Renaming a set of files to 001, 002, ... on Linux

20
votes

I originally had a set of images of the form image_001.jpg, image_002.jpg, ...

I went through them and removed several. Now I'd like to rename the leftover files back to image_001.jpg, image_002.jpg, ...

Is there a Linux command that will do this neatly? I'm familiar with rename but can't see anything to order file names like this. I'm thinking that since ls *.jpg lists the files in order (with gaps), the solution would be to pass the output of that into a bash loop or something?

6
linuxbashshellfilerename

6 Answers

14
votes

If I understand right, you have e.g. image_001.jpg, image_003.jpg, image_005.jpg, and you want to rename to image_001.jpg, image_002.jpg, image_003.jpg.

EDIT: This is modified to put the temp file in the current directory. As Stephan202 noted, this can make a significant difference if temp is on a different filesystem. To avoid hitting the temp file in the loop, it now goes through image*

i=1; temp=$(mktemp -p .); for file in image*
do
mv "$file" $temp;
mv $temp $(printf "image_%0.3d.jpg" $i)
i=$((i + 1))
done
9
votes

A simple loop (test with echo, execute with mv):

I=1
for F in *; do
  echo "$F" `printf image_%03d.jpg $I`
  #mv "$F" `printf image_%03d.jpg $I` 2>/dev/null || true
  I=$((I + 1))
done

(I added 2>/dev/null || true to suppress warnings about identical source and target files. If this is not to your liking, go with Matthew Flaschen's answer.)

2
votes

Try the following script:

numerate.sh

This code snipped should do the job:

./numerate.sh -d <your image folder> -b <start number> -L 3 -p image_ -s .jpg -o numerically -r
1
votes

This does the reverse of what you are asking (taking files of the form *.jpg.001 and converting them to *.001.jpg), but can easily be modified for your purpose:

for file in * 

do

if [[ "$file" =~ "(.*)\.([[:alpha:]]+)\.([[:digit:]]{3,})$" ]]

then

mv "${BASH_REMATCH[0]}" "${BASH_REMATCH[1]}.${BASH_REMATCH[3]}.${BASH_REMATCH[2]}"

fi

done
1
votes

Some good answers here already; but some rely on hiding errors which is not a good idea (that assumes mv will only error because of a condition that is expected - what about all the other reaons mv might error?).

Moreover, it can be done a little shorter and should be better quoted:

for file in *; do
    printf -vsequenceImage 'image_%03d.jpg' "$((++i))"
    [[ -e $sequenceImage ]] || \
        mv "$file" "$sequenceImage"
done

Also note that you shouldn't capitalize your variables in bash scripts.

0
votes

I was going to suggest something like the above using a for loop, an iterator, cut -f1 -d "_", then mv i i.iterator. It looks like it's already covered other ways, though.