Mic*_*ael 23 python numpy multi-index dataframe pandas
我想修改一个pandas MultiIndex DataFrame,使每个索引组包含指定范围之间的日期.我希望每个小组用值0(或NaN
)填写2013-06-11到2013-12-31的缺失日期.
Group A, Group B, Date, Value
loc_a group_a 2013-06-11 22
2013-07-02 35
2013-07-09 14
2013-07-30 9
2013-08-06 4
2013-09-03 40
2013-10-01 18
group_b 2013-07-09 4
2013-08-06 2
2013-09-03 5
group_c 2013-07-09 1
2013-09-03 2
loc_b group_a 2013-10-01 3
Run Code Online (Sandbox Code Playgroud)
我已经看过一些关于reindex
ing的讨论,但那是一个简单的(非分组的)时间序列数据.
是否有捷径可寻?
以下是我在完成此操作时所做的一些尝试.例如:一旦我被拆开['A', 'B']
,我就可以重新索引.
df = pd.DataFrame({'A': ['loc_a'] * 12 + ['loc_b'],
'B': ['group_a'] * 7 + ['group_b'] * 3 + ['group_c'] * 2 + ['group_a'],
'Date': ["2013-06-11",
"2013-07-02",
"2013-07-09",
"2013-07-30",
"2013-08-06",
"2013-09-03",
"2013-10-01",
"2013-07-09",
"2013-08-06",
"2013-09-03",
"2013-07-09",
"2013-09-03",
"2013-10-01"],
'Value': [22, 35, 14, 9, 4, 40, 18, 4, 2, 5, 1, 2, 3]})
df.Date = df['Date'].apply(lambda x: pd.to_datetime(x).date())
df = df.set_index(['A', 'B', 'Date'])
dt_start = dt.datetime(2013,6,1)
all_dates = [(dt_start + dt.timedelta(days=x)).date() for x in range(0,60)]
df2 = df.unstack(['A', 'B'])
df3 = df2.reindex(index=all_dates).fillna(0)
df4 = df3.stack(['A', 'B'])
## df4 is about where I want to get, now I'm trying to get it back in the form of df...
df5 = df4.reset_index()
df6 = df5.rename(columns={'level_0' : 'Date'})
df7 = df6.groupby(['A', 'B', 'Date'])['Value'].sum()
Run Code Online (Sandbox Code Playgroud)
最后几行让我有点难过.我希望df6
我可以简单地set_index
回到['A', 'B', 'Date']
,但是没有对值进行分组,因为它们在初始df
DataFrame中被分组.
有关如何重新索引未堆叠的DataFrame,重新索引以及使DataFrame采用与原始格式相同的格式的任何想法?
Chr*_*ong 16
您可以基于现有多索引级别的笛卡尔积来创建新的多索引.然后,使用新索引重新索引数据框.
new_index = pd.MultiIndex.from_product(df.index.levels)
new_df = df.reindex(new_index)
# Optional: convert missing values to zero, and convert the data back
# to integers. See explanation below.
new_df = new_df.fillna(0).astype(int)
Run Code Online (Sandbox Code Playgroud)
而已!新数据框具有所有可能的索引值.现有数据已正确编入索引.
继续阅读以获得更详细的解释.
import pandas as pd
df = pd.DataFrame({'A': ['loc_a'] * 12 + ['loc_b'],
'B': ['group_a'] * 7 + ['group_b'] * 3 + ['group_c'] * 2 + ['group_a'],
'Date': ["2013-06-11",
"2013-07-02",
"2013-07-09",
"2013-07-30",
"2013-08-06",
"2013-09-03",
"2013-10-01",
"2013-07-09",
"2013-08-06",
"2013-09-03",
"2013-07-09",
"2013-09-03",
"2013-10-01"],
'Value': [22, 35, 14, 9, 4, 40, 18, 4, 2, 5, 1, 2, 3]})
df.Date = pd.to_datetime(df.Date)
df = df.set_index(['A', 'B', 'Date'])
Run Code Online (Sandbox Code Playgroud)
以下是示例数据的样子
Value
A B Date
loc_a group_a 2013-06-11 22
2013-07-02 35
2013-07-09 14
2013-07-30 9
2013-08-06 4
2013-09-03 40
2013-10-01 18
group_b 2013-07-09 4
2013-08-06 2
2013-09-03 5
group_c 2013-07-09 1
2013-09-03 2
loc_b group_a 2013-10-01 3
Run Code Online (Sandbox Code Playgroud)
使用from_product,我们可以创建一个新的多索引.这个新索引是来自旧索引所有级别的所有值的笛卡尔积.
new_index = pd.MultiIndex.from_product(df.index.levels)
Run Code Online (Sandbox Code Playgroud)
使用新索引重新索引现有数据框.
new_df = df.reindex(new_index)
Run Code Online (Sandbox Code Playgroud)
现在存在所有可能的组合.缺失值为null(NaN).
扩展的重新索引数据框如下所示:
Value
loc_a group_a 2013-06-11 22.0
2013-07-02 35.0
2013-07-09 14.0
2013-07-30 9.0
2013-08-06 4.0
2013-09-03 40.0
2013-10-01 18.0
group_b 2013-06-11 NaN
2013-07-02 NaN
2013-07-09 4.0
2013-07-30 NaN
2013-08-06 2.0
2013-09-03 5.0
2013-10-01 NaN
group_c 2013-06-11 NaN
2013-07-02 NaN
2013-07-09 1.0
2013-07-30 NaN
2013-08-06 NaN
2013-09-03 2.0
2013-10-01 NaN
loc_b group_a 2013-06-11 NaN
2013-07-02 NaN
2013-07-09 NaN
2013-07-30 NaN
2013-08-06 NaN
2013-09-03 NaN
2013-10-01 3.0
group_b 2013-06-11 NaN
2013-07-02 NaN
2013-07-09 NaN
2013-07-30 NaN
2013-08-06 NaN
2013-09-03 NaN
2013-10-01 NaN
group_c 2013-06-11 NaN
2013-07-02 NaN
2013-07-09 NaN
2013-07-30 NaN
2013-08-06 NaN
2013-09-03 NaN
2013-10-01 NaN
Run Code Online (Sandbox Code Playgroud)
您可以看到新数据框中的数据已从整数转换为浮点数.Pandas在整数列中不能有空值.或者,我们可以将所有空值转换为0,并将数据转换回整数.
new_df = new_df.fillna(0).astype(int)
Run Code Online (Sandbox Code Playgroud)
结果
Value
loc_a group_a 2013-06-11 22
2013-07-02 35
2013-07-09 14
2013-07-30 9
2013-08-06 4
2013-09-03 40
2013-10-01 18
group_b 2013-06-11 0
2013-07-02 0
2013-07-09 4
2013-07-30 0
2013-08-06 2
2013-09-03 5
2013-10-01 0
group_c 2013-06-11 0
2013-07-02 0
2013-07-09 1
2013-07-30 0
2013-08-06 0
2013-09-03 2
2013-10-01 0
loc_b group_a 2013-06-11 0
2013-07-02 0
2013-07-09 0
2013-07-30 0
2013-08-06 0
2013-09-03 0
2013-10-01 3
group_b 2013-06-11 0
2013-07-02 0
2013-07-09 0
2013-07-30 0
2013-08-06 0
2013-09-03 0
2013-10-01 0
group_c 2013-06-11 0
2013-07-02 0
2013-07-09 0
2013-07-30 0
2013-08-06 0
2013-09-03 0
2013-10-01 0
Run Code Online (Sandbox Code Playgroud)
Tom*_*ger 15
你的问题并不清楚你到底错过了哪些日期; 我只是假设你想填写NaN
你在其他地方观察过的任何日期.如果这个假设有问题,我的解决方案将不得不修改.
旁注:包含一条线来创建它可能会很不错 DataFrame
In [55]: df = pd.DataFrame({'A': ['loc_a'] * 12 + ['loc_b'],
....: 'B': ['group_a'] * 7 + ['group_b'] * 3 + ['group_c'] * 2 + ['group_a'],
....: 'Date': ["2013-06-11",
....: "2013-07-02",
....: "2013-07-09",
....: "2013-07-30",
....: "2013-08-06",
....: "2013-09-03",
....: "2013-10-01",
....: "2013-07-09",
....: "2013-08-06",
....: "2013-09-03",
....: "2013-07-09",
....: "2013-09-03",
....: "2013-10-01"],
....: 'Value': [22, 35, 14, 9, 4, 40, 18, 4, 2, 5, 1, 2, 3]})
In [56]:
In [56]: df.Date = pd.to_datetime(df.Date)
In [57]: df = df.set_index(['A', 'B', 'Date'])
In [58]:
In [58]: print(df)
Value
A B Date
loc_a group_a 2013-06-11 22
2013-07-02 35
2013-07-09 14
2013-07-30 9
2013-08-06 4
2013-09-03 40
2013-10-01 18
group_b 2013-07-09 4
2013-08-06 2
2013-09-03 5
group_c 2013-07-09 1
2013-09-03 2
loc_b group_a 2013-10-01 3
Run Code Online (Sandbox Code Playgroud)
为了填充未观察到的值,我们将使用unstack
和stack
方法.拆散将创建NaN
我们感兴趣的s,然后我们将它们叠加起来使用.
In [71]: df.unstack(['A', 'B'])
Out[71]:
Value
A loc_a loc_b
B group_a group_b group_c group_a
Date
2013-06-11 22 NaN NaN NaN
2013-07-02 35 NaN NaN NaN
2013-07-09 14 4 1 NaN
2013-07-30 9 NaN NaN NaN
2013-08-06 4 2 NaN NaN
2013-09-03 40 5 2 NaN
2013-10-01 18 NaN NaN 3
In [59]: df.unstack(['A', 'B']).fillna(0).stack(['A', 'B'])
Out[59]:
Value
Date A B
2013-06-11 loc_a group_a 22
group_b 0
group_c 0
loc_b group_a 0
2013-07-02 loc_a group_a 35
group_b 0
group_c 0
loc_b group_a 0
2013-07-09 loc_a group_a 14
group_b 4
group_c 1
loc_b group_a 0
2013-07-30 loc_a group_a 9
group_b 0
group_c 0
loc_b group_a 0
2013-08-06 loc_a group_a 4
group_b 2
group_c 0
loc_b group_a 0
2013-09-03 loc_a group_a 40
group_b 5
group_c 2
loc_b group_a 0
2013-10-01 loc_a group_a 18
group_b 0
group_c 0
loc_b group_a 3
Run Code Online (Sandbox Code Playgroud)
根据需要重新排序索引级别.
我不得不fillna(0)
在中间滑动,以免NaN
s掉落. stack
确实dropna
有争议.我认为将其设置为false将保持所有NaN
行.可能有一个bug?
归档时间: |
|
查看次数: |
7060 次 |
最近记录: |