首先,我已阅读 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>

它会呈现像
+---+---+---+
| 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 |       |
+---+---+---+---+

我上一个答案中的代码只能解析在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解析为一个二维数组,该数组表示 恰好 它如何在浏览器中呈现。或者有人可以解释浏览器如何呈现表格也很好。

最佳答案

你不能只计算 tdth 单元格,不。您必须扫描整个表格以获取每行的列数,并将前一行的任何事件行跨度添加到该计数中。

different scenario parsing a table with rowspans 中,我跟踪了每列编号的行跨度计数,以确保来自不同单元格的数据最终出现在正确的列中。这里可以使用类似的技术。

首先计数列;只保留最高的数字。保留 2 个或更大的行跨度数列表,并为您处理的每一行列从每个行数中减去 1。这样你就知道每行有多少“额外”列。取最高的列数来构建输出矩阵。

接下来,再次循环遍历行和单元格,这次跟踪字典中从列号到事件计数的映射的行跨度。再次,将任何值为 2 或转移到下一行的值。然后移动列号以考虑任何事件的行跨度;如果第 0 列上有事件的行跨度等,则行中的第一个 td 实际上是第二个。

您的代码将跨列和跨行的值重复复制到输出中;我通过在给定单元格的 colspanrowspan 数字(每个默认为 1)上创建一个循环来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 声明重叠单元格是一个错误,由用户代理解决冲突。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

这将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]]

最后但并非最不重要的是,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 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,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']]

关于python - 如何使用 rowspan 和 colspan 解析表,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48393253/

10-12 18:29