2 回答

TA貢獻(xiàn)1831條經(jīng)驗(yàn) 獲得超9個(gè)贊
將此添加到我的觀點(diǎn)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 %}
解決了我的問(wèn)題。

TA貢獻(xiàn)1804條經(jīng)驗(yàn) 獲得超2個(gè)贊
據(jù)我了解您的問(wèn)題以及 dict_lyrics 的進(jìn)展情況,如果您想以這種形式顯示問(wèn)題 0 的答案 0 和問(wèn)題 1 的答案 1:
word1 - word3
word2 - word4
你可以這樣做:
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舉行['word1', 'word3']和second_qa舉行['word2', 'word4']
以更一般的方式,例如,這個(gè) dict_lyricsdict_lyrics = {'question': ['word1', 'word2', 'wordX'], 'answer': ['word3', 'word4', 'wordY']}
并假設(shè)len(dict_lyrics["question"]) == len(dict_lyrics["answer"])
for i in range(len(li["question"])):
print(f"{li['question'][i]} {li['answer'][i]}")
應(yīng)該做的伎倆:
word1 word3
word2 word4
wordX wordY
編輯:發(fā)表評(píng)論后,我猜您正在尋找某種 HTML 輸出。我會(huì)像這樣更改您在問(wèn)題中顯示的循環(huán):
{% 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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