如何在熊貓中一起使用distinct和where子句？

Question

我有一個數據框和列表如下

op1 = pd.DataFrame({
'subject_id':[1,1,2,3,4,4,5],
'iid': [21,22,23,24,26,26,27],
'los':[121,122,123,124,111,111,131],
'area':['a','a','b','c','d','d','f'],
'date' : ['1/1/2017','1/2/2017','1/3/2017','1/4/2017','1/6/2017','1/6/2017','1/8/2109'],
'val' :[5,10,5,16,26,26,7]
 })

sub_list = [1,2,3,4]

我想檢查sub_list的subject_id是否存在於op1 。 如果存在，則從los ， iid ， area獲取該subject_id的distinct值（查找subject_id 1 and 4之間的差異（重復）

我嘗試了以下內容，但不能有多列

op1[op1['subject_id'].isin(sub_list)] # how to use distinct records here?

我必須將此應用於一百萬條記錄。 因此，任何優雅高效的解決方案都是有幫助的

我正在尋找類似的東西

select distinct subject_id, iid,los, area from op1
where subject_id in [sub_list] 

我希望我的輸出如下所示

Answer 1

如果打算僅返回選定的列，請執行以下操作：

result = op1.loc[op1["subject_id"].isin(sub_list), ["subject_id", "los", "iid", "area"]].drop_duplicates()

Answer 2

我不確定這有多快，但是您可以嘗試：

(op1[['subject_id','iid','los','area']]
     .drop_duplicates(['subject_id','iid','los','area'])
     .set_index('subject_id')
     .loc[sub_list]
)

Answer 3

op1[op1['subject_id'].isin(sub_list)].drop_duplicates(subset=list_columns_to_distinct)

Answer 4

這實際上是先前答案的混合

distCols = ["subject_id", "iid",
            "los", "area"]

op1[op1['subject_id'].isin(sub_list)].drop_duplicates(distCols)