保存和加载对象并使用pickle

Question

我正在尝试使用pickle模块保存和加载对象.
首先,我声明我的对象:

>>> class Fruits:pass
...
>>> banana = Fruits()

>>> banana.color = 'yellow'
>>> banana.value = 30

之后我打开一个名为'Fruits.obj'的文件(之前我创建了一个新的.txt文件,我重命名为'Fruits.obj'):

>>> import pickle
>>> filehandler = open(b"Fruits.obj","wb")
>>> pickle.dump(banana,filehandler)

执行此操作后,我关闭了会话,然后我开始了一个新的会话,然后我将下一个(尝试访问它应该保存的对象):

file = open("Fruits.obj",'r')
object_file = pickle.load(file)

但我有这样的信息:

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python31\lib\pickle.py", line 1365, in load
encoding=encoding, errors=errors).load()
ValueError: read() from the underlying stream did notreturn bytes

我不知道该怎么办,因为我不明白这个消息.有谁知道如何加载我的对象'香蕉'？谢谢!

编辑: 因为你们有些人已经吮吸我放:

>>> import pickle
>>> file = open("Fruits.obj",'rb')

没有问题,但我接下来的是:

>>> object_file = pickle.load(file)

我有错误:

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python31\lib\pickle.py", line 1365, in load
encoding=encoding, errors=errors).load()
EOFError

Answer 1

至于你的第二个问题:

 Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 File "C:\Python31\lib\pickle.py", line
 1365, in load encoding=encoding,
 errors=errors).load() EOFError

读完文件内容后,文件指针将位于文件的末尾 - 将无法读取更多数据.您必须回放文件,以便再次从头开始读取:

file.seek(0)

您通常想要做的是使用上下文管理器打开文件并从中读取数据.这样,文件将在块完成执行后自动关闭,这也将帮助您将文件操作组织成有意义的块.

最后,cPickle是C中pickle模块的更快实现.所以:

In [1]: import cPickle

In [2]: d = {"a": 1, "b": 2}

In [4]: with open(r"someobject.pickle", "wb") as output_file:
   ...:     cPickle.dump(d, output_file)
   ...:

# pickle_file will be closed at this point, preventing your from accessing it any further

In [5]: with open(r"someobject.pickle", "rb") as input_file:
   ...:     e = cPickle.load(input_file)
   ...:

In [7]: print e
------> print(e)
{'a': 1, 'b': 2}

Answer 2

以下适用于我:

class Fruits: pass

banana = Fruits()

banana.color = 'yellow'
banana.value = 30

import pickle

filehandler = open("Fruits.obj","wb")
pickle.dump(banana,filehandler)
filehandler.close()

file = open("Fruits.obj",'rb')
object_file = pickle.load(file)
file.close()

print(object_file.color, object_file.value, sep=', ')
# yellow, 30

Answer 3

你忘了把它读成二进制文件.

在你的写作部分,你有:

open(b"Fruits.obj","wb") # Note the wb part (Write Binary)

在阅读部分中,您有:

file = open("Fruits.obj",'r') # Note the r part, there should be a b too

所以用以下代替:

file = open("Fruits.obj",'rb')

它会工作:)

至于你的第二个错误,很可能是因为没有正确关闭/同步文件.

试试这段代码来写:

>>> import pickle
>>> filehandler = open(b"Fruits.obj","wb")
>>> pickle.dump(banana,filehandler)
>>> filehandler.close()

这(未更改)阅读:

>>> import pickle
>>> file = open("Fruits.obj",'rb')
>>> object_file = pickle.load(file)

一个更整洁的版本将使用该with声明.

写作:

>>> import pickle
>>> with open('Fruits.obj', 'wb') as fp:
>>>     pickle.dump(banana, fp)

阅读:

>>> import pickle
>>> with open('Fruits.obj', 'rb') as fp:
>>>     banana = pickle.load(fp)

Answer 4

在这种情况下,始终以二进制模式打开

file = open("Fruits.obj",'rb')

Answer 5

您可以使用anycache为您完成这项工作。假设您有一个myfunc创建实例的函数：

from anycache import anycache

class Fruits:pass

@anycache(cachedir='/path/to/your/cache')    
def myfunc()
    banana = Fruits()
    banana.color = 'yellow'
    banana.value = 30
return banana

Anycachemyfunc在第一次调用并将结果腌制到文件中，cachedir使用唯一标识符（取决于函数名和参数）作为文件名。在任何连续运行中，都会加载腌制对象。

如果cachedir在 python 运行之间保留了 ，pickled 对象将从之前的 python 运行中获取。

函数参数也被考虑在内。重构的实现同样有效：

from anycache import anycache

class Fruits:pass

@anycache(cachedir='/path/to/your/cache')    
def myfunc(color, value)
    fruit = Fruits()
    fruit.color = color
    fruit.value = value
return fruit