Python 熊猫:基于日期的缺失值插补
我有一个熊猫数据框,如下所示:Python 熊猫:基于日期的缺失值插补,python,pandas,dataframe,imputation,Python,Pandas,Dataframe,Imputation,我有一个熊猫数据框,如下所示: df_first = pd.DataFrame({"id": [102, 102, 102, 102, 103, 103], "val1": [np.nan, 4, np.nan, np.nan, 1, np.nan], "val2": [5, np.nan, np.nan, np.nan, np.nan, 5], "rand": [np.nan, 3, 7, 8, np.nan, 4], "val3": [5, np.nan, np.nan, np.nan, 3
df_first = pd.DataFrame({"id": [102, 102, 102, 102, 103, 103], "val1": [np.nan, 4, np.nan, np.nan, 1, np.nan], "val2": [5, np.nan, np.nan, np.nan, np.nan, 5], "rand": [np.nan, 3, 7, 8, np.nan, 4], "val3": [5, np.nan, np.nan, np.nan, 3, np.nan], "unique_date": [pd.Timestamp(2002, 3, 3), pd.Timestamp(2002, 3, 5), pd.Timestamp(2003, 4, 5), pd.Timestamp(2003, 4, 9), pd.Timestamp(2003, 8, 7), pd.Timestamp(2003, 9, 7)], "end_date": [pd.Timestamp(2005, 3, 3), pd.Timestamp(2003, 4, 7), np.nan, np.nan, pd.Timestamp(2003, 10, 7), np.nan]})
df_first
id val1 val2 rand val3 unique_date end_date
0 102 NaN 5.0 NaN 5.0 2002-03-03 2005-03-03
1 102 4.0 NaN 3.0 NaN 2002-03-05 2003-04-07
2 102 NaN NaN 7.0 NaN 2003-04-05 NaT
3 102 NaN NaN 8.0 NaN 2003-04-09 NaT
4 103 1.0 NaN NaN 3.0 2003-08-07 2003-10-07
5 103 NaN 5.0 4.0 NaN 2003-09-07 NaT
缺失值插补应以如下方式进行:从具有end\u date
值的数据框中向前填充每行中出现的值
只要相同的id
的唯一日期早于结束日期
,则执行正向填充
根据上面最后一段所述,应按照id
进行正向填充
最后,缺失值插补应仅针对某些名称中包含val
的列进行。一个重要的注意事项是,没有其他列的名称中有这种模式。如果我没有说得足够清楚,上面发布的数据帧的解决方案如下所示:
id val1 val2 rand val3 unique_date
0 102 NaN 5.0 NaN 5.0 2002-03-03
1 102 4.0 5.0 3.0 5.0 2002-03-05
2 102 4.0 5.0 7.0 5.0 2003-04-05
3 102 NaN 5.0 8.0 5.0 2003-04-09
4 103 1.0 NaN NaN 3.0 2003-08-07
5 103 1.0 5.0 4.0 3.0 2003-08-07
如果你需要进一步澄清,请告诉我,因为乍一看整个事情似乎相当复杂
期待你的回答 对不起,这个问题和解释都很混乱。最后,我能够通过以下方式实现我想要的
df_first = pd.DataFrame({"id": [102, 102, 102, 102, 103, 103],
"val1": [np.nan, 4, np.nan, np.nan, 1, np.nan],
"val2": [5, np.nan, np.nan, np.nan, np.nan, 5],
"val3": [np.nan, 3, np.nan, np.nan, np.nan, 4],
"val4": [5, np.nan, np.nan, np.nan, 3, np.nan],
"rand": [3, np.nan, 1, np.nan, 5, 6],
"unique_date": [pd.Timestamp(2002, 3, 3),
pd.Timestamp(2002, 3, 5),
pd.Timestamp(2003, 4, 5),
pd.Timestamp(2003, 4, 9),
pd.Timestamp(2003, 8, 7),
pd.Timestamp(2003, 9, 7)],
"end_date": [pd.Timestamp(2005, 3, 3),
pd.Timestamp(2003, 4, 7),
np.nan,
np.nan,
pd.Timestamp(2003, 10, 7),
np.nan]})
display(df_first)
indexes = []
columns = df_first.filter(like="val").columns
for column in columns:
indexes.append(df_first.columns.get_loc(column))
elements = df_first.values[:,indexes]
ids = df_first.values[:,df_first.columns.get_loc("id")]
start_dates = df_first.values[:,df_first.columns.get_loc("unique_date")]
end_dates = df_first.values[:,df_first.columns.get_loc("end_date")]
for i in range(len(elements)):
if pd.notnull(end_dates[i]):
not_nan_indexes = np.argwhere(~pd.isnull(elements[i])).ravel()
elements_prop = elements[i,not_nan_indexes]
j = i
while (j < len(elements) and start_dates[j] < end_dates[i] and ids[i] == ids[j]):
elements[j, not_nan_indexes] = elements_prop
j+=1
df_first[columns] = elements
df_first = df_first.drop(columns="end_date")
display(df_first)
df_first=pd.DataFrame({“id”:[102102102102103103],
“val1:[np.nan,4,np.nan,np.nan,1,np.nan],
“val2:[5,np.nan,np.nan,np.nan,np.nan,5],
“val3”:[np.nan,3,np.nan,np.nan,np.nan,4],
“val4:[5,np.nan,np.nan,np.nan,3,np.nan],
“兰德”:[3,np.nan,1,np.nan,5,6],
“唯一日期”:[pd.时间戳(2002,3,3),
pd.时间戳(2002,3,5),
pd.时间戳(2003,4,5),
pd.时间戳(2003,4,9),
pd.时间戳(2003,8,7),
pd.时间戳(2003年9月7日)],
“结束日期”:[pd.时间戳(2005,3,3),
pd.时间戳(2003,4,7),
np.nan,
np.nan,
pd.时间戳(2003,10,7),
np.nan]})
显示(df_优先)
索引=[]
columns=df_first.filter(like=“val”).columns
对于列中的列:
index.append(df_first.columns.get_loc(column))
元素=df_first.值[:,索引]
ids=df_first.values[:,df_first.columns.get_loc(“id”)]
开始日期=df_first.values[:,df_first.columns.get_loc(“唯一日期”)]
结束日期=df_first.values[:,df_first.columns.get_loc(“结束日期”)]
对于范围内的i(len(元素)):
如果pd.notnull(结束日期[i]):
not_nan_index=np.argwhere(~pd.isnull(elements[i])).ravel()
elements\u prop=元素[i,而非索引]
j=i
而(j
可能解决方案有些过火,但我找不到任何具体的方法来实现我想要的。您已经尝试了什么?您能告诉我这个示例数据的“val3”是如何向前填充的吗?@coldspeed与其他“val”列相同。这样看,“id”的人在“唯一日期”开始服用“val”药物,一直服用到“结束日期”。“唯一_日期”还与该人员“id”的其他事件相关联。其他事件是本例中的其他列“rand”。这就是为什么有几行具有相同的“id”和不同的“unique_date”(与此人相关的其他事件发生在其他日期)。@sundance如果我从数据以及要编辑的列的索引中提取numpy矩阵,我知道如何执行此操作。然后,进行正向插补,并将数据转换回数据帧。