How to find duplicate directories

Let create some testing directory tree:

#!/bin/bash

top="./testdir"
[[ -e "$top" ]] && { echo "$top already exists!" >&2; exit 1; }

mkfile() { printf "%s\n" $(basename "$1") > "$1"; }

mkdir -p "$top"/d1/d1{1,2}
mkdir -p "$top"/d2/d1some/d12copy
mkfile "$top/d1/d12/a"
mkfile "$top/d1/d12/b"
mkfile "$top/d2/d1some/d12copy/a"
mkfile "$top/d2/d1some/d12copy/b"
mkfile "$top/d2/x"
mkfile "$top/z"

The structure is: find testdir \( -type d -printf "%p/\n" , -type f -print \)

testdir/
testdir/d1/
testdir/d1/d11/
testdir/d1/d12/
testdir/d1/d12/a
testdir/d1/d12/b
testdir/d2/
testdir/d2/d1some/
testdir/d2/d1some/d12copy/
testdir/d2/d1some/d12copy/a
testdir/d2/d1some/d12copy/b
testdir/d2/x
testdir/z

I need find the duplicate directories, but I need consider only files (e.g. I should ignore (sub)directories without files). So, from the above test-tree the wanted result is:

duplicate directories:
testdir/d1
testdir/d2/d1some

because in both (sub)trees are only two identical files a and b. (and several directories, without files).

Of course, I could md5deep -Zr ., also could walk the whole tree using perl script (using File::Find+Digest::MD5 or using Path::Tiny or like.) and calculate the file's md5-digests, but this doesn't helps for finding the duplicate directories... :(

Any idea how to do this? Honestly, I haven't any idea.

EDIT

• I don't need working code. (I'm able to code myself)
• I "just" need some ideas "how to approach" the solution of the problem. :)

Edit2

The rationale behind - why need this: I have approx 2.5 TB data copied from many external HDD's as a result of wrong backup-strategy. E.g. over the years, the whole $HOME dirs are copied into (many different) external HDD's. Many sub-directories has the same content, but they're in different paths. So, now I trying to eliminate the same-content directories.

And I need do this by directories, because here are directories, which has some duplicates files, but not all. Let say:

/some/path/project1/a
/some/path/project1/b

and

/some/path/project2/a
/some/path/project2/x

e.g. the a is a duplicate file (not only the name, but by the content too) - but it is needed for the both projects. So i want keep the a in both directories - even if they're duplicate files. Therefore me looking for a "logic" how to find duplicate directories.

Some key points:

• If I understand right (from your comment, where you said: "(Also, when me saying identical files I mean identical by their content, not by their name)" , you want find duplicate directories, e.g. where their content is exactly the same as in some other directory, regardless of the file-names.
• for this you must calculate some checksum or digest for the files. Identical digest = identical file. (with great probability). :) As you already said, the md5deep -Zr -of /top/dir is a good starting point.
• I added the -of, because for such job you don't want calculate the contents of the symlinks-targets, or other special files like fifo - just plain files.
• calculating the md5 for each file in 2.5TB tree, sure will take few hours of work, unless you have very fast machine. The md5deep runs a thread for each cpu-core. So, while it runs, you can make some scripts.
• Also, consider run the md5deep as sudo, because it could be frustrating if after a long run-time you will get some error-messages about unreadable files, only because you forgot to change the files-ownerships...(Just a note) :) :)

For the "how to":

• For comparing "directories" you need calculate some "directory-digest", for easy compare and finding duplicates.
• The one most important thing is realize the following key points:
  • you could exclude directories, where are files with unique digests. If the file is unique, e.g. has not any duplicates, that's mean that is pointless checking it's directory. Unique file in some directory means, that the directory is unique too. So, the script should ignore every directory where are files with unique MD5 digests (from the md5deep's output.)
  • You don't need calculate the "directory-digest" from the files itself. (as you trying it in your followup question). It is enough to calculate the "directory digest" using the already calculated md5 for the files, just must ensure that you sort them first!

e.g. for example if your directory /path/to/some containing only two files a and b and

if file "a" has md5 : 0cc175b9c0f1b6a831c399e269772661
and file "b" has md5: 92eb5ffee6ae2fec3ad71c777531578f

you can calculate the "directory-digest" from the above file-digests, e.g. using the Digest::MD5 you could do:

perl -MDigest::MD5=md5_hex -E 'say md5_hex(sort qw( 92eb5ffee6ae2fec3ad71c777531578f 0cc175b9c0f1b6a831c399e269772661))'

and will get 3bc22fb7aaebe9c8c5d7de312b876bb8 as your "directory-digest". The sort is crucial(!) here, because the same command, but without the sort:

perl -MDigest::MD5=md5_hex -E 'say md5_hex(qw( 92eb5ffee6ae2fec3ad71c777531578f 0cc175b9c0f1b6a831c399e269772661))'

produces: 3a13f2408f269db87ef0110a90e168ae.

Note, even if the above digests aren't the digests of your files, but they're will be unique for every directory with different files and will be the same for the identical files. (because identical files, has identical md5 file-digest). The sorting ensures, that you will calculate the digest always in the same order, e.g. if some other directory will contain two files

file "aaa" has md5 : 92eb5ffee6ae2fec3ad71c777531578f
file "bbb" has md5 : 0cc175b9c0f1b6a831c399e269772661

using the above sort and md5 you will again get: 3bc22fb7aaebe9c8c5d7de312b876bb8 - e.g. the directory containing same files as above...

So, in such way you can calculate some "directory-digest" for every directory you have and could be ensured that if you get another directory digest 3bc22fb7aaebe9c8c5d7de312b876bb8 thats means: this directory has exactly the above two files a and b (even if their names are different).

This method is fast, because you will calculate the "directory-digests" only from small 32bytes strings, so you avoids excessive multiple file-digest-caclulations.

The final part is easy now. Your final data should be in form:

3a13f2408f269db87ef0110a90e168ae /some/directory
16ea2389b5e62bc66b873e27072b0d20 /another/directory
3a13f2408f269db87ef0110a90e168ae /path/to/other/directory

so, from this is easy to get: the

/some/directory and the /path/to/other/directory are identical, because they has identical "directory-digests".

Hm... All the above is only a few lines long perl script. Probably would be faster to write here directly the perl-script as the above long textual answer - but, you said - you don't want code... :) :)

A traversal can identify directories which are duplicates in the sense you describe. I take it that this is: if all files in a directory are equal to all files of another then their paths are duplicates.

Find all files in each directory and form a string with their names. You can concatenate the names with a comma, say (or some other sequence that is certainly not in any names). This is to be compared. Prepend the path to this string, so to identify directories.

Comparison can be done for instance by populating a hash with keys being strings with filenames and path their values. Once you find that a key already exists you can check the content of files, and add the path to the list of duplicates.

The strings with path don't have to be actually formed, as you can build the hash and dupes list during the traversal. Having the full list first allows for other kinds of accounting, if desired.

This is altogether very little code to write.

An example. Let's say that you have

dir1/subdir1/{a,b}  # duplicates (files 'a' and 'b' are considered equal)
dir2/subdir2/{a,b}

and

proj1/subproj1/{a,b,X}  # NOT duplicates, since there are different files
proj2/subproj2/{a,b,Y}

The above prescription would give you strings

'dir1/subdir1/a,b',
'dir2/subdir2/a,b',
'proj1/subproj1/a,b,X',
'proj2/subproj2/a,b,Y';

where the (sub)string 'a,b' identifies dir1/subdir1 and dir2/subdir2 as duplicates.

I don't see how you can avoid a traversal to build a system that accounts for all files.

The procedure above is the first step, not handling directories with files and subdirectories.

Consider

   dirA/          dirB/
a b sdA/       a X sdB/
    c d            c d

Here the paths dirA/sdA/ and dirB/sdB/ are duplicates by the problem description but the whole dirA/ and dirB/ are distinct. This isn't shown in the question but I'd expect it to be of interest.

The procedure from the first part can be modified for this. Iterate through directories, forming a path component at every step. Get all files in each, and all subdirectories (if none we are done). Append the comma-separated file list to the path component (/sdA/). So the representation of the above is

'dirA/sdA,a,b/c,d',  'dirB/sdB,a,X/c,d'

For each file-list substring (c,d) found to already exist we can check its path against the existing one, component by component. Now a hash with keys like c,d won't do since this example has the same file-list for distinct hierarchies, but a modified (or other) data structure is needed.

Finally, there may be more subdirectories parallel to sdA (say sdA2). We care only for its own path, but except for the parallel files (a,b, in that component of the path dirA/sdaA2,a,b/). So keep in mind all bottom-level file-lists (c,d) with their paths and, if file-lists are equal and paths are of same length, check whether their paths have a,b file-lists equal in each path component.

I don't know whether this is a workable solution for you, but I'd expect "near-duplicates" to be rare -- the backup is either a duplicate or not. So there may not be much need to handle futher edge-cases in complex sprawling hierarchies. This procedure should be at least a useful pre-selection mechanism, that would greatly reduce the need for further work.

This assumes that equal file-names very likely indicate equal files. A part of that is my expectation that if a file was even just renamed it still cannot be considered a duplicate. If this is not so this approach won't work and one would need something along the lines of the answer by jm666.