<pre><code>l = [ [ 1, "EEBC92F4-DA8C-4B58-8730-3119F6B1C045", 1, 1386882230, "399231", 1386882230, "399231", "{\n}", "2010", "NON-HISPANIC BLACK", "MALE", "HUMAN IMMUNODEFICIENCY VIRUS DISEASE", "297", "5" ],
[ 2, "84C91A4A-19E2-4AD2-9493-17B84707CA4E", 2, 1386882230, "399231", 1386882230, "399231", "{\n}", "2012", "NON-HISPANIC BLACK", "MALE", "INFLUENZA AND PNEUMONIA", "2011", "3" ]]
#get index of the year
l[0].index('2010') #8
l[1].index('2012') #8
</code></pre>
<p>如果您只想打印,您可以:</p>
^{pr2}$
<p>这将给出以下输出:</p>
<pre><code>2010
NON-HISPANIC BLACK
MALE
HUMAN IMMUNODEFICIENCY VIRUS DISEASE
297
5
2012
NON-HISPANIC BLACK
MALE
INFLUENZA AND PNEUMONIA
2011
3
</code></pre>
<p>不完全确定这是否是你想要的,但是你写了你需要在相应年份之后的所有信息(“我指的是每一行的年份之后的所有信息”,你在评论中对James的回复),这样你就可以首先提取这些元素并将它们存储在一个新的列表中(以防你想做其他的事情)除印刷外):</p>
<pre><code>lmod = [x[8:] for x in l]
</code></pre>
<p><code>lmod</code>然后如下所示:</p>
<pre><code>[['2010',
'NON-HISPANIC BLACK',
'MALE',
'HUMAN IMMUNODEFICIENCY VIRUS DISEASE',
'297',
'5'],
['2012',
'NON-HISPANIC BLACK',
'MALE',
'INFLUENZA AND PNEUMONIA',
'2011',
'3']]
</code></pre>
<p>现在可以像您那样打印:</p>
<pre><code>for sl in lmod:
for el in sl:
print el
</code></pre>
<p>输出:</p>
<pre><code>2010
NON-HISPANIC BLACK
MALE
HUMAN IMMUNODEFICIENCY VIRUS DISEASE
297
5
2012
NON-HISPANIC BLACK
MALE
INFLUENZA AND PNEUMONIA
2011
3
</code></pre>
<p>这就是你要找的吗?在</p>