我有一个带有日期时间和两列的数据框，我必须在'X'列的'特定日期'中找到最大的空值范围，并在该特定日期的两列中将其替换为零。除此之外，我还必须创建名称为'flag'的第三列，该列在其他两列中的每个零插补将携带1的值，否则将为0。在下面的示例中，1月1日的最大拉伸空值是3倍，因此我必须将其替换为零。同样，我必须复制1月2日的流程。
以下是我的示例数据：
Datetime X Y
01-01-2018 00:00 1 1
01-01-2018 00:05 nan 2
01-01-2018 00:10 2 nan
01-01-2018 00:15 3 4
01-01-2018 00:20 2 2
01-01-2018 00:25 nan 1
01-01-2018 00:30 nan nan
01-01-2018 00:35 nan nan
01-01-2018 00:40 4 4
02-01-2018 00:00 nan nan
02-01-2018 00:05 2 3
02-01-2018 00:10 2 2
02-01-2018 00:15 2 5
02-01-2018 00:20 2 2
02-01-2018 00:25 nan nan
02-01-2018 00:30 nan 1
02-01-2018 00:35 3 nan
02-01-2018 00:40 nan nan
“下面是我期望的结果”
Datetime X Y Flag
01-01-2018 00:00 1 1 0
01-01-2018 00:05 nan 2 0
01-01-2018 00:10 2 nan 0
01-01-2018 00:15 3 4 0
01-01-2018 00:20 2 2 0
01-01-2018 00:25 0 0 1
01-01-2018 00:30 0 0 1
01-01-2018 00:35 0 0 1
01-01-2018 00:40 4 4 0
02-01-2018 00:00 nan nan 0
02-01-2018 00:05 2 3 0
02-01-2018 00:10 2 2 0
02-01-2018 00:15 2 5 0
02-01-2018 00:20 2 2 0
02-01-2018 00:25 nan nan 0
02-01-2018 00:30 nan 1 0
02-01-2018 00:35 3 nan 0
02-01-2018 00:40 nan nan 0
此问题是上一个问题的扩展。这是链接Python - Find maximum null values in stretch and replacing with 0
答案 0 :(得分：2)
首先为每个由唯一值填充的列创建连续的组：
df1 = df.isna()
df2 = df1.ne(df1.groupby(df1.index.date).shift()).cumsum().where(df1)
df2['Y'] *= len(df2)
print (df2)
X Y
Datetime
2018-01-01 00:00:00 NaN NaN
2018-01-01 00:05:00 2.0 NaN
2018-01-01 00:10:00 NaN 36.0
2018-01-01 00:15:00 NaN NaN
2018-01-01 00:20:00 NaN NaN
2018-01-01 00:25:00 4.0 NaN
2018-01-01 00:30:00 4.0 72.0
2018-01-01 00:35:00 4.0 72.0
2018-01-01 00:40:00 NaN NaN
2018-02-01 00:00:00 6.0 108.0
2018-02-01 00:05:00 NaN NaN
2018-02-01 00:10:00 NaN NaN
2018-02-01 00:15:00 NaN NaN
2018-02-01 00:20:00 NaN NaN
2018-02-01 00:25:00 8.0 144.0
2018-02-01 00:30:00 8.0 NaN
2018-02-01 00:35:00 NaN 180.0
2018-02-01 00:40:00 10.0 180.0
然后获取具有最大数量的组-这里是组4：
a = df2.stack().value_counts().index[0]
print (a)
4.0
获取集合0的匹配行和Flag列的转换掩码，将掩码转换为从Tru/False到1/0的整数：
mask = df2.eq(a).any(axis=1)
df.loc[mask,:] = 0
df['Flag'] = mask.astype(int)
print (df)
X Y Flag
Datetime
2018-01-01 00:00:00 1.0 1.0 0
2018-01-01 00:05:00 NaN 2.0 0
2018-01-01 00:10:00 2.0 NaN 0
2018-01-01 00:15:00 3.0 4.0 0
2018-01-01 00:20:00 2.0 2.0 0
2018-01-01 00:25:00 0.0 0.0 1
2018-01-01 00:30:00 0.0 0.0 1
2018-01-01 00:35:00 0.0 0.0 1
2018-01-01 00:40:00 4.0 4.0 0
2018-02-01 00:00:00 NaN NaN 0
2018-02-01 00:05:00 2.0 3.0 0
2018-02-01 00:10:00 2.0 2.0 0
2018-02-01 00:15:00 2.0 5.0 0
2018-02-01 00:20:00 2.0 2.0 0
2018-02-01 00:25:00 NaN NaN 0
2018-02-01 00:30:00 NaN 1.0 0
2018-02-01 00:35:00 3.0 NaN 0
2018-02-01 00:40:00 NaN NaN 0
编辑：
为列表中的匹配日期添加了新条件：
dates = df.index.floor('d')
filtered = ['2018-01-01','2019-01-01']
m = dates.isin(filtered)
df1 = df.isna() & m[:, None]
df2 = df1.ne(df1.groupby(dates).shift()).cumsum().where(df1)
df2['Y'] *= len(df2)
print (df2)
X Y
Datetime
2018-01-01 00:00:00 NaN NaN
2018-01-01 00:05:00 2.0 NaN
2018-01-01 00:10:00 NaN 36.0
2018-01-01 00:15:00 NaN NaN
2018-01-01 00:20:00 NaN NaN
2018-01-01 00:25:00 4.0 NaN
2018-01-01 00:30:00 4.0 72.0
2018-01-01 00:35:00 4.0 72.0
2018-01-01 00:40:00 NaN NaN
2018-02-01 00:00:00 NaN NaN
2018-02-01 00:05:00 NaN NaN
2018-02-01 00:10:00 NaN NaN
2018-02-01 00:15:00 NaN NaN
2018-02-01 00:20:00 NaN NaN
2018-02-01 00:25:00 NaN NaN
2018-02-01 00:30:00 NaN NaN
2018-02-01 00:35:00 NaN NaN
2018-02-01 00:40:00 NaN NaN
a = df2.stack().value_counts().index[0]
#solution working also if no NaNs per filtered rows (prevent IndexError: index 0 is out of bounds)
#a = next(iter(df2.stack().value_counts().index), -1)
mask = df2.eq(a).any(axis=1)
df.loc[mask,:] = 0
df['Flag'] = mask.astype(int)
print (df)
X Y Flag
Datetime
2018-01-01 00:00:00 1.0 1.0 0
2018-01-01 00:05:00 NaN 2.0 0
2018-01-01 00:10:00 2.0 NaN 0
2018-01-01 00:15:00 3.0 4.0 0
2018-01-01 00:20:00 2.0 2.0 0
2018-01-01 00:25:00 0.0 0.0 1
2018-01-01 00:30:00 0.0 0.0 1
2018-01-01 00:35:00 0.0 0.0 1
2018-01-01 00:40:00 4.0 4.0 0
2018-02-01 00:00:00 NaN NaN 0
2018-02-01 00:05:00 2.0 3.0 0
2018-02-01 00:10:00 2.0 2.0 0
2018-02-01 00:15:00 2.0 5.0 0
2018-02-01 00:20:00 2.0 2.0 0
2018-02-01 00:25:00 NaN NaN 0
2018-02-01 00:30:00 NaN 1.0 0
2018-02-01 00:35:00 3.0 NaN 0