python - 如何获取 X 列为 False 的同一条船上连续安装的最旧部件安装？

Question

我有一个很好的算法，可以返回同一条船上连续安装的最旧的部分安装:How to get oldest part installation of consecutive installations the same boat in pandas?

我只想将此算法应用于具有 col['x'] == False 的行。我怎样才能做到这一点？请参阅下面的预期输出。

算法:print (df[['boat','part']].ne(df[['boat','part']].shift()).any(axis=1))

输入

| boat | part | date           | x     |
|------|------|----------------|-------|
| A    | B    | 12/19/13 08:19 | TRUE  |<-- ignore and keep this
| A    | C    | 3/24/14 10:26  | TRUE  |<-- ignore and keep this
| A    | D    | 7/21/14 09:46  | TRUE  |<-- ignore and keep this
| A    | D    | 3/5/18 22:35   | FALSE |<--evaluate this with algorithm
| A    | D    | 7/27/19 00:43  | FALSE |<--evaluate this with algorithm
| A    | D    | 8/9/19 00:53   | FALSE |<--evaluate this with algorithm
| A    | D    | 9/6/19 23:10   | FALSE |<--evaluate this with algorithm

预期输出

| boat | part | date           | x     |
|------|------|----------------|-------|
| A    | B    | 12/19/13 08:19 | TRUE  |<-- keep because x == True
| A    | C    | 3/24/14 10:26  | TRUE  |<-- keep because x == True
| A    | D    | 7/21/14 09:46  | TRUE  |<-- keep because x == True
| A    | D    | 3/5/18 22:35   | FALSE |<-- Keep because first install of consecutive installs

输入2

| boat | part | date             | x     |
|------|------|------------------|-------|
| A    | E    | 12/20/13   03:27 | TRUE  |<-- ignore and keep this
| B    | E    | 7/21/14   09:46  | FALSE |<--evaluate this with algorithm
| C    | E    | 1/8/16   17:08   | TRUE  |<-- ignore and keep this
| C    | E    | 3/23/16   17:15  | TRUE  |<-- ignore and keep this
| B    | E    | 3/5/18   22:35   | FALSE |<--evaluate this with algorithm
| B    | E    | 7/27/19   00:43  | FALSE |<--evaluate this with algorithm
| B    | E    | 8/9/19   00:53   | FALSE |<--evaluate this with algorithm
| B    | E    | 9/6/19   23:10   | FALSE |<--evaluate this with algorithm

预期输出 2

| boat | part | date             | x     |
|------|------|------------------|-------|
| A    | E    | 12/20/13   03:27 | TRUE  |<-- keep because x == True
| B    | E    | 7/21/14   09:46  | FALSE |<-- Keep because first install of consecutive installs
| C    | E    | 1/8/16   17:08   | TRUE  |<-- keep because x == True
| C    | E    | 3/23/16   17:15  | TRUE  |<-- keep because x == True
| B    | E    | 3/5/18   22:35   | FALSE |<-- Keep because first install of consecutive installs

Answer 1

一个想法是，如果 DataFrame.assign True 为 true，则创建由唯一值填充的新辅助列和 Index.where并处理所有 3 列:

df1 = df.assign(tmp = df.index.where(df['x'], -1))[['boat','part','tmp']]
df = df[df1.ne(df1.shift()).any(axis=1)]
print (df)
  boat part            date      x
0    A    E  12/20/13 03:27   True
1    C    E    1/8/16 17:08   True
2    C    E   3/23/16 17:15   True
3    B    E    3/5/18 22:35  False

第二个:

df1 = df.assign(tmp = df.index.where(df['x'], -1))[['boat','part','tmp']]
df = df[df1.ne(df1.shift()).any(axis=1)]
print (df)
  boat part            date      x
0    A    E  12/20/13 03:27   True
1    B    E   7/21/14 09:46  False
2    C    E    1/8/16 17:08   True
3    C    E   3/23/16 17:15   True
4    B    E    3/5/18 22:35  False