在Python中遍历XML时如何采用前面的元素? [英] How to take preceding element when iterating over XML in Python?

问题描述

我有一个这样的XML:

I have an XML structured like this:

 <?xml version="1.0" encoding="utf-8"?>
<pages>
    <page id="1" bbox="0.000,0.000,462.047,680.315" rotate="0">
        <textbox id="0" bbox="179.739,592.028,261.007,604.510">
            <textline bbox="179.739,592.028,261.007,604.510">
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">C</text>
                <text font="NUMPTY+ImprintMTnum-it"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.333">A</text>
                <text font="NUMPTY+ImprintMTnum-it"  bbox="192.745,592.218,199.339,603.578" ncolour="0" size="12.333">P</text>
                <text font="NUMPTY+ImprintMTnum-it"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.333">I</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">T</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">O</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">L</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">O</text>
                <text></text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">I</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">I</text>
                <text font="NUMPTY+ImprintMTnum"  bbox="191.745,592.218,199.339,603.578" ncolour="0" size="12.482">I</text>
                <text></text>
            </textline>
        </textbox>
    </page>
</pages>

文本标记中的属性bbox具有四个值，我需要确定元素的第一个bbox值与其前一个值的差.换句话说，前两个bbox之间的距离是1.

Attribute bbox in text tag has four values, and I need to have the difference of the first bbox value of an element and its preceding one. In other words, the distance between the first two bboxes is 1.

到目前为止，我的代码是:

So far my code is:

def wrap(line, idxList):
    if len(idxList) == 0:
        return    # No elements to wrap
    # Take the first element from the original location
    idx = idxList.pop(0)     # Index of the first element
    elem = removeByIdx(line, idx) # The indicated element
    # Create "newline" element with "elem" inside
    nElem = E.newline(elem)
    line.insert(idx, nElem)  # Put it in place of "elem"
    while len(idxList) > 0:  # Process the rest of index list
        # Value not used, but must be removed
        idxList.pop(0)
        # Remove the current element from the original location
        currElem = removeByIdx(line, idx + 1)
        nElem.append(currElem)  # Append it to "newline"

for line in root.iter('textline'):
    idxList = []
    for elem in line:
        bbox = elem.attrib.get('bbox')
        if bbox is not None:
            tbl = bbox.split(',')

            distance = float(tbl[2]) - float(tbl[0])
        else:
            distance = 100  # "Too big" value
        if distance > 10:
            par = elem.getparent()
            idx = par.index(elem)
            idxList.append(idx)
        else:  # "Wrong" element, wrap elements "gathered" so far
            wrap(line, idxList)
            idxList = []
    # Process "good" elements without any "bad" after them, if any
    wrap(line, idxList)

但是感兴趣的问题是:

for line in root.iter('textline'):
idxList = []
for elem in line:
    bbox = elem.attrib.get('bbox')
    if bbox is not None:
        tbl = bbox.split(',')

        distance = float(tbl[2]) - float(tbl[0])

我尝试了很多，但真的不知道该怎么做.

I tried a lot and really don't know how to do it.

推荐答案

如果我完全了解您的需求，则希望选择符合以下条件的文本节点:

If I fully understand your needs, you want to select text nodes which respect the following condition :

文本节点的bbox值-前面的文本节点的bbox值不大于10.

bbox value of the text node - bbox value of the preceding text nodes not greater than 10.

您可以尝试使用XSL和XPath.首先是XSL代码(在下一步将bbox值与XPath进行比较的强制步骤):

You could try with XSL and XPath. First the XSL code (mandatory step to compare bbox value with XPath in the next step) :

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="no" indent="yes"/>

<xsl:template match="@bbox">
  <xsl:attribute name="{name()}">
  <xsl:value-of select="substring(.,1,3)" />
  </xsl:attribute>
</xsl:template>

<xsl:template match="@font">
  <xsl:attribute name="{name()}">
  <xsl:text>NUMPTY+ImprintMTnum</xsl:text>
  </xsl:attribute>
</xsl:template>

<xsl:template match="*[not(node())]"/> 
<xsl:strip-space  elements="*"/>

 <xsl:template match="@*|node()">
  <xsl:copy>
  <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
 </xsl:template>

</xsl:stylesheet>

然后:

import lxml.etree as IP

xml = IP.parse(xml_filename)
xslt = IP .parse(xsl_filename)
transform = IP.XSLT(xslt)

然后请求:

tree = IP.parse(transform)
for nodes in tree.xpath("//text[@bbox<preceding::text[1]/@bbox+11]"):
    print(nodes)

用//text[@bbox>preceding::text[1]/@bbox]替换//text[@bbox<preceding::text[1]/@bbox+11]以测试示例数据(将选择bbox值大于先前文本bbox值的文本节点).

Replace //text[@bbox<preceding::text[1]/@bbox+11] with //text[@bbox>preceding::text[1]/@bbox] to test with your sample data (will select text nodes with greater bbox value than the preceding text bbox value).

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