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<p>这是我的方法。我很难把整本字典还回去</p>
<pre><code>def get_col(amount):
letter = 0
value = []
values = {}
for i in range(amount):
letter = get_column_letter(i + 1)
[value.append(row.value) for row in ws[letter]]
values = dict(zip(letter, [value]))
value = []
return values
</code></pre>
<p>我希望它像这样输出:</p>
<pre><code>{'A': ['ID', 'value is 1', 'value is 2', 'value is 3', 'value is 4', 'value is 5', 'value is 6']}
{'B': ['Name', 'value is 1', 'value is 2', 'value is 3', 'value is 4', 'value is 5', 'value is 6']}
{'C': ['Math', 'value is 1', 'value is 2', 'value is 3', 'value is 4', 'value is 5', 'value is 6']}
</code></pre>
<p>但当返回位于“for”一侧时,它只返回</p>
<pre><code>{'A': ['ID', 'value is 1', 'value is 2', 'value is 3', 'value is 4', 'value is 5', 'value is 6']}
</code></pre>
<p>当返回在“for”循环之外时,它返回</p>
<pre><code>{'C': ['Math', 'value is 1', 'value is 2', 'value is 3', 'value is 4', 'value is 5', 'value is 6']}
</code></pre>
<p>任何帮助都将不胜感激。谢谢大家!</p>