Pandas read_xml()方法测试策略

时间:2017-01-21 06:08:01

标签: python xml pandas xslt xpath

有趣的是,pandas I/O tools不保留read_xml()方法和对应to_xml()。但是,read_json证明可以为数据帧导入实现树状结构,并为标记格式实现read_html

现在,如果pandas团队确实为未来的pandas版本考虑了这样的read_xml方法,那么他们将采用什么实现:使用内置xml.etree.ElementTree及其iterfind()或{进行解析{1}}函数或第三方模块iterparse()及其XPath 1.0和XSLT 1.0方法?

下面是我在一个简单,扁平,以元素为中心的XML输入上的四种方法类型的测试运行。所有这些都设置为root的任何二级子级的基因化解析,并且每个方法应该产生完全相同的pandas数据帧。除字典列表中的最后一次调用lxml之外的所有字符。 XSLT方法将XML转换为CSV,以便pd.Dataframe()中的已转换StringIO()

问题 (多部分)

  • 性能:当您对文件进行迭代解析时,您如何解释为较大文件经常推荐的较慢pd.read_csv()?是否部分归因于iterparse逻辑检查?

  • 内存:CPU内存是否与I / O调用的时间相关? XSLT和XPath 1.0往往不能很好地扩展XML文档,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if来电的最佳策略吗?请参阅以下有趣答案:generator版本和iterwalk user-defined版本。两个上传列表到数据帧。

输入数据(Stackoverflow的当前top users by year ,其中包含我们的熊猫朋友)

Dataframe()

Python 方法

<?xml version="1.0" encoding="utf-8"?>
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计时 (当前XML和XML的子项为25倍(即900个StackOverflow用户记录)

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)))

1 个答案:

答案 0 :(得分:0)

  

性能:随着文件被迭代解析,您如何解释通常建议对较大文件使用的较慢的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?

如果我删除i并仅使用first_tag,它似乎要快很多,所以可以,部分原因是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一次if检查的目的,但是我也不确定为什么要丢失仅包含空格的元素。始终删除最后一个if可以节省一点时间:

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特大的数据框。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调用中的时序相关? 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。

我不太确定您所说的“ 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。这两个上传列表都已添加到数据帧。

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。 DataFrame()调用也将需要做更少的工作,因为它不必在每一行的dict中查找键,从而可以找出哪一列具有什么值。