擅长:python、mysql、java
<p>您可以使用<code>itertools.groupby</code>:</p>
<pre><code>from itertools import groupby as gb
myDict = {0: {37.0: {'C23H27O9': 'C22H23O10'}}, 1: {74.0: {'C23H27O9': 'C21H19O11'}}, 2: {111.0: {'C23H27O9': 'C20H15O12'}}, 3: {148.0: {'C23H27O9': 'C19H11O13'}}, 4: {37.0: {'C22H23O10': 'C21H19O11'}}, 5: {74.0: {'C22H23O10': 'C20H15O12'}}, 6: {111.0: {'C22H23O10': 'C19H11O13'}}, 7: {37.0: {'C21H19O11': 'C20H15O12'}}, 8: {74.0: {'C21H19O11': 'C19H11O13'}}, 9: {37.0: {'C20H15O12': 'C19H11O13'}}}
d = sorted([(a, [j for k in b.items() for j in k]) for i in myDict.values() for a, b in i.items()], key=lambda x:x[0])
r = [(a, [*{i for _, j in b for i in j}]) for a, b in gb(d, key=lambda x:x[0])]
</code></pre>
<p>输出:</p>
<pre><code>[(37.0, ['C19H11O13', 'C20H15O12', 'C21H19O11', 'C22H23O10', 'C23H27O9']), (74.0, ['C19H11O13', 'C20H15O12', 'C21H19O11', 'C22H23O10', 'C23H27O9']), (111.0, ['C20H15O12', 'C19H11O13', 'C23H27O9', 'C22H23O10']), (148.0, ['C23H27O9', 'C19H11O13'])]
</code></pre>