如何将列表列表转换为相对频率字典？

Question

我有一个列表列表，我想将其转换为相对频率字典。

列表列表：

[['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
 ['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
 ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
 ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d', 'SBS38'],
 ['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d']]

列表列表的相对频率字典

{['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d']: 0.6,
 ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d']: 0.2,
 ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d', 'SBS38']: 0.2}

我怎样才能做到这一点？

Answer 1

您需要使用元组而不是列表将它们用作字典键（键需要是可散列的），但如果这不是问题，您可以使用collections.counter来应对变化，然后是字典理解：

from collections import Counter

l = [['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
     ['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
     ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'],
     ['SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d', 'SBS38'],
     ['SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d']]

counts = Counter(map(tuple,l))
result = {k:count/len(l) for k, count in counts.items()}

结果：

{('SBS1', 'SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'): 0.6,
 ('SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d'): 0.2,
 ('SBS5', 'SBS7a', 'SBS7b', 'SBS7c', 'SBS7d', 'SBS38'): 0.2}