熊猫-基于条件的重复行

Question

如果该行符合条件，我将尝试创建一个重复行。在下表中，我根据groupby创建了一个累积计数，然后对groupby的MAX进行了另一次计算。

df['PathID'] = df.groupby(DateCompleted).cumcount() + 1
df['MaxPathID'] = df.groupby(DateCompleted)['PathID'].transform(max)

Date Completed    PathID    MaxPathID
1/31/17           1         3
1/31/17           2         3
1/31/17           3         3
2/1/17            1         1
2/2/17            1         2
2/2/17            2         2

在这种情况下，我只想复制2/1/17的记录，因为该日期只有一个实例（即MaxPathID == 1）。

所需输出：

Date Completed    PathID    MaxPathID
1/31/17           1         3
1/31/17           2         3
1/31/17           3         3
2/1/17            1         1
2/1/17            1         1
2/2/17            1         2
2/2/17            2         2

提前致谢！

Answer 1

我认为您需要unique先获取行Date Completed，然后concat获取原始行：

df1 = df.loc[~df['Date Completed'].duplicated(keep=False), ['Date Completed']]
print (df1)
  Date Completed
3         2/1/17

df = pd.concat([df,df1], ignore_index=True).sort_values('Date Completed')
df['PathID'] = df.groupby('Date Completed').cumcount() + 1
df['MaxPathID'] = df.groupby('Date Completed')['PathID'].transform(max)
print (df)
  Date Completed  PathID  MaxPathID
0        1/31/17       1          3
1        1/31/17       2          3
2        1/31/17       3          3
3         2/1/17       1          2
6         2/1/17       2          2
4         2/2/17       1          2
5         2/2/17       2          2

编辑：

print (df)
  Date Completed  a  b
0        1/31/17  4  5
1        1/31/17  3  5
2        1/31/17  6  3
3         2/1/17  7  9
4         2/2/17  2  0
5         2/2/17  6  7

df1 = df[~df['Date Completed'].duplicated(keep=False)]
#alternative - boolean indexing by numpy array
#df1 = df[~df['Date Completed'].duplicated(keep=False).values]
print (df1)
  Date Completed  a  b
3         2/1/17  7  9

df = pd.concat([df,df1], ignore_index=True).sort_values('Date Completed')
print (df)
  Date Completed  a  b
0        1/31/17  4  5
1        1/31/17  3  5
2        1/31/17  6  3
3         2/1/17  7  9
6         2/1/17  7  9
4         2/2/17  2  0
5         2/2/17  6  7