<p>您可以使用<a href="https://docs.python.org/3.7/library/stdtypes.html#str.startswith" rel="nofollow noreferrer">startswith</a>:</p>
<pre><code>selected_ipc = ['H01L']
df = ['F24J3/02 ', 'G01N31/10 ', 'H01L27/14 ', 'G21H1/10 ', 'H01L21/36 ']
for item in selected_ipc:
for item1 in df:
if item1.startswith(item):
print(item1)
else:
print("No match")
</code></pre>
<p><strong>输出</strong></p>
<pre><code>No match
No match
H01L27/14
No match
H01L21/36
</code></pre>
<p><strong>更新</p>
<p>对于嵌套列表,可以使用<a href="https://docs.python.org/3/tutorial/datastructures.html#list-comprehensions" rel="nofollow noreferrer">list comprehension</a>:</p>
<pre><code>selected_ipc = ['H01L']
df = [['F24J3/02 ', 'A123'], ['G01N31/10 ', 'A124'], ['H01L27/14 ', 'A125'], ['G21H1/10 ', 'A126'],
['H01L21/36 ', 'A127']]
result = [lst for lst in df if any(lst[0].startswith(e) for e in selected_ipc)]
print(result)
</code></pre>
<p><strong>输出</strong></p>
<pre><code>[['H01L27/14 ', 'A125'], ['H01L21/36 ', 'A127']]
</code></pre>
<p>作为一种替代方法,您可以使用带有两个循环的<em>pythonic</em>方式:</p>
<pre><code>selected_ipc = ['H01L']
df = [['F24J3/02 ', 'A123'], ['G01N31/10 ', 'A124'], ['H01L27/14 ', 'A125'], ['G21H1/10 ', 'A126'],
['H01L21/36 ', 'A127']]
result = []
for lst in df:
found = False
for e in selected_ipc:
if lst[0].startswith(e):
found = True
result.append(lst)
break
if not found:
print("No match")
print(result)
</code></pre>
<p><strong>输出</strong></p>
<pre><code>No match
No match
No match
[['H01L27/14 ', 'A125'], ['H01L21/36 ', 'A127']]
</code></pre>