搁置,Python,更新字典

搁置,Python,更新字典,python,dictionary,shelve,Python,Dictionary,Shelve,我在Python中使用Shelve,我遇到了一个问题: 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'}

我在Python中使用Shelve,我遇到了一个问题:

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'}}
我不明白为什么它没有更新?有什么想法吗?

这在本手册中有介绍。基本上,关键字参数
writeback
to
shelve.open
负责:

如果可选的
writeback
参数设置为
True
,则所有条目 访问的数据也缓存在内存中,并在
sync()
close()
;这可以使在 持久字典,但是,如果访问了许多条目,它可以 为缓存消耗大量内存,它可以使 关闭操作非常慢,因为所有访问的条目都被写回 (无法确定哪些访问的条目是可变的,或者 哪些基因实际上发生了变异)

同一页中的示例:

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

@莫加纳伦:没问题:)