计算字典列表中的相似值

创建一个元组列表,其中每个元组的第一个元素是原始单词,其余元素是相应的 corrected_word 元组中的元素。然后将此列表放入 计数器

来自集合导入计数器
ctr=计数器((mylist中项目的(item['original_word'],*item['corrected_word']))

这提供了:

计数器({(‘test3’,‘test3‘,33):3,(‘test1’,‘est12’,3):2,(‘est2’,‘test 22’,2):1})

然后,构建结果列表,并根据所需值进行排序:

result=排序([
{'original_word':ow,'corrected_word':(*cw,count)}对于(ow,*cw),在ctr.items()中计数
],key=lambda项:项['corrected_word'][1])

这给出了所需的结果:

[
{'original_word':'test2','corrected_word':('test22',2,1)},
{'original_word':'test1','corrected_word':('test12',3,2)},
{‘original_word’:‘test3’,‘corrected_word’:(‘test3‘,33,3)}
]

在线试用

然后,构建结果列表,并根据所需值进行排序:

result = sorted([
          {'original_word': ow, 'corrected_word': (*cw, count)} for (ow, *cw), count in ctr.items()
          ], key=lambda item: item['corrected_word'][1])

这给出了所需的结果:

[
 {'original_word': 'test2', 'corrected_word': ('test22', 2, 1)},
 {'original_word': 'test1', 'corrected_word': ('test12', 3, 2)},
 {'original_word': 'test3', 'corrected_word': ('test3', 33, 3)}
]

在线试用!

1. 转换为计数器之前的元组
2. 转换回dicts并添加计数
3. 根据数字排序

def dict_and_add_count(item):
    original_data, count = item
    original_dict = dict(original_data)
    original_dict['corrected_word'] = (*original_dict['corrected_word'], count)
    return original_dict

counted_unique_tuples = Counter(tuple(d.items()) for d in mylist)
dict_with_count = map(dict_and_add_count, counted_unique_tuples.items())
sorted_dicts = sorted(dict_with_count, key=lambda x: x['corrected_word'][1])

sorted_dicts将是

[{'original_word': 'test2', 'corrected_word': ('test22', 2, 1)},
{'original_word': 'test1', 'corrected_word': ('test12', 3, 2)},
{'original_word': 'test3', 'corrected_word': ('test3', 33, 3)}]