Linux: Move 1 million files into prefix-based crea

问题:

I have a directory called "images" filled with about one million images. Yep.

I want to write a shell command to rename all of those images into the following format:

original: filename.jpg
new: /f/i/l/filename.jpg

Any suggestions?

Thanks,
Dan

回答1:

for i in *.*; do mkdir -p ${i:0:1}/${i:1:1}/${i:2:1}/; mv $i ${i:0:1}/${i:1:1}/${i:2:1}/; done;

The ${i:0:1}/${i:1:1}/${i:2:1} part could probably be a variable, or shorter or different, but the command above gets the job done. You'll probably face performance issues but if you really want to use it, narrow the *.* to fewer options (a*.*, b*.* or what fits you)

edit: added a $ before i for mv, as noted by Dan

回答2:

You can generate the new file name using, e.g., sed:

$ echo "test.jpg" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/'
t/e/s/test.jpg

So, you can do something like this (assuming all the directories are already created):

for f in *; do
   mv -i "$f" "$(echo "$f" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/')"
done

or, if you can't use the bash $( syntax:

for f in *; do
   mv -i "$f" "`echo "$f" | sed -e 's/^\(\(.\)\(.\)\(.\).*\)$/\2\/\3\/\4\/\1/'`"
done

However, considering the number of files, you may just want to use perl as that's a lot of sed and mv processes to spawn:

#!/usr/bin/perl -w
use strict;

# warning: untested
opendir DIR, "." or die "opendir: $!";
my @files = readdir(DIR); # can't change dir while reading: read in advance
closedir DIR;
foreach my $f (@files) {
    (my $new_name = $f) =~ s!^((.)(.)(.).*)$!$2/$3/$4/$1/;
    -e $new_name and die "$new_name already exists";
    rename($f, $new_name);
}

That perl is surely limited to same-filesystem only, though you can use File::Copy::move to get around that.

回答3:

You can do it as a bash script:

#!/bin/bash

base=base

mkdir -p $base/shorts

for n in *
do
    if [ ${#n} -lt 3 ]
    then
        mv $n $base/shorts
    else
        dir=$base/${n:0:1}/${n:1:1}/${n:2:1}
        mkdir -p $dir
        mv $n $dir
    fi
done

Needless to say, you might need to worry about spaces and the files with short names.

回答4:

I suggest a short python script. Most shell tools will balk at that much input (though xargs may do the trick). Will update with example in a sec.

#!/usr/bin/python
import os, shutil

src_dir = '/src/dir'
dest_dir = '/dest/dir'

for fn in os.listdir(src_dir):
  os.makedirs(dest_dir+'/'+fn[0]+'/'+fn[1]+'/'+fn[2]+'/')
  shutil.copyfile(src_dir+'/'+fn, dest_dir+'/'+fn[0]+'/'+fn[1]+'/'+fn[2]+'/'+fn)

回答5:

Any of the proposed solutions which use a wildcard syntax in the shell will likely fail due to the sheer number of files you have. Of the current proposed solutions, the perl one is probably the best.

However, you can easily adapt any of the shell script methods to deal with any number of files thus:

ls -1 | \
while read filename
do
  # insert the loop body of your preference here, operating on "filename"
done

I would still use perl, but if you're limited to only having simple unix tools around, then combining one of the above shell solutions with a loop like I've shown should get you there. It'll be slow, though.