一种方法是创建一个包含所有worker和skills的dataframe（通过交叉连接它们），然后将其连接到worker df以查找哪些组合不存在;这是每个工人都不具备的技能

workers = pd.DataFrame({
    'worker': ['worker1', 'worker1', 'worker1', 'worker2', 'worker2', 'worker2', 'worker3', 'worker3', 'worker3'],
    'skills_of_worker': ['skill1', 'skill2', 'skill3', 'skill2', 'skill3', 'skill4', 'skill1', 'skill3', 'skill4']
})
skills = pd.DataFrame({'skills': ['skill1', 'skill2', 'skill3', 'skill4']})

all_workers = pd.Series(workers['worker'].unique(), name='worker')
res = skills.merge(all_workers, how='cross').merge(workers, how='left', left_on=['worker', 'skills'], right_on=['worker', 'skills_of_worker'])
res = res[res['skills_of_worker'].isna()].drop('skills_of_worker', axis=1).rename(columns={'skills':'skills_they_dont_have'}).reset_index(drop=True)

输出：

skills_they_dont_have   worker
0                skill1  worker2
1                skill2  worker3
2                skill4  worker1

我会使用set操作（difference），然后explode：

skills = {'skill1', 'skill2', 'skill3', 'skill4'}

out = (
  df.groupby('worker', as_index=False)['skills_of_worker']
   .agg(skills.difference)
   .explode('skills_of_worker')
   .rename(columns={'skills_of_worker': 'skills_they_dont_have'})
)

输出：

worker skills_they_dont_have
0  worker1                skill4
1  worker2                skill1
2  worker3                skill2