Python 多个groupby后如何将pandas数据从索引移动到列

How to move pandas data from index to column after multiple groupby

提问by prooffreader

I have the following pandas dataframe:

我有以下熊猫数据框：

dfalph.head()

token    year    uses  books
  386   xanthos  1830    3     3
  387   xanthos  1840    1     1
  388   xanthos  1840    2     2
  389   xanthos  1868    2     2
  390   xanthos  1875    1     1

I aggregate the rows with duplicate tokenand yearslike so:

我用聚合重复行token和years像这样：

dfalph = dfalph[['token','year','uses','books']].groupby(['token', 'year']).agg([np.sum])
dfalph.columns = dfalph.columns.droplevel(1)
dfalph.head()

               uses  books
token    year       
xanthos  1830    3     3
         1840    3     3
         1867    2     2
         1868    2     2
         1875    1     1

Instead of having the 'token' and 'year' fields in the index, I would like to return them to columns and have an integer index.

我想将它们返回到列并具有整数索引，而不是在索引中包含“令牌”和“年份”字段。

采纳答案by DSM

Method #1: reset_index()

方法#1：reset_index()

>>> g
              uses  books
               sum    sum
token   year             
xanthos 1830     3      3
        1840     3      3
        1868     2      2
        1875     1      1

[4 rows x 2 columns]
>>> g = g.reset_index()
>>> g
     token  year  uses  books
                   sum    sum
0  xanthos  1830     3      3
1  xanthos  1840     3      3
2  xanthos  1868     2      2
3  xanthos  1875     1      1

[4 rows x 4 columns]

Method #2: don't make the index in the first place, using as_index=False

方法#2：首先不要创建索引，使用as_index=False

>>> g = dfalph[['token', 'year', 'uses', 'books']].groupby(['token', 'year'], as_index=False).sum()
>>> g
     token  year  uses  books
0  xanthos  1830     3      3
1  xanthos  1840     3      3
2  xanthos  1868     2      2
3  xanthos  1875     1      1

[4 rows x 4 columns]

回答by Adarsh Madrecha

I defer form the accepted answer. While there are 2 ways to do this, these will not necessarily result in same output. Specially when you are using Grouperin groupby

我推迟了接受的答案。虽然有两种方法可以做到这一点，但这些方法不一定会产生相同的输出。特别是当您使用Grouper在groupby

• index=False
• reset_index()

• index=False
• reset_index()

example df

例子 df

+---------+---------+-------------+------------+
| column1 | column2 | column_date | column_sum |
+---------+---------+-------------+------------+
| A       | M       | 26-10-2018  |          2 |
| B       | M       | 28-10-2018  |          3 |
| A       | M       | 30-10-2018  |          6 |
| B       | M       | 01-11-2018  |          3 |
| C       | N       | 03-11-2018  |          4 |
+---------+---------+-------------+------------+

They do not work the same way.

它们的工作方式不同。

df = df.groupby(
    by=[
        'column1',
        'column2',
        pd.Grouper(key='column_date', freq='M')
    ],
    as_index=False
).sum()

The above will give

以上会给

+---------+---------+------------+
| column1 | column2 | column_sum |
+---------+---------+------------+
| A       | M       |          8 |
| B       | M       |          3 |
| B       | M       |          3 |
| C       | N       |          4 |
+---------+---------+------------+

While,

尽管，

df = df.groupby(
    by=[
        'column1',
        'column2',
        pd.Grouper(key='column_date', freq='M')
    ]
).sum().reset_index()

Will give

会给

+---------+---------+-------------+------------+
| column1 | column2 | column_date | column_sum |
+---------+---------+-------------+------------+
| A       | M       | 31-10-2018  |          8 |
| B       | M       | 31-10-2018  |          3 |
| B       | M       | 30-11-2018  |          3 |
| C       | N       | 30-11-2018  |          4 |
+---------+---------+-------------+------------+

回答by user1809802

You need to add drop=True:

您需要添加drop=True：

df.reset_index(drop=True)

df = df.groupby(
    by=[
        'column1',
        'column2',
        pd.Grouper(key='column_date', freq='M')
    ]
).sum().reset_index(drop=True)