檢查一個 dataframe 列是否是另一列的子集

Question

我有一個 dataframe 列Enrolled_Months和Eligible_Months ，描述如下：

month_list1 = [
    [(1, 2018), (2, 2018), (3, 2019)],
    [(7, 2018), (8, 2018), (10, 2018)],
    [(4, 2018), (5, 2018), (7, 2018)],
    [(1, 2019), (2, 2019), (4, 2019)]
]

month_list2 = [
    [(2, 2018), (3, 2019)],
    [(7, 2018), (8, 2018)],
    [(2, 2018), (3, 2019)],
    [(10, 2018), (11, 2019)]
]

EID = [1, 2, 3, 4]

df = pd.DataFrame({
    'EID': EID,
    'Enrolled_Months': month_list1,
    'Eligible_Months': month_list2
})
df

Out[6]: 
   EID                     Enrolled_Months           Eligible_Months
0    1   [(1, 2018), (2, 2018), (3, 2019)]    [(2, 2018), (3, 2019)]
1    2  [(7, 2018), (8, 2018), (10, 2018)]    [(7, 2018), (8, 2018)]
2    3   [(4, 2018), (5, 2018), (7, 2018)]    [(2, 2018), (3, 2019)]
3    4   [(1, 2019), (2, 2019), (4, 2019)]  [(10, 2018), (11, 2019)]

我想創建一個名為Check的新列，如果Enrolled_Months包含Eligible_Months的所有元素，則該列為真。 我想要的 output 如下：

Out[8]: 
   EID                     Enrolled_Months           Eligible_Months  Check
0    1   [(1, 2018), (2, 2018), (3, 2019)]    [(2, 2018), (3, 2019)]   True
1    2  [(7, 2018), (8, 2018), (10, 2018)]    [(7, 2018), (8, 2018)]   True
2    3   [(4, 2018), (5, 2018), (7, 2018)]    [(2, 2018), (3, 2019)]  False
3    4   [(1, 2019), (2, 2019), (4, 2019)]  [(10, 2018), (11, 2019)]  False

我試過以下方法：

df['Check'] = set(df['Eligible_Months']).issubset(df['Enrolled_Months'])

但最終得到錯誤TypeError: unhashable type: 'list' 。

關於如何實現這一目標的任何想法？

旁注： Enrolled_Months數據最初采用非常不同的格式，每個月都有自己的二進制列，以及一個單獨的Year列指定年份（imo 設計非常糟糕）。 我創建了列表列，因為我認為它更容易使用，但如果原始格式更適合我想要實現的目標，請告訴我。

Answer 1

您可以使用一些explodes然后eval和any ：

df['Check'] = df.explode('Eligible_Months').explode('Enrolled_Months').eval('Enrolled_Months == Eligible_Months').groupby(level=0).any()

Output：

>>> df
   EID                     Enrolled_Months           Eligible_Months  Check
0    1   [(1, 2018), (2, 2018), (3, 2019)]    [(2, 2018), (3, 2019)]   True
1    2  [(7, 2018), (8, 2018), (10, 2018)]    [(7, 2018), (8, 2018)]   True
2    3   [(4, 2018), (5, 2018), (7, 2018)]    [(2, 2018), (3, 2019)]  False
3    4   [(1, 2019), (2, 2019), (4, 2019)]  [(10, 2018), (11, 2019)]  False

Answer 2

您可以使用df.apply()創建新列：

df['Check'] = df.apply(
    lambda row: set(row['Eligible_Months']).issubset(row['Enrolled_Months']), axis=1
)

這輸出：

   EID                     Enrolled_Months           Eligible_Months  Check
0    1   [(1, 2018), (2, 2018), (3, 2019)]    [(2, 2018), (3, 2019)]   True
1    2  [(7, 2018), (8, 2018), (10, 2018)]    [(7, 2018), (8, 2018)]   True
2    3   [(4, 2018), (5, 2018), (7, 2018)]    [(2, 2018), (3, 2019)]  False
3    4   [(1, 2019), (2, 2019), (4, 2019)]  [(10, 2018), (11, 2019)]  False

Answer 3

列表理解工作正常：

df.assign(check = [set(l).issuperset(r) 
                   for l, r in 
                   zip(df.Enrolled_Months, df.Eligible_Months)])

   EID                     Enrolled_Months           Eligible_Months  check
0    1   [(1, 2018), (2, 2018), (3, 2019)]    [(2, 2018), (3, 2019)]   True
1    2  [(7, 2018), (8, 2018), (10, 2018)]    [(7, 2018), (8, 2018)]   True
2    3   [(4, 2018), (5, 2018), (7, 2018)]    [(2, 2018), (3, 2019)]  False
3    4   [(1, 2019), (2, 2019), (4, 2019)]  [(10, 2018), (11, 2019)]  False