首先，你可以对列表进行排序，因为它们是数字，并将它们转换为字符串。然后，对于每个字符串，只需检查它是否是任何其他行的子字符串，如果是，它就是子集。由于所有内容都进行了排序，我们可以确保数字的顺序不会影响这一步。
最后，只过滤掉那些没有被标识为子集的。

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'cycles': [[9, 5, 4, 3], [9, 5, 4], [2, 4, 3], [2, 3]],
    'members': [4, 3, 3, 2]
})
print(df)

         cycles  members
0  [9, 5, 4, 3]        4
1     [9, 5, 4]        3
2     [2, 4, 3]        3
3        [2, 3]        2

df['cycles'] = df['cycles'].map(np.sort) 
df['cycles_str'] = [','.join(map(str, c)) for c in df['cycles']]
# Here we check if matches are >1, because it will match with itself once!
df['is_subset'] = [df['cycles_str'].str.contains(c_str).sum() > 1 for c_str in df['cycles_str']]
df = df.loc[df['is_subset'] == False]
df = df.drop(['cycles_str', 'is_subset'], axis=1)

         cycles  members
0  [3, 4, 5, 9]        4
2     [2, 3, 4]        3

• • 编辑-以上操作不适用于[1，2，4]和[1，2，3，4]**

重写了代码。使用2个循环和set检查列表解析的子集：

# check if >1 True, as it will match with itself once!
df['is_subset'] = [[set(y).issubset(set(x)) for x in df['cycles']].count(True)>1 for y in df['cycles']]
df = df.loc[df['is_subset'] == False]
df = df.drop('is_subset', axis=1)
print(df)

         cycles  members
0  [9, 5, 4, 3]        4
2     [2, 4, 3]        3