在Python 2.7中高效读取800 GB XML文件 [英] Efficient reading of 800 GB XML file in Python 2.7

问题描述

我正在读取python 2.7中的800 GB xml文件并使用etree迭代解析器解析它。

I am reading an 800 GB xml file in python 2.7 and parsing it with an etree iterative parser.

目前，我只是使用 open（'foo.txt'），没有缓冲参数。我有点困惑这是否是我应该采取的方法，或者我应该使用缓冲参数或使用io中的东西，如io.BufferedReader或io.open或io.TextIOBase。

Currently, I am just using open('foo.txt') with no buffering argument. I am a little confused whether this is the approach I should take or I should use a buffering argument or use something from io like io.BufferedReader or io.open or io.TextIOBase.

非常感谢正确方向上的一点。

A point in the right direction would be much appreciated.

推荐答案

标准 open（）函数已经返回一个缓冲文件（如果有的话）在你的平台上）。对于通常完全缓冲的文件对象。

The standard open() function already, by default, returns a buffered file (if available on your platform). For file objects that is usually fully buffered.

通常这里意味着Python将此保留给C stdlib实现;它使用 fopen（）来电（ wfopen（）在Windows上支持UTF-16文件名），这意味着选择了文件的默认缓冲;在Linux上我相信这将是8kb。对于像XML解析这样的纯读取操作，这种类型的缓冲完全你想要的。

Usually here means that Python leaves this to the C stdlib implementation; it uses a fopen() call (wfopen() on Windows to support UTF-16 filenames), which means that the default buffering for a file is chosen; on Linux I believe that would be 8kb. For a pure-read operation like XML parsing this type of buffering is exactly what you want.

由<$ c $完成的XML解析c> iterparse 以16384字节（16kb）的块读取文件。

The XML parsing done by iterparse reads the file in chunks of 16384 bytes (16kb).

如果要控制buffersize，请使用缓冲关键字参数：

If you want to control the buffersize, use the buffering keyword argument:

open('foo.xml', buffering=(2<<16) + 8)  # buffer enough for 8 full parser reads

将覆盖默认值缓冲区大小（我期望匹配文件块大小或其倍数）。根据本文增加读取缓冲区应该 help，并且使用至少4倍的预期读取块大小加上8个字节的大小将提高读取性能。在上面的示例中，我将其设置为ElementTree读取大小的8倍。

which will override the default buffer size (which I'd expect to match the file block size or a multiple thereof). According to this article increasing the read buffer should help, and using a size at least 4 times the expected read block size plus 8 bytes is going to improve read performance. In the above example I've set it to 8 times the ElementTree read size.

io.open（） function 表示对象的新Python 3 I / O结构，其中I / O已被拆分为新的类类型层次结构，以提供更大的灵活性。价格更加间接，数据需要通过更多层，而Python C代码本身更有效，而不是将其留给操作系统。

The io.open() function represents the new Python 3 I/O structure of objects, where I/O has been split up into a new hierarchy of class types to give you more flexibility. The price is more indirection, more layers for the data to have to travel through, and the Python C code does more work itself instead of leaving that to the OS.

你可以尝试查看 io.open（'foo.xml'，'rb'，buffering = 2<< 16）是否会表现更好。以 rb 模式打开将为您提供 io.BufferedReader 实例。

You could try and see if io.open('foo.xml', 'rb', buffering=2<<16) is going to perform any better. Opening in rb mode will give you a io.BufferedReader instance.

你不想要使用 io.TextIOWrapper ;底层的expat解析器需要原始数据，因为它将解码您的XML文件编码本身。它只会增加额外的开销;如果你在 r （textmode）中打开，你会得到这种类型。

You do not want to use io.TextIOWrapper; the underlying expat parser wants raw data as it'll decode your XML file encoding itself. It would only add extra overhead; you get this type if you open in r (textmode) instead.

使用 io .open（）可以为您提供更多灵活性和更丰富的API，但是使用 open（）而不是<$来打开底层C文件对象c $ c> fopen（），所有缓冲都由Python io.BufferedIOBase 实现处理。

Using io.open() may give you more flexibility and a richer API, but the underlying C file object is opened using open() instead of fopen(), and all buffering is handled by the Python io.BufferedIOBase implementation.

我认为你的问题将是处理这个野兽，而不是文件读取。在读取800GB文件时，磁盘缓存无论如何都会被拍摄。

Your problem will be processing this beast, not the file reads, I think. The disk cache will be pretty much shot anyway when reading a 800GB file.

这篇关于在Python 2.7中高效读取800 GB XML文件的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！