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Home/ Questions/Q 8129381
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T08:17:25+00:00 2026-06-06T08:17:25+00:00

I will let the following terminal session speak for itself: >>> import shelve >>>

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I will let the following terminal session speak for itself:

>>> import shelve
>>> s = shelve.open('TestShelve')
>>> from collections import deque
>>> s['store'] = deque()
>>> d = s['store']
>>> print s['store']
deque([])
>>> print d
deque([])
>>> s['store'].appendleft('Teststr')
>>> d.appendleft('Teststr')
>>> print s['store']
deque([])
>>> print d
deque(['Teststr'])

Shouldn’t d and s['store'] point to the same object? Why does appendleft work on d but not on s['store']?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T08:17:27+00:00Added an answer on June 6, 2026 at 8:17 am

    It turns out they’re not the same so any operations you perform on them won’t match:

    >>> import shelve
    >>> s = shelve.open('TestShelve')
    >>> from collections import deque
    >>> s['store'] = deque()
    >>> d = s['store']
    >>> id(s['store'])
    27439296
    >>> id(d)
    27439184
    

    To modify items as you coded, you need to pass the parameter writeback=True:

    s = shelve.open('TestShelve', writeback=True)
    

    See the documentation:

    If the writeback parameter is True, the object will hold a cache of
    all entries accessed and write them back to the dict at sync and close
    times. This allows natural operations on mutable entries, but can
    consume much more memory and make sync and close take a long time.

    You can also do it with writeback=False but then you need to write the code exactly as in the provided example:

    # having opened d without writeback=True, you need to code carefully:
    temp = d['xx']      # extracts the copy
    temp.append(5)      # mutates the copy
    d['xx'] = temp      # stores the copy right back, to persist it
    
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