python : how to speed up this file loading

Question

我正在寻找一种加速文件加载的方法，如下所示:

数据包含约100万行，制表符以“\t”(制表符)分隔，utf8编码，使用下面的代码解析完整文件大约需要9秒。然而，我希望能达到大约一秒!

def load(filename):
    features = []
    with codecs.open(filename, 'rb',  'utf-8') as f:
        previous = ""
        for n, s in enumerate(f):
            splitted = tuple(s.rstrip().split("\t"))
            if len(splitted) != 2:
                sys.exit("wrong format!")
            if previous >= splitted:
                sys.exit("unordered feature")
            previous = splitted
            features.append(splitted)
    return features   

我想知道是否有任何二进制格式数据可以加快速度？或者，如果我可以从一些 NumPy 或任何其他库中受益，以获得更快的加载速度。

也许你可以给我关于另一个速度瓶颈的建议？

编辑:所以我尝试了你的一些想法，谢谢!顺便说一句，我真的需要巨大列表中的元组(字符串，字符串)...这是结果，我获得了 50% 的时间:) 现在我要照顾 NumPy 二进制数据，正如我所注意到的另一个大文件的加载速度真的非常快......

import codecs

def load0(filename): 
    with codecs.open(filename, 'rb',  'utf-8') as f: 
    return f.readlines() 

def load1(filename): 
    with codecs.open(filename, 'rb',  'utf-8') as f: 
    return [tuple(x.rstrip().split("\t")) for x in f.readlines()]

def load3(filename):
    features = []
    with codecs.open(filename, 'rb',  'utf-8') as f:
    for n, s in enumerate(f):
        splitted = tuple(s.rstrip().split("\t"))
        features.append(splitted)
    return features

def load4(filename): 
    with codecs.open(filename, 'rb',  'utf-8') as f: 
    for s in f: 
        yield tuple(s.rstrip().split("\t")) 

a = datetime.datetime.now()
r0 = load0(myfile)
b = datetime.datetime.now()
print "f.readlines(): %s" % (b-a)

a = datetime.datetime.now()
r1 = load1(myfile)
b = datetime.datetime.now()
print """[tuple(x.rstrip().split("\\t")) for x in f.readlines()]: %s""" % (b-a)

a = datetime.datetime.now()
r3 = load3(myfile)
b = datetime.datetime.now()
print """load3: %s""" % (b-a)
if r1 == r3: print "OK: speeded and similars!"

a = datetime.datetime.now()
r4 = [x for x in load4(myfile)] 
b = datetime.datetime.now()
print """load4: %s""" % (b-a)
if r4 == r3: print "OK: speeded and similars!"

结果:

f.readlines(): 0:00:00.208000
[tuple(x.rstrip().split("\t")) for x in f.readlines()]: 0:00:02.310000
load3: 0:00:07.883000
OK: speeded and similars!
load4: 0:00:07.943000
OK: speeded and similars!

非常奇怪的是，我注意到我可以在连续两次运行中获得几乎双倍的时间(但不是每次):

>>> ================================ RESTART ================================
>>> 
f.readlines(): 0:00:00.220000
[tuple(x.rstrip().split("\t")) for x in f.readlines()]: 0:00:02.479000
load3: 0:00:08.288000
OK: speeded and similars!
>>> ================================ RESTART ================================
>>> 
f.readlines(): 0:00:00.279000
[tuple(x.rstrip().split("\t")) for x in f.readlines()]: 0:00:04.983000
load3: 0:00:10.404000
OK: speeded and similars!

编辑最新:好吧，我尝试修改以使用numpy.load...这对我来说很奇怪...从“正常”文件与我的1022860 个字符串，10 KB。 执行完 numpy.save(numpy.array(load1(myfile))) 后，我的大小达到了 895 MB!然后使用 numpy.load() 重新加载它，我在连续运行中得到这种计时:

  >>> ================================ RESTART ================================
  loading: 0:00:11.422000 done.
  >>> ================================ RESTART ================================
  loading: 0:00:00.759000 done.

可能 numpy 会做一些内存操作以避免将来重新加载吗？

Answer 1

尝试这个版本，因为你提到检查并不重要，所以我已经消除了它。

def load(filename):
    with codecs.open(filename, 'rb',  'utf-8') as f:
        for s in f:
            yield tuple(s.rstrip().split("\t"))

results = [x for x in load('somebigfile.txt')]