Subtract 1 from all file names (rename them) in a directory.

Question

I have a directory which contains image files with names like

image1.jpg 
image2.jpg 
image3.jpg
...

Unfortunately, the image names must be zero based, so image1.jpg should be image0.jpg, image2.jpg should be image1.jpg and so on.

I can write a script to generate mv commands like these, put them in a shell script, and then execute them -

mv image1.jpg image0.jpg
mv image2.jpg image1.jpg
mv image3.jpg image2.jpg
...

But I suppose there is a neater way to do it in Unix. So what is it?

Answer 1

The good old perl rename:

rename 's/(\d+)(\.jpg)/($1-1).$2/e' *

[Remarks]

Image numbers should be greater than 0.

In case images are greater than 9 and have not leading 0s, use $(ls -v1 *) to avoid clobbering. Proposed by @arielf and noticed by @Graeme.

When in doubt use also -v for verbose and -n for no-action.

Answer 2

You can pipe the generated mv commands to bash. So you don't have to copy them in a script and execute that. See:

command_that_generates_mv_commands | bash

And everything will be executed that is piped to bash.

Answer 3

You can iterate on ls output, this one works for your example:

i=0 
for file in $(ls *.jpg | sort) ; do 
     mv $file $(echo $file | sed 's/[0-9]*.jpg$/'${i}'.jpg/')
     i=$((++i))
done

You must be on the same path of your files

Answer 4

The following seems to work for anything that fits the pattern imageNUMBER.jpg. I've placed echo before the mv command to first show what the command would do; to actually perform the renaming, just remove the echo

for i in `ls image*.jpg|sort -V` ; do 
    x=`echo $i|sed -e "s/image\(.*\).jpg/\1/"`
    y=$(( $x - 1 ))
    echo mv -i $i image$y.jpg
done

In the first line, the ls image*.jpg|sort -V will cause the JPG files to be listed with ascending numbers in the filename. The x= line extracts the number from the filename. The y= line then decrements the number by one. The input filename and the y number are then used in the mv command, where the -i flag will notify you before overwriting a file.

For my own small test, this produced the output:

mv -i image1.jpg image0.jpg
mv -i image2.jpg image1.jpg
mv -i image123.jpg image122.jpg

Personally, I'd suggest renaming to a more different filename, since now the order in which the files are processed can make a big difference.

Answer 5

Using the perl script prename, which is symlinked to rename on Debian based distros, also needs GNU find/sort. The files are put in ascending order to prevent any overwriting.

find . -regex '\./image[0-9]+\.jpg' -print0 | sort -zV |
  xargs -0 rename -n 's/(\d+)\.jpg$/@{[$1-1]}.jpg/'

Remove the -n once you are sure it does what you want. Will warn about files already existing before doing so. However as long as it shows the files being renamed in ascending order, there will be no conflicts when run for real.

Answer 6

With zsh:

autoload zmv # (in ~/.zshrc)
zmv -Qf -n 'image(<->).jpg(n)' 'image$(($1-1)).jpg'

(remove -n when happy).

(n) is to sort the list numerically so image9.jpg is renamed before image10.jpg.