Question

我有一个python dict，其中每个键对应一个标题，与每个标题关联的列表包含任意数量的值：

data = { 
    "heading1": ['h1-val1', 'h1-val2', 'h1-val3', ],
    "heading2": ['h2-val1', ],
    "heading3": ['h3-val1', 'h3-val2', 'h3-val3', 'h3-val4', ],
}

我需要在Django模板中将其渲染为表格，其中值在每个标题下方垂直列出，任何缺失值都呈现为空表格单元格：

<table>
<thead>
    <tr>
    <th>heading1</th>
    <th>heading2</th>
    <th>heading3</th>
    </tr>
</thead>
<tbody>
    <tr>
    <td>h1-val1</td>
    <td>h2-val1</td>
    <td>h3-val1</td>
    </tr>
    <tr>
    <td>h1-val2</td>
    <td></td>
    <td>h3-val2</td>
    </tr>
    <tr>
    <td>h1-val3</td>
    <td></td>
    <td>h3-val3</td>
    </tr>
    <tr>
    <td></td>
    <td></td>
    <td>h3-val4</td>
    </tr>
</tbody>
</table>

实现这一目标的最佳方式是什么？

我的第一个倾向是将原始字典重新排列成2D矩阵，然后将其传递到模板中。我确信我不是第一个遇到这种问题的人，我很好奇其他人是如何解决这个问题的。

更新：仅供参考，这是我对此问题的原始解决方案（我对此并不满意）。

# Using the data dict from the question:
size = max(len(data['heading1']), len(data['heading2']), len(data['heading3']))
matrix = [[None, None, None] for i in range(size)] # initialize an empty matrix

# manually copy the data into the appropriate column :(
i = 0
for item in data['heading1']:
    matrix[i][0] = item
    i += 1
i = 0
for item in data['heading2']:
    matrix[i][1] = item
    i += 1
i = 0
for item in data['heading3']:
    matrix[i][2] = item
    i += 1

然后我将矩阵传递到模板中，如下所示：

<table>
<thead><tr>
    <th>heading1</th>
    <th>heading2</th>
    <th>heading3</th>
</tr></thead>
<tbody>
{% for row in matrix %}
    <tr>
    {% for col in row %}
        <td>{% if col %}{{ col }}{% else %}&nbsp;{% endif %}</td>
    {% endfor %}
    </tr>
{% endfor %}
</tbody>
</table>

Answer 1

如果我们稍微更改一下游戏，那么实际上很容易转过来（只要你的列表没有填充......）

from django.template import Context, Template

data = {
    "heading1": ['h1-val1', 'h1-val2', 'h1-val3', ],
    "heading2": ['h2-val1', ],
    "heading3": ['h3-val1', 'h3-val2', 'h3-val3', 'h3-val4', ],
}

# we'll need to split the headings from the data
# rather than using keys() I'm just hard coding so I can control the order
headings = ["heading1", "heading2", "heading3"]

columns = [data[heading] for heading in headings]

# get the length of the longest column
max_len = len(max(columns, key=len))

for col in columns:
    # padding the short columns with None
    col += [None,] * (max_len - len(col))

# Then rotate the structure...
rows = [[col[i] for col in columns] for i in range(max_len)]


dj_template ="""
<table>
{# headings #}
    <tr>
    {% for heading in headings %}
        <th>{{ heading }}</th>
    {% endfor %}
    </tr>
{# data #}
{% for row in data %}
    <tr>
        {% for val in row %}
        <td>{{ val|default:'' }}</td>
        {% endfor %}
    </tr>
{% endfor %}
</table>
"""

# finally, the code I used to render the template:
tmpl = Template(dj_template)
tmpl.render(Context(dict(data=rows, headings=headings)))

对我来说，这会产生以下内容（空白行被删除）：

<table>
    <tr>
        <th>heading1</th>
        <th>heading2</th>
        <th>heading3</th>
    </tr>
    <tr>
        <td>h1-val1</td>
        <td>h2-val1</td>
        <td>h3-val1</td>
    </tr>
    <tr>
        <td>h1-val2</td>
        <td></td>
        <td>h3-val2</td>
    </tr>
    <tr>
        <td>h1-val3</td>
        <td></td>
        <td>h3-val3</td>
    </tr>
    <tr>
        <td></td>
        <td></td>
        <td>h3-val4</td>
    </tr>
</table>

Answer 2

Kenneth Reitz建议你也可以使用tablib来解决这个问题，所以我想我也会在这里加入：

import tablib

d = tablib.Dataset()
d.append_col(['h1-val1', 'h1-val2', 'h1-val3', ''], header="heading1")
d.append_col(['h2-val1', 'h2-val2', '', ''], header="heading2")  
d.append_col(['h3-val1', 'h3-val2', 'h3-val3', 'h3-val4', ], header="heading3") 
d.headers = ['heading1', 'heading2', 'heading3']

这会转储数据集中的所有相关数据，然后您可以在模板中使用以下内容进行渲染：

{{ d.html }}

哪个会生成如下所示的html：

<table>
<thead>
<tr><th>heading1</th>
<th>heading2</th>
<th>heading3</th></tr>
</thead>
<tr><td>h1-val1</td>
<td>h2-val1</td>
<td>h3-val1</td></tr>
<tr><td>h1-val2</td>
<td>h2-val2</td>
<td>h3-val2</td></tr>
<tr><td>h1-val3</td>
<td></td>
<td>h3-val3</td></tr>
<tr><td></td>
<td></td>
<td>h3-val4</td></tr>
</table>

Answer 3

尝试以下代码。请注意，由于您对数据使用关联数组，因此无法保证标题的显示顺序。

print "<table>"

print "<thead><tr>"
order = []
for k in data.keys():
    print "<td>" + k + "</td>"
    order.append(k)
print "</tr></thead>"

print "<tbody>"
for k in order:
    print "<tr>"
    for v in data[k]:
        print "<td>" + v + "</td>"
    print "</tr>"
print "</tbody>"
print "</table>"

如何将python dict呈现为具有垂直列的html表

3 个答案: