我已经浏览了Python文档提供的信息,但我还是有点困惑。有人可以发布一个示例代码,编写一个新文件,然后使用pickle将字典转储到其中吗?


当前回答

如果你只是想把字典存储在一个文件中,可以像这样使用pickle

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

如果您想在多个文件中保存和恢复多个字典 缓存和存储更复杂的数据, 使用anycache。 泡菜周围的其他东西它都有

from anycache import anycache

@anycache(cachedir='path/to/files')
def myfunc(hello):
    return {'hello', hello}

的参数来存储不同的myfunc结果 不同的文件在cachedir并重新加载它们。

有关更多详细信息,请参阅文档。

其他回答

一般来说,除非字典中只有简单的对象,如字符串和整数,否则pickle字典将会失败。

Python 2.7.9 (default, Dec 11 2014, 01:21:43) 
[GCC 4.2.1 Compatible Apple Clang 4.1 ((tags/Apple/clang-421.11.66))] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from numpy import *
>>> type(globals())     
<type 'dict'>
>>> import pickle
>>> pik = pickle.dumps(globals())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 306, in save
    rv = reduce(self.proto)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/copy_reg.py", line 70, in _reduce_ex
    raise TypeError, "can't pickle %s objects" % base.__name__
TypeError: can't pickle module objects
>>> 

即使是一个非常简单的字典也经常会失败。这取决于内容。

>>> d = {'x': lambda x:x}
>>> pik = pickle.dumps(d)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 649, in save_dict
    self._batch_setitems(obj.iteritems())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 663, in _batch_setitems
    save(v)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 748, in save_global
    (obj, module, name))
pickle.PicklingError: Can't pickle <function <lambda> at 0x102178668>: it's not found as __main__.<lambda>

然而,如果你使用一个更好的序列化器,如dill或cloudpickle,那么大多数字典都可以被pickle:

>>> import dill
>>> pik = dill.dumps(d)

或者如果你想保存你的字典到一个文件…

>>> with open('save.pik', 'w') as f:
...   dill.dump(globals(), f)
... 

后一个例子与这里发布的任何其他好的答案是一样的(除了忽略了字典内容的可选择性之外,这些答案都是好的)。

>>> import pickle
>>> with open("/tmp/picklefile", "wb") as f:
...     pickle.dump({}, f)
... 

通常情况下,最好使用cPickle实现

>>> import cPickle as pickle
>>> help(pickle.dump)
Help on built-in function dump in module cPickle:

dump(...)
    dump(obj, file, protocol=0) -- Write an object in pickle format to the given file.

    See the Pickler docstring for the meaning of optional argument proto.

试试这个:

import pickle

a = {'hello': 'world'}

with open('filename.pickle', 'wb') as handle:
    pickle.dump(a, handle, protocol=pickle.HIGHEST_PROTOCOL)

with open('filename.pickle', 'rb') as handle:
    b = pickle.load(handle)

print(a == b)

上面的解决方案没有任何特定于dict对象的内容。同样的方法也适用于许多Python对象,包括任意类的实例和任意复杂的数据结构嵌套。例如,将第二行替换为以下几行:

import datetime
today = datetime.datetime.now()
a = [{'hello': 'world'}, 1, 2.3333, 4, True, "x", 
     ("y", [[["z"], "y"], "x"]), {'today', today}]

也会产生True的结果。

由于某些对象本身的性质,它们不能被pickle。例如,pickle包含打开文件句柄的结构是没有意义的。

供你参考,熊猫现在有办法拯救泡菜了。

我发现它更容易。

pd.to_pickle(object_to_save,'/temp/saved_pkl.pickle' )
import pickle

dictobj = {'Jack' : 123, 'John' : 456}

filename = "/foldername/filestore"

fileobj = open(filename, 'wb')

pickle.dump(dictobj, fileobj)

fileobj.close()