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我正在使用 Python 中的 Shelve，但遇到了一个问题：

In [391]: x
Out[391]: {'broken': {'position': 25, 'page': 1, 'letter': 'a'}}

In [392]: x['broken'].update({'page':1,'position':25,'letter':'b'})

In [393]: x
Out[393]: {'broken': {'position': 25, 'page': 1, 'letter': 'a'}}

不明白为什么不更新？有什么想法吗？

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1 回答 1

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这在文档中有所介绍。基本上，关键字参数writebacktoshelve.open负责：

如果可选writeback参数设置为True，则所有访问的条目也都缓存在内存中，并写回sync()and close()；这可以更方便地改变持久字典中的可变条目，但是，如果访问了许多条目，它会消耗大量内存用于缓存，并且由于所有访问的条目都被写回，它会使关闭操作非常慢（无法确定哪些访问的条目是可变的，也无法确定哪些条目实际上是突变的）。

来自同一页面的示例：

d = shelve.open(filename) # open -- file may get suffix added by low-level
                          # library
# as d was opened WITHOUT writeback=True, beware:
d['xx'] = range(4)  # this works as expected, but...
d['xx'].append(5)   # *this doesn't!* -- d['xx'] is STILL range(4)!

# 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

# or, d=shelve.open(filename,writeback=True) would let you just code
# d['xx'].append(5) and have it work as expected, BUT it would also
# consume more memory and make the d.close() operation slower.
d.close()       # close it
于 2012-12-21T20:41:41.287 回答