<p>另一个答案不处理<code>needle</code>是<code>in</code>一个字符串，而<code>in</code>是一个列表的情况。也就是说，另一个答案只在<code>needle</code>等于该值或该值直接包含时处理。因此，另一种解决方案在问题中包含的测试用例中不起作用。你知道吗</p>
<p>一般来说，可以递归地检查值是否包含<code>needle</code>。你知道吗</p>
<pre><code>from typing import Any
from typing import Dict
from typing import Iterable
def keys_containing_value(d: Dict, needle: Any):
def recursively_in(x: Any, n: Any):
if isinstance(x, str) and isinstance(n, str):
return n in x
if type(x) is type(n):
return x == n
if isinstance(x, Iterable) and not isinstance(x, str):
return any(recursively_in(item, n) for item in x)
return [k for k, v in d.items() if recursively_in(v, needle)]
</code></pre>
<p>下面是该函数如何使用您的数据和测试用例：</p>
<pre><code>mydict = {
'romeo': 16,
'juliette': 16,
'zena': 'red',
'gabby': 'red',
'lauren': 'smart',
'betty': 'smart',
'kristen': 'sweet',
'lisa': 'sweet',
'lucy': ['smart', 'sweet'],
'shawna': 'green field',
'terry': 'the green field is on fire',
'dorothy': 'shelly plays on the green field',
'jenny': ['intelligent', 'cuts the smart green field grass'],
}
needles = [
16,
'red',
'smart',
'sweet',
'green field',
]
for needle in needles:
print(keys_containing_value(mydict, needle))
</code></pre>
<p><strong>输出：</strong></p>
<pre><code>['romeo', 'juliette']
['zena', 'gabby']
['lauren', 'betty', 'lucy', 'jenny']
['kristen', 'lisa', 'lucy']
['shawna', 'terry', 'dorothy', 'jenny']
</code></pre>
