搁置,Python,更新字典
我在Python中使用Shelve,我遇到了一个问题: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'}
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
toshelve.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()
和
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:)