item1 $item2
}

Basically, the loop reads: for each item1 taken from list1, and item2 taken from list2. You can then replace command_name with your own command.

Upvotes: 4

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