如何减少在python中加载pickle文件所花费的时间

时间:2014-11-11 07:51:15

标签: python performance pickle

我在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)

加载此次泡菜需要 15秒如何缩短这段时间?

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需要更少的时间来转储和加载,但加载文件仍然需要很长时间

3 个答案:

答案 0 :(得分:23)

尝试使用json library代替pickle。这应该是你的案例中的一个选项,因为你正在处理一个相对简单的对象字典。

根据this website

  

JSON的读取速度(加载速度)快25倍,速度提高15倍   写作(转储)。

另请参阅此问题:What is faster - Loading a pickled dictionary object or Loading a JSON file - to a dictionary?

升级Python或使用带有固定Python版本的the marshal module也有助于提高速度(code adapted from here):

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 uses pickle protocol 3 as default,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(如果为转储提供负值,则选择),这是协议3的两倍慢。

答案 1 :(得分:7)

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)方面取得了很好的效果。这是通过简单地完成提到here

的pickle加载调用来实现的

您案例中的示例代码段如下:

$parameters[0]

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)

答案 2 :(得分:3)

如果您尝试将字典存储到单个文件中,那么大文件的加载时间会降低您的速度。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单个文件。然后,您可以在多个线程中(或使用多处理)对文件进行pickle和unpickled。对于一个非常大的字典,无论您选择哪种序列化,这都应该比读取单个文件快得多。有一些像kleptojoblib这样的软件包已经为你做了很多(如果不是全部的话)。我会检查这些包裹。 (注意:我是klepto作者。请参阅https://github.com/uqfoundation/klepto)。