如何减少在python中加载pickle文件所花费的时间

Question

我已经在python中创建了一个字典并将其转换为pickle.它的大小达到了300MB.现在,我想加载相同的泡菜.

output = open('myfile.pkl', 'rb')
mydict = pickle.load(output)

加载这个泡菜大约需要15秒.我怎样才能减少这个时间？

硬件规格:Ubuntu 14.04,4GB RAM

下面的代码显示了使用json,pickle,cPickle转储或加载文件所需的时间.

转储后,文件大小约为300MB.

import json, pickle, cPickle
import os, timeit
import json

mydict= {all values to be added}

def dump_json():    
    output = open('myfile1.json', 'wb')
    json.dump(mydict, output)
    output.close()    

def dump_pickle():    
    output = open('myfile2.pkl', 'wb')
    pickle.dump(mydict, output,protocol=cPickle.HIGHEST_PROTOCOL)
    output.close()

def dump_cpickle():    
    output = open('myfile3.pkl', 'wb')
    cPickle.dump(mydict, output,protocol=cPickle.HIGHEST_PROTOCOL)
    output.close()

def load_json():
    output = open('myfile1.json', 'rb')
    mydict = json.load(output)
    output.close()

def load_pickle():
    output = open('myfile2.pkl', 'rb')
    mydict = pickle.load(output)
    output.close()

def load_cpickle():
    output = open('myfile3.pkl', 'rb')
    mydict = pickle.load(output)
    output.close()


if __name__ == '__main__':
    print "Json dump: "
    t = timeit.Timer(stmt="pickle_wr.dump_json()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

    print "Pickle dump: "
    t = timeit.Timer(stmt="pickle_wr.dump_pickle()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

    print "cPickle dump: "
    t = timeit.Timer(stmt="pickle_wr.dump_cpickle()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

    print "Json load: "
    t = timeit.Timer(stmt="pickle_wr.load_json()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

    print "pickle load: "
    t = timeit.Timer(stmt="pickle_wr.load_pickle()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

    print "cPickle load: "
    t = timeit.Timer(stmt="pickle_wr.load_cpickle()", setup="import pickle_wr")  
    print t.timeit(1),'\n'

输出:

Json dump: 
42.5809804916 

Pickle dump: 
52.87407804489 

cPickle dump: 
1.1903790187836 

Json load: 
12.240660209656 

pickle load: 
24.48748306274 

cPickle load: 
24.4888298893

我已经看到cPickle转储和加载所需的时间较少,但加载文件仍需要很长时间.

Answer 1

尝试使用json库而不是pickle.在您的情况下,这应该是一个选项,因为您正在处理一个相对简单的对象字典.

根据这个网站,

JSON在读取(加载)方面快25倍,在写入(转储)方面快15倍.

另请参阅此问题:什么是更快 - 将pickle字典对象或加载JSON文件加载到字典？

升级Python或使用具有固定Python版本的marshal模块也有助于提高速度(代码从此处改编):

try: import cPickle
except: import pickle as cPickle
import pickle
import json, marshal, random
from time import time
from hashlib import md5

test_runs = 1000

if __name__ == "__main__":
    payload = {
        "float": [(random.randrange(0, 99) + random.random()) for i in range(1000)],
        "int": [random.randrange(0, 9999) for i in range(1000)],
        "str": [md5(str(random.random()).encode('utf8')).hexdigest() for i in range(1000)]
    }
    modules = [json, pickle, cPickle, marshal]

    for payload_type in payload:
        data = payload[payload_type]
        for module in modules:
            start = time()
            if module.__name__ in ['pickle', 'cPickle']:
                for i in range(test_runs): serialized = module.dumps(data, protocol=-1)
            else:
                for i in range(test_runs): serialized = module.dumps(data)
            w = time() - start
            start = time()
            for i in range(test_runs):
                unserialized = module.loads(serialized)
            r = time() - start
            print("%s %s W %.3f R %.3f" % (module.__name__, payload_type, w, r))

结果:

C:\Python27\python.exe -u "serialization_benchmark.py"
json int W 0.125 R 0.156
pickle int W 2.808 R 1.139
cPickle int W 0.047 R 0.046
marshal int W 0.016 R 0.031
json float W 1.981 R 0.624
pickle float W 2.607 R 1.092
cPickle float W 0.063 R 0.062
marshal float W 0.047 R 0.031
json str W 0.172 R 0.437
pickle str W 5.149 R 2.309
cPickle str W 0.281 R 0.156
marshal str W 0.109 R 0.047

C:\pypy-1.6\pypy-c -u "serialization_benchmark.py"
json int W 0.515 R 0.452
pickle int W 0.546 R 0.219
cPickle int W 0.577 R 0.171
marshal int W 0.032 R 0.031
json float W 2.390 R 1.341
pickle float W 0.656 R 0.436
cPickle float W 0.593 R 0.406
marshal float W 0.327 R 0.203
json str W 1.141 R 1.186
pickle str W 0.702 R 0.546
cPickle str W 0.828 R 0.562
marshal str W 0.265 R 0.078

c:\Python34\python -u "serialization_benchmark.py"
json int W 0.203 R 0.140
pickle int W 0.047 R 0.062
pickle int W 0.031 R 0.062
marshal int W 0.031 R 0.047
json float W 1.935 R 0.749
pickle float W 0.047 R 0.062
pickle float W 0.047 R 0.062
marshal float W 0.047 R 0.047
json str W 0.281 R 0.187
pickle str W 0.125 R 0.140
pickle str W 0.125 R 0.140
marshal str W 0.094 R 0.078

Python 3.4使用pickle协议3作为默认值,与协议4相比没有任何区别.Python 2将协议2作为最高的pickle协议(如果为dump提供负值,则选择),这是协议3的两倍.

Answer 2

我在使用cPickle本身读取大文件(例如:~750 MB igraph对象 - 二进制pickle文件)方面取得了不错的成绩.这是通过简单地结束这里提到的pickle加载调用来实现的

您的案例中的示例代码段将类似于:

import timeit
import cPickle as pickle
import gc


def load_cpickle_gc():
    output = open('myfile3.pkl', 'rb')

    # disable garbage collector
    gc.disable()

    mydict = pickle.load(output)

    # enable garbage collector again
    gc.enable()
    output.close()


if __name__ == '__main__':
    print "cPickle load (with gc workaround): "
    t = timeit.Timer(stmt="pickle_wr.load_cpickle_gc()", setup="import pickle_wr")
    print t.timeit(1),'\n'

当然,可能有更合适的方法来完成任务,但是,这种解决方法确实可以大大减少所需的时间.(对我来说,它从843.04s减少到41.28s,大约20x)

Answer 3

如果您尝试将字典存储到单个文件中，则大文件的加载时间会减慢您的速度。您可以做的最简单的事情之一是将字典写入磁盘上的目录，每个字典条目都是一个单独的文件。然后，您可以在多个线程（或使用多重处理）中对文件进行 pickle 和 unpickled。对于非常大的字典，这应该比读取单个文件要快得多，无论您选择什么序列化器。有一些类似klepto和 的软件包joblib已经为您做了很多事情（如果不是上述全部）。我会检查那些包裹。（注：我是klepto作者。请参阅https://github.com/uqfoundation/klepto）。