擅长:python、mysql、java
<p>您可以尝试@LevLevitsky演示的<code>defaultdict</code>,也可以尝试这种可行但复杂的方法:</p>
<pre><code>result = {}
for i in liste:
if i[0] not in result.keys():
result[i[0]] = {i[1]: {i[2]: tuple(i[3:4])}}
else:
if i[1] not in result[i[0]].keys():
result[i[0]][i[1]] = {i[2]: tuple(i[3:4])}
else:
if i[2] not in result[i[0]][i[1]].keys():
result[i[0]][i[1]][i[2]] = tuple(i[3:4])
else:
result[i[0]][i[1]][i[2]]+=tuple(i[3:4])
print result #{'JACKET': {'MOD-2': {'BLACK': ('L', 'XL')}, 'MOD-1': {'GREEN': ('S', 'M')}}, 'VEST': {'MODEL-1': {'BEIGE': ('M', 'S')}}}
</code></pre>