<p>我的解决方案是列表列表,但很简单。你知道吗</p>
<pre><code>data = ["15674 24000 Manager Gregory the 1st John", "15674 24000 Manager Gregory the 1st John",
"15674 24000 Manager Gregory the 1st John", "15674 24000 Manager Gregory the 1st John"]
data1 = []
for items in data:
splitNames = items.split()
number, salary, position, first, des1, des2, last = splitNames
data1.append([last, first + ' ' + des1 + ' ' + des2, number, position, salary])
for items in data1:
print items
</code></pre>
<p><strong>结果:</strong></p>
<pre><code>['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']
['John', 'Gregory the 1st', '15674', 'Manager', '24000']
</code></pre>
<p>这不是最好的解决方案,我是这样写的,所以很容易理解,因为这是基本的操作。你知道吗</p>
<p><strong>元组:</strong></p>
<p>如果您想让它成为元组列表,只需重写这一行:</p>
<pre><code>data1.append(tuple([last, first + ' ' + des1 + ' ' + des2, number, position, salary]))
</code></pre>
<p>通过使用tuple(),请参阅python文档中的解释<a href="http://docs.python.org/2/library/functions.html#tuple" rel="nofollow">here</a>。你知道吗</p>
<p><strong>结果:</strong></p>
<pre><code>('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')
('John', 'Gregory the 1st', '15674', 'Manager', '24000')
</code></pre>
<p><strong>适应您的代码:</strong></p>
<pre><code>for items in line:
breakup= items.split()
number, salary, position, first, des1, des2, last = breakup
data.append(tuple([last, first + ' ' + des1 + ' ' + des2, number, position, salary]))
print data
</code></pre>
<p><strong>结果:</strong></p>
<pre><code>[('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000'), ('John', 'Gregory the 1st', '15674', 'Manager', '24000')]
</code></pre>