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Home/ Questions/Q 3628258
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T23:59:34+00:00 2026-05-18T23:59:34+00:00

I’m using zipfile and tarfile Python modules to open, extract and compress archives. I

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I’m using zipfile and tarfile Python modules to open, extract and compress archives.
I need to display the archive structure in a QTreeWidget, and I don’t know how to go on. To get the infos I use the function infos(path) from this file.
I would like to obtain something like this (from Ark):
alt text

For example, if I receive this filenames: ('GCI/PyFiles/prova3.py', 'GCI/', 'GCI/PyFiles/', 'GCI/Screenshots/', 'GCI/prova2.py', 'prova.py'), I’d like to obtain this:

- prova.py
- GCI/
    |
    |- prova2.py
    |- PyFiles/
             |- prova3.py
    |- Screenshots/

in my QTreeWidget.

Thank you,
rubik

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T23:59:35+00:00Added an answer on May 18, 2026 at 11:59 pm

    I’m not sure how QTreeWidget wants it’s data offhand, but here is a (possibly bad way) to build the structure in memory.

    x = ('GCI/PyFiles/prova3.py', 'GCI/', \
         'GCI/PyFiles/', 'GCI/Screenshots/', \
         'GCI/prova2.py', 'prova.py')
    
    structure = {}
    for fn in x:
        path = fn.split('/')
    
        tmpd = structure
        for p in path[:-1]:
            try:
                tmpd = tmpd[p]
            except KeyError:
                tmpd = tmpd[p] = {}
    
        tmpd[path[-1]] = None
    

    This will give you a dictionary structure that for each key is either another dictionary (representing a folder) or None representing that the key is a file.

    The better way to do this would create a class like this:

    class Node(object):
        def __init__(self):
            self.dirs = {}
            self.files = []
    

    or something like that which you can populate. If I remember correctly from my QT programming days, the QTreeWidget wants a datasource so you basically could figure out what that source looks like and populate it appropriately. There’s also probably the option to do this,

    [sp for _,sp in sorted(
             (len(splitpath),splitpath) for splitpath in
                (path.split('/') for path in x)
             )
        ]
    

    which would return you:

    [['prova.py'], ['GCI', ''], ['GCI', 'prova2.py'], 
     ['GCI', 'PyFiles', ''], ['GCI', 'PyFiles', 'prova3.py'], 
     ['GCI', 'Screenshots', '']]
    
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