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
import pickle
dictobj = {'Jack' : 123, 'John' : 456}
filename = "/foldername/filestore"
fileobj = open(filename, 'wb')
pickle.dump(dictobj, fileobj)
fileobj.close()
其他回答
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。
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.
# Save a dictionary into a pickle file.
import pickle
favorite_color = {"lion": "yellow", "kitty": "red"} # create a dictionary
pickle.dump(favorite_color, open("save.p", "wb")) # save it into a file named save.p
# -------------------------------------------------------------
# Load the dictionary back from the pickle file.
import pickle
favorite_color = pickle.load(open("save.p", "rb"))
# favorite_color is now {"lion": "yellow", "kitty": "red"}
import pickle
your_data = {'foo': 'bar'}
# Store data (serialize)
with open('filename.pickle', 'wb') as handle:
pickle.dump(your_data, handle, protocol=pickle.HIGHEST_PROTOCOL)
# Load data (deserialize)
with open('filename.pickle', 'rb') as handle:
unserialized_data = pickle.load(handle)
print(your_data == unserialized_data)
HIGHEST_PROTOCOL的优点是文件变得更小。这使得解腌有时要快得多。
重要提示:pickle的最大文件大小约为2GB。
替代方法
import mpu
your_data = {'foo': 'bar'}
mpu.io.write('filename.pickle', data)
unserialized_data = mpu.io.read('filename.pickle')
选择格式
CSV:超简单格式(读写) JSON:适合编写人类可读的数据;非常常用(读和写) YAML: YAML是JSON的超集,但更容易阅读(读写,JSON和YAML的比较) pickle: Python序列化格式(读和写) MessagePack (Python包):更紧凑的表示(读和写) HDF5 (Python包):适合矩阵(读和写) XML:也存在*叹*(读和写)
对于您的应用程序,以下内容可能很重要:
其他编程语言的支持 读写能力 紧凑性(文件大小)
请参见:数据序列化格式的比较
如果您正在寻找一种创建配置文件的方法,您可能想要阅读我的简短文章Python中的配置文件
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