Pandas groupby /按日期在多列上进行数据透视

Question

我正试图从这个df获得以下输出。它是从一个django查询构造的，它被转换为df：

messages = Message.objects.all()
df = pd.DataFrame.from_records(messages.values())

+---+-----------------+------------+---------------------+
|   |    date_time    | error_desc |        text         |
+---+-----------------+------------+---------------------+
| 0 | 3/31/2019 12:35 | Error msg  | Hello there         |
| 1 | 3/31/2019 12:35 |            | Nothing really here |
| 2 | 4/1/2019 12:35  | Error msg  | What if I told you  |
| 3 | 4/1/2019 12:35  |            | Yes                 |
| 4 | 4/1/2019 12:35  | Error Msg  | Maybe               |
| 5 | 4/2/2019 12:35  |            | Sure I could        |
| 6 | 4/2/2019 12:35  |            | Hello again         |
+---+-----------------+------------+---------------------+

输出：

+-----------+-------------+--------+-----------------------------+--------------+
|   date    | Total count | Errors | Greeting (start with hello) | errors/total |
+-----------+-------------+--------+-----------------------------+--------------+
| 3/31/2019 |           2 |      1 |                           1 | 50%          |
| 4/1/2019  |           3 |      2 |                           0 | 66.67%       |
| 4/2/2019  |           2 |      0 |                           1 | 0%           |
+-----------+-------------+--------+-----------------------------+--------------+

我部分能够使用以下代码到达那里，但它似乎有点迂回的做法。我根据他们是否符合条件然后分组来标记每个'是'/'否'。

df['date'] = df['date_time'].dt.date
df['greeting'] = np.where(df["text"].str.lower().str.startswith('hello'), "Yes", "No")
df['error'] = np.where(df["error_desc"].notnull(), "Yes", "No")

df.set_index("date")
    .groupby(level="date")
    .apply(lambda g: g.apply(pd.value_counts))
    .unstack(level=1)
    .fillna(0)

这会产生计数，但会产生多个是/否列。

在这之后我可以做一些操作，但有没有更有效的方法来提出我之后的输出？

Answer 1

您可以在多列上使用lambda：

df.groupby('date').apply(lambda x: 
                         pd.Series({'total_count': len(x),
                                    'error_count': (x['error'] == 'Yes').sum(),
                                    'hello_count': (x['greeting'] == 'Yes').sum()}))

要计算比率：

df['errors/total'] = df['error_count'] / df['total_count']

Answer 2

这是我试过的，给了我你想要的答案：

---

df['date_time'] = pd.to_datetime(df['date_time']).dt.date
df1=pd.DataFrame()
df1['total count'] = df['date_time'].groupby(df['date_time']).count()
df1['errors'] = df['error_desc'].groupby(df['date_time']).count()
df1['Greeting'] = df['text'].groupby(df['date_time']).apply(lambda x: x[x.str.lower().str.startswith('hello')].count())
df1['errors/total'] = round(df1['errors']/df1['total count']*100,2)