Question

I have read many questions in stackoverflow, related to python compare directory. However, my current question is bit different.
I have two directories, which contains two different version release package contents. Now I want to compare to ensure the contents are same. However few files have version name embedded into them. Now which is the best possible way to compare them and conclude (except the version difference, all files match).

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For example:
Version V1R1C1 contains directory structure as below

pmt> find . -name "*"
.
./c1
./c1/c2
./c1/c1_V1R1C1.cfg
./a1
./a1/a1_V1R1C1.cfg
./a1/a2
./a1/a2/a1a2_V1R1C1.cfg
./b1/a_best_file.txt
./b1/b2/a_test_file.txt
./b1/b2/b1b2_V1R1C1.cfg
./a_V1R1C1.cfg

Version V2R3C1 may contain below structure

pmt> find . -name "*"
.
./c1
./c1/c2
./c1/c1_V2R3C1.cfg
./a1
./a1/a1_V2R3C1.cfg
./a1/a2
./a1/a2/a1a2_V2R3C1.cfg
./b1/a_best_file.txt
./b1/b2/a_test_file.txt
./b1/b2/b1b2_V2R3C1.cfg
./a_V2R3C1.cfg

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In the above case, the program must flag it as equivalent structure.

I can think of few solutions – like for example, read both the directory structure recursively into cache (dict), rip the version information and compare etc. But looks like not a completely effective mechanism because of two reason 1. It does not utilize the inbuilt directory compare 2.The multiple read/rip/compare is bound to cost (especially with huge directory tree structure).

I am looking for ideas, which are simple and efficient than the one above.

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PS :
1. In case of any difference (except the version unlike the above example), I would like to use the left/right etc to get diffed list.
2. We can assume before hand which is the version name in both directories (like V1R1C1 in first case and V2R3C1 in second case).

Answer 1

how about use set compare?

set((remove_version(filepath) for filepath in iter_file(dic1))) == set((remove_version(filepath) for filepath in iter_file(dic2)))