我怎样才能用泡菜来保存字典？

我已经浏览了Python文档给出的信息，但是我仍然有点困惑。 有人可以发布示例代码，将写入一个新的文件，然后使用pickle转储字典到它？

Answer 1

尝试这个：

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 

Answer 2

由于Blender似乎不想将HIGHEST_PROTOCOL添加到他的答案中，所以这里有一个：

 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的优点是文件变小了。 这使得有时候快速取出。

重要提示 ：腌菜的最大文件大小约为2GB。

备择scheme

• CSV：超简单格式（ 读写 ）
• JSON：写出人类可读的数据很好， 非常常用（ 读写 ）
• YAML：YAML是JSON的超集，但更容易阅读（ 读写 ， 比较JSON和YAML ）
• pickle：Python序列化格式（ 读写 ）
• MessagePack （ Python包 ）：更紧凑的表示（ 读写 ）
• HDF5 （ Python包 ）：很好的matrix（ 读写 ）
• XML：也存在*叹*（ 读写 ）

对于您的应用程序，以下可能是重要的：

• 支持其他编程语言
• 阅读/写作performance
• 紧凑（文件大小）

另请参阅： 数据序列化格式的比较

如果你正在寻找一种configuration文件的方式，你可能想阅读我的简短文章Pythonconfiguration文件

Answer 3

 # 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"} 

Answer 4

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

通常最好使用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. 

Answer 5

一般来说，酸洗dict将会失败，除非你只有简单的对象，比如string和整数。

 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dict往往会失败。 这只取决于内容。

 >>> 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> 

但是，如果使用更好的序列化程序（如dill或cloudpickle ，则可以对大多数字典进行腌制：

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

或者，如果你想保存你的dict到一个文件…

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

后面的例子与其他任何在这里发布的好答案是相同的（除了忽略dict的内容的可拣性以外）。

Answer 6

 import pickle dictobj = {'Jack' : 123, 'John' : 456} filename = "/foldername/filestore" fileobj = open(filename, 'wb') pickle.dump(dictobj, fileobj) fileobj.close() 

Answer 7

我发现酸洗混乱（可能是因为我很厚）。 我发现这个工作，但：

 myDictionaryString=str(myDictionary) 

然后你可以写入一个文本文件。 我放弃了尝试使用泡菜，因为我得到的错误告诉我写整数.dat文件。 我为不使用泡菜道歉。