如何用rowspan和colspan解析表

Question

首先，我阅读了Parsing a table with rowspan and colspan。我什至回答了这个问题。请在将其标记为重复之前阅读。

<table border="1">
  <tr>
    <th>A</th>
    <th>B</th>
  </tr>
  <tr>
    <td rowspan="2">C</td>
    <td rowspan="1">D</td>
  </tr>
  <tr>
    <td>E</td>
    <td>F</td>
  </tr>
  <tr>
    <td>G</td>
    <td>H</td>
  </tr>
</table>

它会像这样渲染

+---+---+---+
| A | B |   |
+---+---+   |
|   | D |   |
+ C +---+---+
|   | E | F |
+---+---+---+
| G | H |   |
+---+---+---+

<table border="1">
  <tr>
    <th>A</th>
    <th>B</th>
  </tr>
  <tr>
    <td rowspan="2">C</td>
    <td rowspan="2">D</td>
  </tr>
  <tr>
    <td>E</td>
    <td>F</td>
  </tr>
  <tr>
    <td>G</td>
    <td>H</td>
  </tr>
</table>

但是，这将呈现为这样。

+---+---+-------+
| A | B |       |
+---+---+-------+
|   |   |       |
| C | D +---+---+
|   |   | E | F |
+---+---+---+---+
| G | H |       |
+---+---+---+---+

我上一个答案的代码只能解析第一行中定义了所有列的表。

def table_to_2d(table_tag):
    rows = table_tag("tr")
    cols = rows[0](["td", "th"])
    table = [[None] * len(cols) for _ in range(len(rows))]
    for row_i, row in enumerate(rows):
        for col_i, col in enumerate(row(["td", "th"])):
            insert(table, row_i, col_i, col)
    return table


def insert(table, row, col, element):
    if row >= len(table) or col >= len(table[row]):
        return
    if table[row][col] is None:
        value = element.get_text()
        table[row][col] = value
        if element.has_attr("colspan"):
            span = int(element["colspan"])
            for i in range(1, span):
                table[row][col+i] = value
        if element.has_attr("rowspan"):
            span = int(element["rowspan"])
            for i in range(1, span):
                table[row+i][col] = value
    else:
        insert(table, row, col + 1, element)

soup = BeautifulSoup('''
    <table>
        <tr><th>1</th><th>2</th><th>5</th></tr>
        <tr><td rowspan="2">3</td><td colspan="2">4</td></tr>
        <tr><td>6</td><td>7</td></tr>
    </table>''', 'html.parser')
print(table_to_2d(soup.table))

我的问题是如何将表格解析为二维数组，该数组准确地表示它在浏览器中的呈现方式。或者有人可以解释一下浏览器如何呈现表格也可以。

Answer 1

您不能只计算td 或th 单元格，不。您必须对表进行扫描以获取每行的列数，并将前一行的任何活动行跨度添加到该计数中。

在different scenario parsing a table with rowspans 中，我跟踪了每个列号的行跨度计数，以确保来自不同单元格的数据最终位于正确的列中。这里可以使用类似的技术。

第一个计数列；只保留最高的数字。保留 2 或更大的行跨数列表，并为您处理的每一行列减去 1。这样你就知道每行有多少“额外”列。取最高的列数来构建您的输出矩阵。

接下来，再次循环遍历行和单元格，这一次在字典中跟踪从列号到活动计数的行跨度。再一次，将任何值为 2 或超过下一行的东西转移到下一行。然后移动列号以考虑任何活动的行跨度；如果在第 0 列上有活动的行跨度，则连续的第一个 td 实际上是第二个，等等。

您的代码将跨越的列和行的值重复复制到输出中；我通过在给定单元格的 colspan 和 rowspan 数字（每个默认为 1）上创建循环以多次复制该值来实现相同的目的。我忽略了重叠的单元格； HTML table specifications 表示重叠单元格是错误的，由用户代理来解决冲突。在下面的代码中 colspan 胜过 rowspan 单元格。

from itertools import product

def table_to_2d(table_tag):
    rowspans = []  # track pending rowspans
    rows = table_tag.find_all('tr')

    # first scan, see how many columns we need
    colcount = 0
    for r, row in enumerate(rows):
        cells = row.find_all(['td', 'th'], recursive=False)
        # count columns (including spanned).
        # add active rowspans from preceding rows
        # we *ignore* the colspan value on the last cell, to prevent
        # creating 'phantom' columns with no actual cells, only extended
        # colspans. This is achieved by hardcoding the last cell width as 1. 
        # a colspan of 0 means “fill until the end” but can really only apply
        # to the last cell; ignore it elsewhere. 
        colcount = max(
            colcount,
            sum(int(c.get('colspan', 1)) or 1 for c in cells[:-1]) + len(cells[-1:]) + len(rowspans))
        # update rowspan bookkeeping; 0 is a span to the bottom. 
        rowspans += [int(c.get('rowspan', 1)) or len(rows) - r for c in cells]
        rowspans = [s - 1 for s in rowspans if s > 1]

    # it doesn't matter if there are still rowspan numbers 'active'; no extra
    # rows to show in the table means the larger than 1 rowspan numbers in the
    # last table row are ignored.

    # build an empty matrix for all possible cells
    table = [[None] * colcount for row in rows]

    # fill matrix from row data
    rowspans = {}  # track pending rowspans, column number mapping to count
    for row, row_elem in enumerate(rows):
        span_offset = 0  # how many columns are skipped due to row and colspans 
        for col, cell in enumerate(row_elem.find_all(['td', 'th'], recursive=False)):
            # adjust for preceding row and colspans
            col += span_offset
            while rowspans.get(col, 0):
                span_offset += 1
                col += 1

            # fill table data
            rowspan = rowspans[col] = int(cell.get('rowspan', 1)) or len(rows) - row
            colspan = int(cell.get('colspan', 1)) or colcount - col
            # next column is offset by the colspan
            span_offset += colspan - 1
            value = cell.get_text()
            for drow, dcol in product(range(rowspan), range(colspan)):
                try:
                    table[row + drow][col + dcol] = value
                    rowspans[col + dcol] = rowspan
                except IndexError:
                    # rowspan or colspan outside the confines of the table
                    pass

        # update rowspan bookkeeping
        rowspans = {c: s - 1 for c, s in rowspans.items() if s > 1}

    return table

这会正确解析您的示例表：

>>> from pprint import pprint
>>> pprint(table_to_2d(soup.table), width=30)
[['1', '2', '5'],
 ['3', '4', '4'],
 ['3', '6', '7']]

并处理您的其他示例；第一张表：

>>> table1 = BeautifulSoup('''
... <table border="1">
...   <tr>
...     <th>A</th>
...     <th>B</th>
...   </tr>
...   <tr>
...     <td rowspan="2">C</td>
...     <td rowspan="1">D</td>
...   </tr>
...   <tr>
...     <td>E</td>
...     <td>F</td>
...   </tr>
...   <tr>
...     <td>G</td>
...     <td>H</td>
...   </tr>
... </table>''', 'html.parser')
>>> pprint(table_to_2d(table1.table), width=30)
[['A', 'B', None],
 ['C', 'D', None],
 ['C', 'E', 'F'],
 ['G', 'H', None]]

第二个：

>>> table2 = BeautifulSoup('''
... <table border="1">
...   <tr>
...     <th>A</th>
...     <th>B</th>
...   </tr>
...   <tr>
...     <td rowspan="2">C</td>
...     <td rowspan="2">D</td>
...   </tr>
...   <tr>
...     <td>E</td>
...     <td>F</td>
...   </tr>
...   <tr>
...     <td>G</td>
...     <td>H</td>
...   </tr>
... </table>
... ''', 'html.parser')
>>> pprint(table_to_2d(table2.table), width=30)
[['A', 'B', None, None],
 ['C', 'D', None, None],
 ['C', 'D', 'E', 'F'],
 ['G', 'H', None, None]]

最后但同样重要的是，代码正确处理超出实际表的跨度和"0" 跨度（延伸到末端），如下例所示：

<table border="1">
  <tr>
    <td rowspan="3">A</td>
    <td rowspan="0">B</td>
    <td>C</td>
    <td colspan="2">D</td>
  </tr>
  <tr>
    <td colspan="0">E</td>
  </tr>
</table>

有两行每行 4 个单元格，即使 rowspan 和 colspan 值会让您相信可能有 3 和 5：

+---+---+---+---+
|   |   | C | D |
| A | B +---+---+
|   |   |   E   |
+---+---+-------+

像浏览器一样处理这种跨越；它们被忽略，并且 0 跨度扩展到剩余的行或列：

>>> span_demo = BeautifulSoup('''
... <table border="1">
...   <tr>
...     <td rowspan="3">A</td>
...     <td rowspan="0">B</td>
...     <td>C</td>
...     <td colspan="2">D</td>
...   </tr>
...   <tr>
...     <td colspan="0">E</td>
...   </tr>
... </table>''', 'html.parser')
>>> pprint(table_to_2d(span_demo.table), width=30)
[['A', 'B', 'C', 'D'],
 ['A', 'B', 'E', 'E']]

Answer 2

需要注意的是，Martijn Pieters 解决方案不考虑同时具有 rowspan 和 colspan 属性的单元格的情况。 例如

<table border="1">
    <tr>
        <td rowspan="3" colspan="3">A</td>
        <td>B</td>
        <td>C</td>
        <td>D</td>
    </tr>
    <tr>
        <td colspan="3">E</td>
    </tr>
    <tr>
        <td colspan="1">E</td>
        <td>C</td>
        <td>C</td>
    </tr>
    <tr>
        <td colspan="1">E</td>
        <td>C</td>
        <td>C</td>
        <td>C</td>
        <td>C</td>
        <td>C</td>
    </tr>
</table>

此表呈现给

+-----------+---+---+---+
| A         | B | C | D |
|           +---+---+---+
|           | E         |
|           +---+---+---+
|           | E | C | C |
+---+---+---+---+---+---+
| E | C | C | C | C | C |
+---+---+---+---+---+---+

但是如果我们应用我们得到的函数

[['A', 'A', 'A', 'B', 'C', 'D'],
 ['A', 'E', 'E', 'E', None, None],
 ['A', 'E', 'C', 'C', None, None],
 ['E', 'C', 'C', 'C', 'C', 'C']]

可能存在一些边缘情况，但将行跨度簿记扩展到行跨度和列跨度的product 中的单元格，即

   for drow, dcol in product(range(rowspan), range(colspan)):
            try:
                table[row + drow][col + dcol] = value
                rowspans[col + dcol] = rowspan
            except IndexError:
                # rowspan or colspan outside the confines of the table
                pass

似乎在这个线程中的例子上工作，对于上面的表格它会输出

[['A', 'A', 'A', 'B', 'C', 'D'],
 ['A', 'A', 'A', 'E', 'E', 'E'],
 ['A', 'A', 'A', 'E', 'C', 'C'],
 ['E', 'C', 'C', 'C', 'C', 'C']]

Answer 3

使用通常的遍历方法，只需将解析器类型更改为lxml即可。

soup = BeautifulSoup(resp.text, "lxml")

现在用通常的方式解析它。