Question

我正在尝试将两个列表与同一个字典键相关联；然后在html表中显示列表。我不知道是否选择了一种错误的方法来存储列表，或者是否弄乱了显示列表所需的for循环的配置。想知道是否有人可以提供帮助。

到目前为止我所拥有的：

我想联系的是：KeyA 两个列表：ListA和ListB。

存储：

listA = [1,2,3,4]
listB = [A,B]
keys = {}
lists = []
lists.append([ListA,ListB])
keys.setdefault(KeyA, []).append(lists)

显示：

{% for option in keys %}
    <tr class = 'collapse level1' data-depth="1" >
        <td class ='site_device'> </td>
        <td class = 'site_device' > {{option}}</td>
        <td class ='site_device'></td>
        <td class ='site_device'></td>

        {% for list_holder in keys[option] %}
            {% for lists in list_holder %}
                {% for v1 in lists %}
                    <tr class = 'collapse level2' data-depth="2" >
                        <td class ='site_device'></td>
                        <td class ='site_device'></td>
                        <td class ='site_device'>{{ v1 }}</td>
                        <td class ='site_device'></td>
                    </tr>
                {% endfor %}
            {% endfor %}
        {% endfor %}
    </tr>
{% endfor %}

此方法给出：

<tr class = 'collapse level1' data-depth="1" >
    <td class ='site_device'> </td>
    <td class = 'site_device' > KeyA</td>
    <td class ='site_device'></td>
    <td class ='site_device'></td>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'>[1, 2, 3, 4]</td>
        <td class ='site_device'></td>
    </tr>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'>['A','B']</td>
        <td class ='site_device'></td>
    </tr>
</tr>

我要针对的目标：

<tr class = 'collapse level1' data-depth="1" >
    <td class ='site_device'> </td>
    <td class = 'site_device' > KeyA</td>
    <td class ='site_device'></td>
    <td class ='site_device'></td>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'> 1 </td>
        <td class ='site_device'> A </td>
    </tr>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'> 2 </td>
        <td class ='site_device'> B </td>
    </tr>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'> 3 </td>
        <td class ='site_device'></td>
    </tr>
    <tr class = 'collapse level2' data-depth="2" >
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        <td class ='site_device'> 4 </td>
        <td class ='site_device'></td>
    </tr>
</tr>

Answer 1

append向列表中添加了一个新元素，因此最终得到一个三级列表（lists是[[[1, 2, 3, 4], ['A', 'B']]]而不是[[1, 2, 3, 4], ['A', 'B']]）

尝试使用extend代替：

lists.extend([ListA,ListB])

（或选择一一追加）

lists.append(ListA)
lists.append(ListB)

Answer 2

我没有测试它，但是您将列表嵌入了一个级别。

因此，替换

lists.append([listA,listB])

通过

lists.append(listA)
lists.append(listB)

还有什么，为什么要使用它？：

keys.setdefault(KeyA, []).append(lists)

代替

keys.setdefault(KeyA, lists)
# or even
keys['KeyA']=lists

更容易阅读。

对于另一个问题，可能会出现类似这样的循环：


{% for option in keys %}
    <tr class = 'collapse level1' data-depth="1" >
        <td class ='site_device'> </td>
        <td class = 'site_device' > {{option}}</td>
        <td class ='site_device'></td>
        <td class ='site_device'></td>
        {% for list_holder in keys[option] %}
            {% for i in range(list_holder[0]|length) %}
                    <tr class = 'collapse level2' data-depth="2" >
                        <td class ='site_device'></td>
                        <td class ='site_device'></td>
                        <td class ='site_device'>{{ list_holder[0][i] }}</td>
                        <td class ='site_device'>{{ list_holder[1][i] }}</td>
                    </tr>
            {% endfor %}
        {% endfor %}
    {% endfor %}

当然，由于两个列表的大小都不相同，因此会引发IndexError（s）。因此，请确保两个列表的大小相同。

将两个列表与字典键相关

2 个答案: