擅长:python、mysql、java
<p>下面是一个使用<code>collections.defaultdict</code>的O(n)解:</p>
<pre><code>from collections import defaultdict
myD = {'key1': {'x' : 123, 'y' : 432},
'key2': {'x' : 456, 'y' : 565},
'key3': {'x' : 789, 'y' : 420}}
# initialise defaultdict of lists
d = defaultdict(list)
# iterate input dictionary and add values to lists
for v1 in myD.values():
for k2, v2 in v1.items():
d[k2].append(v2)
# calculate minimum
res = {k: min(v) for k, v in d.items()}
print(res)
{'x': 123, 'y': 420}
</code></pre>