如何在Python中使用带有xPath的XML的同级兄弟？

Question

我有这样的XML结构：

<?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="192.745,592.218,199.339,603.578" ncolour="0" size="12.333">A</text>
                <text font="NUMPTY+ImprintMTnum-it"  bbox="193.745,592.218,199.339,603.578" 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>

文本标签中的属性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之间的距离为1。在下面的循环中，我需要找到我所使用的bbox属性值的前面同级，以便计算两个之间的距离。

   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)

    #print(etree.tostring(root, encoding='unicode', pretty_print=True))

我尝试使用xPath这样：

for x in tree.xpath("//text[@bbox<preceding::text[1]/@bbox+11]"):
print(x)

但是它什么也不返回。我的路径是否错误，如何将其插入循环中？

Answer 1

您的代码失败的原因是，与先前同级有关的轴名称是

preceding-sibling（不是preceding）。

但是在这里您不需要使用

XPath

Answer 2

//text[@bbox < preceding::text[1]/@bbox + 11]

您正在对@bbox值执行数字微分和数字加法。

但是@bbox不是数字，它是四个数字的逗号分隔列表：