2 回答
![?](http://img1.sycdn.imooc.com/5458657e000125a302200220-100-100.jpg)
TA贡献1831条经验 获得超9个赞
将此添加到我的观点中:
context['combined'] = list(zip(dict_lyrics['question'], dict_lyrics['answer']))
然后在模板中使用它:
{% for i, j in combined %}
{{ i }} -- {{ j }}
{% endfor %}
解决了我的问题。
![?](http://img1.sycdn.imooc.com/5458478b0001f01502200220-100-100.jpg)
TA贡献1804条经验 获得超2个赞
据我了解您的问题以及 dict_lyrics 的进展情况,如果您想以这种形式显示问题 0 的答案 0 和问题 1 的答案 1:
word1 - word3
word2 - word4
你可以这样做:
answers = [v for subl in li.values() for v in subl]
frist_qa = answers[0::2]
second_qa = answers[1::2]
print(f"{' '.join(first_qa)}")
print(f"{' '.join(second_qa)}")
在哪里first_qa举行['word1', 'word3']和second_qa举行['word2', 'word4']
以更一般的方式,例如,这个 dict_lyricsdict_lyrics = {'question': ['word1', 'word2', 'wordX'], 'answer': ['word3', 'word4', 'wordY']}
并假设len(dict_lyrics["question"]) == len(dict_lyrics["answer"])
for i in range(len(li["question"])):
print(f"{li['question'][i]} {li['answer'][i]}")
应该做的伎俩:
word1 word3
word2 word4
wordX wordY
编辑:发表评论后,我猜您正在寻找某种 HTML 输出。我会像这样更改您在问题中显示的循环:
{% for i in range(len(li["question"])): %}
<tr>
<td> Key: {{ li['question'][i] }} </td>
<td> Value: {{ li['answer'][i] }} </td>
</tr>
{% endfor %}
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