python 将具有重复元素的两个列表组合成字典

现有两个列表

aa = ['a','a','b','a','b']
bb = [1,2,3,4,3]

想要得到结果为：

{'a': [1, 2, 4], 'b': [3, 3]}

代码如下：

from itertools import groupby
import numpy as np

aa = ['a','a','b','a','b']
bb = [1,2,3,4,3]

aa_sorted = sorted(aa) # ['a', 'a', 'a', 'b', 'b']
index = sorted(range(len(aa)),key = lambda  k:aa[k] ) # array([0, 1, 3, 2, 4], dtype=int64)
bb_sorted = np.array(bb)[index].tolist() # [1, 2, 4, 3, 3]

aa_sorted_binded = [list(j) for i,j in groupby(aa_sorted)] # [['a', 'a', 'a'], ['b', 'b']]
index = [[0]]
index.append([len(ele) for ele in aa_sorted_binded]) # [[0], [3, 2]]
index = sum(index,[]) # [0, 3, 2]
bb_sorted_binded = [bb_sorted[index[i]:index[i]+index[i+1]] for i in range(len(index)-1)] # [[1, 2, 4], [3, 3]]
result = dict(zip(list(set(aa)),bb_sorted_binded)) # {'a': [1, 2, 4], 'b': [3, 3]}

如果要对result进行键值互换(inverse)

则有

new_result = {value:key for key in result for value in result[key]}

结果为

{1: 'a', 2: 'a', 4: 'a', 3: 'b'}