算法：

• 基于在正则表达式分隔符上拆分letter列来附加辅助列letter_，所述正则表达式分隔符具有作为值/元素列表的每个组
• 分解/展开letter_，以便将每个值放置在单独的行中
• 将每个letter_值Map到其频率（其出现次数）
• 筛选器letter（初始组），仅包含出现一次的项目（最大计数为1）

df = df.assign(letter_=df['letter'].str.split(r'\s*,\s*')).explode('letter_')
df['letter_'] = df['letter_'].map(df['letter_'].value_counts())
df = df.groupby('letter').filter(lambda x: x['letter_'].max() == 1)\
    .drop_duplicates().drop('letter_', axis=1).reset_index(drop=True)
print(df)

id letter
0  75      T
1  54   K, M

要从your previous question继续my answer，我只需要向 Dataframe 添加一个简单的唯一性测试作为新列：

import pandas as pd

df: pd.DataFrame = pd.DataFrame([
    [1, "A, B, D, E, F"], [3, "B, C"], [2, "B"], [75, "T"], [54, "K, M"],
    [9, "D, B"], [23, "B, D, A"], [34, "X, Y, Z"], [67, "X, Y"],
    [12, "E, D"], [15, "G"], [10, "G"], [11, "F"]
], columns=["id", "letters"])

threshold = 0.5

if __name__ == "__main__":
    sub_ids = []
    sub_scores = []
    sub_unique = []
    for i in range(len(df)):
        temp_sub_ids = []
        temp_sub_scores = []
        temp_unique = True
        curr_letters_i = df.iloc[i]["letters"].replace(" ", "").split(",")
        for j in range(len(df)):
            if i == j:
                continue
            curr_letters_j = df.iloc[j]["letters"].replace(" ", "").split(",")
            if any([letter in curr_letters_i for letter in curr_letters_j]):
                temp_unique = False
            if not all([letter in curr_letters_i for letter in curr_letters_j]):
                continue
            if len(curr_letters_j)/len(curr_letters_i) >= threshold:
                temp_sub_ids.append(f"{df.iloc[j]['id']}")
                temp_sub_scores.append(f"{len(curr_letters_j)/len(curr_letters_i):.2f}")
        sub_ids.append(",".join(temp_sub_ids))
        sub_scores.append(temp_sub_scores)
        sub_unique.append(temp_unique)
    df["sub-ids"] = sub_ids
    df["sub-scores"] = sub_scores
    df["unicity"] = sub_unique
    print(df)

这就产生了

id        letters sub-ids sub-scores  unicity
0    1  A, B, D, E, F      23     [0.60]    False
1    3           B, C       2     [0.50]    False
2    2              B                 []    False
3   75              T                 []     True
4   54           K, M                 []     True
5    9           D, B       2     [0.50]    False
6   23        B, D, A       9     [0.67]    False
7   34        X, Y, Z      67     [0.67]    False
8   67           X, Y                 []    False
9   12           E, D                 []    False
10  15              G      10     [1.00]    False
11  10              G      15     [1.00]    False
12  11              F                 []    False

可以过滤（print(df.loc[df["unicity"] == True]）为

id letters sub-ids sub-scores  unicity
3  75       T                 []     True
4  54    K, M                 []     True