擅长:python、mysql、java
<p>有很多方法可以解决这个问题:</p>
<p>一:</p>
<pre><code>a = ['A121', 'A221', 'A321', 'A421', 'A521', 'A621', 'A721', 'A821', 'A921']
INDENT = ' '*10
NUMBER_OF_COLUMNS = 5
lead = INDENT
for item_number, item in enumerate(a):
print("{lead}{item}".format(lead=lead, item=item), end='')
if item_number % NUMBER_OF_COLUMNS == NUMBER_OF_COLUMNS - 1:
lead = ",\n" + INDENT
else:
lead = ", "
</code></pre>
<p>二:</p>
<pre><code>table = []
row = []
for item in a:
row.append(item)
if len(row) == NUMBER_OF_COLUMNS:
table.append(INDENT + ', '.join(row))
row = []
table.append(INDENT + ', '.join(row))
print(',\n'.join(table))
</code></pre>
<p>三,一行:</p>
<pre><code>print(',\n'.join(INDENT + ', '.join(a[i:i + NUMBER_OR_COLUMNS]) for i in range(0, len(a), NUMBER_OR_COLUMNS)))
</code></pre>
<p>或者也许text包装.填充()将起作用:</p>
<pre><code>from textwrap import fill
print(fill(', '.join(a), initial_indent=10*' ', subsequent_indent=10*' ', width=40))
</code></pre>
<p>所有输出(示例中带有逗号a):</p>
<pre><code> A121, A221, A321, A421, A521,
A621, A721, A821, A921
</code></pre>