你需要在这里使用一个循环：

df['check'] = [a in b[0:2] for a,b in zip(df['number'], df['list_of_numbers'])]

输出：

number      list_of_numbers  check
0       1         [1, 3, 2, 5]   True
1       2     [6, 7, 8, 2, 10]  False
2       3  [13, 12, 13, 14, 3]  False

您的方法失败的原因

np.isin在使用前将test_element数组扁平化，因此您不是针对每个列表测试每个元素，而是针对所有列表的连接测试每个元素
示范：

import pandas as pd
import numpy as np
df = pd.DataFrame({'number': [1, 1, 3], # we changed the 2 in 1
                   'list_of_numbers': [[1, 7, 2, 5],  # we removed the 3
                                       [6, 7, 8, 2, 10],
                                       [13, 12, 13, 14, 3]]})

df['check'] = np.isin(df['number'], [x[0:2] for x in df['list_of_numbers']])
print(df)
   number      list_of_numbers  check
0       1         [1, 7, 2, 5]   True
1       1     [6, 7, 8, 2, 10]   True # we have True as the first list has a 1
2       3  [13, 12, 13, 14, 3]  False # now we have False as the 3 in the first list is gone