熊猫read_xml()方法测试策略


109

当前,pandas I / O工具没有维护read_xml()方法,而相应的工具to_xml()。但是,read_json证明可以为数据帧导入和read_html标记格式实现树状结构。

如果大熊猫团队会考虑这样一个read_xml为未来大熊猫版本的方法,他们会追求什么实现:使用内置的解析xml.etree.ElementTreeiterfind()iterparse()功能或第三方模块,lxml其XPath 1.0和XSLT 1.0的方法呢?

以下是我在简单,扁平,以元素为中心的XML输入上针对四种方法类型的测试运行。所有这些都针对root的任何第二级子级进行了通用解析,并且每种方法都应产生完全相同的pandas数据帧。除最后一次调用外pd.Dataframe(),所有其他功能都在词典列表中。XSLT方法将XML转换为CSV,以便StringIO()在中进行转换pd.read_csv()

问题 (多部分)

  • 性能:您如何解释由于iterparse迭代解析文件而通常建议对较大文件使用的速度较慢的速度?部分原因是由于if逻辑检查吗?

  • 内存:CPU内存是否与I / O调用中的时间相关?XSLT和XPath 1.0在较大的XML文档中往往无法很好地扩展,因为必须在内存中读取整个文件才能进行解析。

  • 策略:词典列表是Dataframe()呼叫的最佳策略吗?请参阅以下有趣的答案:生成器版本和iterwalk用户定义版本。两个上载列表到数据帧。

输入数据(Stack Overflow当前的年度最大用户,其中包括我们的熊猫朋友)

<?xml version="1.0" encoding="utf-8"?>
<stackoverflow>
  <topusers>
    <user>Gordon Linoff</user>
    <link>http://www.stackoverflow.com//users/1144035/gordon-linoff</link>
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    <location>??????</location>
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Python方法

import xml.etree.ElementTree as et
import pandas as pd
from io import StringIO
from lxml import etree as lxet

def read_xml_iterfind():
    tree = et.parse('Input.xml')

    data = []
    inner = {}
    for el in tree.iterfind('./*'):
        for i in el.iterfind('*'):
            inner[i.tag] = i.text
        data.append(inner)
        inner = {}

    df = pd.DataFrame(data)

def read_xml_iterparse():
    data = []
    inner = {}
    i = 1
    for (ev, el) in et.iterparse(path):
        if i <= 2:
           first_tag = el.tag

        if el.tag == first_tag and len(inner) != 0:
            data.append(inner)            
            inner = {}

        if el.text is not None and len(el.text.strip()) > 0:
            inner[el.tag] = el.text
    i += 1

    df = pd.DataFrame(data)    

def read_xml_lxml_xpath():     
    tree = lxet.parse('Input.xml')

    data = []
    inner = {}
    for el in tree.xpath('/*/*'):
        for i in el:
            inner[i.tag] = i.text
        data.append(inner)
        inner = {}

    df = pd.DataFrame(data)

def read_xml_lxml_xsl():     
    xml = lxet.parse('Input.xml')

    xslstr = '''
    <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
        <xsl:output version="1.0" encoding="UTF-8" indent="yes"  method="text"/>
        <xsl:strip-space elements="*"/>

        <!-- HEADERS -->
        <xsl:template match = "/*">
            <xsl:for-each select="*[1]/*">
              <xsl:value-of select="local-name()" />
                <xsl:choose>
                   <xsl:when test="position() != last()">
                      <xsl:text>,</xsl:text>
                   </xsl:when>
                   <xsl:otherwise>
                      <xsl:text>&#xa;</xsl:text>
                   </xsl:otherwise>                              
                </xsl:choose>   
            </xsl:for-each>
            <xsl:apply-templates/>
        </xsl:template>

        <!-- DATA ROWS (COMMA-SEPARATED) -->
        <xsl:template match="/*/*" priority="2">    
            <xsl:for-each select="*">
              <xsl:if test="position() = 1">
                   <xsl:text>&quot;</xsl:text>
              </xsl:if>
              <xsl:value-of select="." />
                <xsl:choose>
                   <xsl:when test="position() != last()">
                      <xsl:text>&quot;,&quot;</xsl:text>
                   </xsl:when>
                   <xsl:otherwise>
                      <xsl:text>&quot;&#xa;</xsl:text>
                   </xsl:otherwise>                              
                </xsl:choose>
            </xsl:for-each>
        </xsl:template>

    </xsl:transform>
    '''
    xsl = lxet.fromstring(xslstr)

    transform = lxet.XSLT(xsl)
    newdom = transform(xml)

    df = pd.read_csv(StringIO(str(newdom)))

时序 (当前的XML和XML的子级是25倍(即900条StackOverflow用户记录)

# SHORTER FILE
python -mtimeit -s'import readxml_test_runs as test' 'test.read_xml_iterfind()'
100 loops, best of 3: 3.87 msec per loop

python -mtimeit -s'import readxml_test_runs as test' 'test.read_xml_iterparse()'
100 loops, best of 3: 5.5 msec per loop

python -mtimeit -s'import readxml_test_runs as test' 'test.read_xml_lxml_xpath()'
100 loops, best of 3: 3.86 msec per loop

python -mtimeit -s'import readxml_test_runs as test' 'test.read_xml_lxml_xsl()'
100 loops, best of 3: 5.68 msec per loop

# LARGER FILE
python -mtimeit -n'100' -s'import readxml_test_runs as test' 'test.read_xml_iterfind()'
100 loops, best of 3: 36 msec per loop

python -mtimeit -n'100' -s'import readxml_test_runs as test' 'test.read_xml_iterparse()'
100 loops, best of 3: 78.9 msec per loop

python -mtimeit -n'100' -s'import readxml_test_runs as test' 'test.read_xml_lxml_xpath()'
100 loops, best of 3: 32.7 msec per loop

python -mtimeit -n'100' -s'import readxml_test_runs as test' 'test.read_xml_lxml_xsl()'
100 loops, best of 3: 51.4 msec per loop

10
据我所知,这也在pandas github上进行了讨论。也许在那里开一个问题?
westr

Answers:


1

性能:如何解释由于迭代解析文件而通常建议对较大文件使用的较慢iterparse?部分原因是由于if逻辑检查?

我认为更多的python代码会使它变慢,因为每次都会评估python代码。您是否尝试过像pypy这样的JIT编译器?

如果仅删除i并使用first_tag,它似乎会快很多,所以是的,部分原因在于if逻辑检查:

def read_xml_iterparse2(path):
    data = []
    inner = {}
    first_tag = None
    for (ev, el) in et.iterparse(path):
        if not first_tag:
           first_tag = el.tag

        if el.tag == first_tag and len(inner) != 0:
            data.append(inner)            
            inner = {}

        if el.text is not None and len(el.text.strip()) > 0:
            inner[el.tag] = el.text

    df = pd.DataFrame(data)    

%timeit read_xml_iterparse(path)
# 10 loops, best of 5: 33 ms per loop
%timeit read_xml_iterparse2(path)
# 10 loops, best of 5: 23 ms per loop

我不确定我是否了解上次if检查的目的,但也不确定为什么您会丢失仅空白元素。持续删除最后一个可以if节省一点时间:

def read_xml_iterparse3(path):
    data = []
    inner = {}
    first_tag = None
    for (ev, el) in et.iterparse(path):
        if not first_tag:
           first_tag = el.tag

        if el.tag == first_tag and len(inner) != 0:
            data.append(inner)            
            inner = {}

        inner[el.tag] = el.text

    df = pd.DataFrame(data)    

%timeit read_xml_iterparse(path)
# 10 loops, best of 5: 34.4 ms per loop
%timeit read_xml_iterparse2(path)
# 10 loops, best of 5: 24.5 ms per loop
%timeit read_xml_iterparse3(path)
# 10 loops, best of 5: 20.9 ms per loop

现在,无论是否进行了这些性能改进,您的iterparse版本似乎都会产生一个更大的数据框。这似乎是一个有效的快速版本:

def read_xml_iterparse5(path):
    data = []
    inner = {}
    for (ev, el) in et.iterparse(path):
        # /ending parents trigger a new row, and in our case .text is \n followed by spaces.  it would be more reliable to pass 'topusers' to our read_xml_iterparse5 as the .tag to check
        if el.text and el.text[0] == '\n':
            # ignore /stackoverflow
            if inner:
                data.append(inner)
                inner = {}
        else:
            inner[el.tag] = el.text

    return pd.DataFrame(data)    

print(read_xml_iterfind(path).shape)
# (900, 8)
print(read_xml_iterparse(path).shape)
# (7050, 8)
print(read_xml_lxml_xpath(path).shape)
# (900, 8)
print(read_xml_lxml_xsl(path).shape)
# (900, 8)
print(read_xml_iterparse5(path).shape)
# (900, 8)
%timeit read_xml_iterparse5(path)
# 10 loops, best of 5: 20.6 ms per loop

内存:CPU内存是否与I / O调用中的时间相关?XSLT和XPath 1.0在较大的XML文档中往往无法很好地扩展,因为必须在内存中读取整个文件才能进行解析。

我不能完全确定“ I / O调用”是什么意思,但是如果您的文档足够小以适合缓存,那么一切都会更快,因为它不会从缓存中逐出其他项目。

策略:词典列表是否是Dataframe()调用的最佳策略?请参阅以下有趣的答案:生成器版本和iterwalk用户定义的版本。两个上载列表到数据帧。

列表使用的内存较少,因此根据您拥有的列数,它可能会产生明显的不同。当然,这然后要求您的XML标记具有一致的顺序,看起来确实如此。该DataFrame()调用也将需要做的工作更少,因为它不必在每一行的dict中查找键,以弄清楚哪一列是什么值。


谢谢您的回答。但是,您的答复似乎是针对Python的一般性答复,而不是针对Pandas建议的特定XML方法。也许一个特定的编码示例可以使用上述可复制示例更好地说明JIT想法或Cython?
冻糕

也许我不明白你的问题?如果某些内容适用于所有Python代码,则适用于您的Python代码。如果您在寻找代码示例,则不清楚您的问题。
詹(Jayen)
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