Pandas read_xml() 方法测试策略
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StackOverFlow
Pandas read_xml() method test strategies
提问by Parfait
Currently, pandas I/O toolsdoes not maintain a read_xml()method and the counterpart to_xml(). However, read_jsonproves tree-like structures can be implemented for dataframe import and read_htmlfor markup formats.
目前，pandas I/O 工具不维护read_xml()方法和对应的to_xml(). 但是，read_json证明可以为数据框导入和read_html标记格式实现树状结构。
If the pandas team does consider such a read_xmlmethod for a future pandas version, what implementation would they pursue: parsing with built-in xml.etree.ElementTreewith its iterfind()or iterparse()functions or the third-party module, lxmlwith its XPath 1.0 and XSLT 1.0 methods?
如果大Pandas团队会考虑这样一个read_xml为未来大Pandas版本的方法，他们会追求什么实现：使用内置的解析xml.etree.ElementTree其iterfind()或iterparse()功能或第三方模块，lxml其XPath 1.0和XSLT 1.0的方法呢？
Below are my test runs for four method types on a simple, flat, element-centric XML input. All are set up for generalized parsing for any second level children of root and each method should yield exact same pandas dataframe. All but the last calls pd.Dataframe()on list of dictionaries. The XSLT method transforms XML to CSV for casted StringIO()in pd.read_csv().
下面是我在简单、扁平、以元素为中心的 XML 输入上对四种方法类型的测试运行。所有这些都设置为对 root 的任何二级子项进行广义解析，并且每种方法都应产生完全相同的 Pandas 数据帧。除了最后一个调用pd.Dataframe()字典列表之外的所有调用。XSLT 方法将 XML 转换为 CSV 以StringIO()在pd.read_csv().
Question(multi-part)
问题（多部分）
PERFORMANCE: How do you explain the slower
iterparseoften recommended for larger files as file is iteratively parsed? Is it partly due to theiflogic checks?MEMORY: Do CPU memory correlate with timings in I/O calls? XSLT and XPath 1.0 tend not to scale well with larger XML documents as entire file must be read in memory to be parsed.
STRATEGY: Is list of dictionaries an optimal strategy for
Dataframe()call? See these interesting answers: generatorversion and a iterwalk user-definedversion. Both upcast lists to dataframe.
性能：
iterparse当文件被迭代解析时，你如何解释通常建议用于较大文件的较慢？是否部分是由于if逻辑检查？内存：CPU 内存是否与 I/O 调用中的时间相关？XSLT 和 XPath 1.0 往往不能很好地适应较大的 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：字典列表是
Dataframe()呼叫的最佳策略吗？查看这些有趣的答案：生成器版本和iterwalk 用户定义版本。两个向上播列表到数据帧。
InputData (Stack Overflow's current top users by year of which our pandas friends are included)
输入数据（包括我们的Pandas朋友在内的Stack Overflow 当前顶级用户）
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PythonMethods
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>
</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>"</xsl:text>
</xsl:if>
<xsl:value-of select="." />
<xsl:choose>
<xsl:when test="position() != last()">
<xsl:text>","</xsl:text>
</xsl:when>
<xsl:otherwise>
<xsl:text>"
</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)))
Timings(with current XML and XML with 25 times the children (i.e., 900 StackOverflow user records)
计时（使用当前 XML 和带有 25 倍子级的 XML（即 900 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
回答by Jayen
PERFORMANCE: How do you explain the slower iterparse often recommended for larger files as file is iteratively parsed? Is it partly due to the if logic checks?
性能：您如何解释通常建议用于较大文件的较慢的 iterparse，因为文件是迭代解析的？是否部分是由于 if 逻辑检查？
I would assume that more python code would make it slower, as the python code is evaluated every time. Have you tried a JIT compiler like pypy?
我会假设更多的 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？
If I remove iand use first_tagonly, it seems to be quite a bit faster, so yes it is partly due to the if logic checks:
如果我只删除i和使用first_tag，它似乎要快得多，所以是的，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
I wasn't sure I understood the purpose of the last ifcheck, but I'm also not sure why you would want to lose whitespace-only elements. Removing the last ifconsistently shaves off a little bit of time:
我不确定我是否理解上次if检查的目的，但我也不确定您为什么要丢失纯空白元素。删除最后一个if始终如一地节省一点时间：
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
Now, with or without those performance improvements, your iterparse version seems to produce an extra-large dataframe. Here seems to be a working, fast version:
现在，无论有没有这些性能改进，您的 iterparse 版本似乎都会产生一个超大的数据帧。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
MEMORY: Do CPU memory correlate with timings in I/O calls? XSLT and XPath 1.0 tend not to scale well with larger XML documents as entire file must be read in memory to be parsed.
内存：CPU 内存是否与 I/O 调用中的时间相关？XSLT 和 XPath 1.0 往往不能很好地适应较大的 XML 文档，因为必须在内存中读取整个文件才能进行解析。
I'm not totally sure what you mean by "I/O calls" but if your document is small enough to fit in cache, then everything will be much faster as it won't evict many other items from the cache.
我不完全确定您所说的“I/O 调用”是什么意思，但如果您的文档小到足以放入缓存，那么一切都会快得多，因为它不会从缓存中驱逐许多其他项目。
STRATEGY: Is list of dictionaries an optimal strategy for Dataframe() call? See these interesting answers: generator version and a iterwalk user-defined version. Both upcast lists to dataframe.
策略：字典列表是 Dataframe() 调用的最佳策略吗？查看这些有趣的答案：生成器版本和 iterwalk 用户定义版本。两个向上播列表到数据帧。
The lists use less memory, so depending on how many columns you have, it could make a noticeable difference. Of course, this then requires your XML tags to be in a consistent order, which they do appear to be. The DataFrame()call would also need to do less work, as it doesn't have to lookup keys in the dict on every row, to figure out what column if for what value.
这些列表使用较少的内存，因此根据您拥有的列数，它可能会产生显着的差异。当然，这要求您的 XML 标记具有一致的顺序，它们看起来确实如此。该DataFrame()调用还需要做更少的工作，因为它不必在每一行的 dict 中查找键，以确定哪个列的值。
