Python 去聚合

Question

我有一个在两个日期之间聚合的数据集，我想通过将总数除以这些日期之间的天数来每天对其进行去聚合。 作为样本

StoreID Date_Start    Date_End     Total_Number_of_sales
78       12/04/2015    17/05/2015    79089
80       12/04/2015    17/05/2015    79089

我想要的数据集是：

StoreID Date         Number_Sales 
78         12/04/2015    79089/38(as there are 38 days in between)
78         13/04/2015    79089/38(as there are 38 days in between) 
78         14/04/2015    79089/38(as there are 38 days in between)
78         ...
78         17/05/2015    79089/38(as there are 38 days in between)

任何帮助都会很有用。 谢谢

Answer 1

我不确定这是否正是你想要的，但你可以试试这个（我添加了另一个假想行）：

import datetime as dt
df = pd.DataFrame({'date_start':['12/04/2015','17/05/2015'],
                   'date_end':['18/05/2015','10/06/2015'],
                   'sales':[79089, 1000]})

df['date_start'] = pd.to_datetime(df['date_start'], format='%d/%m/%Y')
df['date_end'] = pd.to_datetime(df['date_end'], format='%d/%m/%Y')
df['days_diff'] = (df['date_end'] - df['date_start']).dt.days


master_df = pd.DataFrame(None)
for row in df.index:
    new_df = pd.DataFrame(index=pd.date_range(start=df['date_start'].iloc[row],
                          end = df['date_end'].iloc[row],
                          freq='d'))
    new_df['number_sales'] = df['sales'].iloc[row] / df['days_diff'].iloc[row]
    master_df = pd.concat([master_df, new_df], axis=0)

首先将字符串日期转换为日期时间对象（这样您就可以计算范围之间的天数），然后根据日期范围创建一个新索引，然后划分销售额。该循环将数据帧的每一行粘贴到“扩展”数据帧中，然后将它们连接到一个主数据帧中。

Answer 2

创建一个新的数据框怎么样？

start = pd.to_datetime(df['Date_Start'].values[0], dayfirst=True)
end = pd.to_datetime(df['Date_End'].values[0], dayfirst=True)
idx = pd.DatetimeIndex(start=start, end=end, freq='D')
res = pd.DataFrame(df['Total_Number_of_sales'].values[0]/len(idx), index=idx, columns=['Number_Sales'])

产量

In[42]: res.head(5)
Out[42]: 
            Number_Sales
2015-04-12   2196.916667
2015-04-13   2196.916667
2015-04-14   2196.916667
2015-04-15   2196.916667
2015-04-16   2196.916667

如果您有多个商店（根据您的评论和编辑），那么您可以遍历所有行，计算销售额并在之后连接生成的数据框。

df = pd.DataFrame({'Store_ID': [78, 78, 80],
    'Date_Start': ['12/04/2015', '18/05/2015', '21/06/2015'],
                   'Date_End': ['17/05/2015', '10/06/2015', '01/07/2015'],
                   'Total_Number_of_sales': [79089., 50000., 25000.]})

to_concat = []
for _, row in df.iterrows():
    start = pd.to_datetime(row['Date_Start'], dayfirst=True)
    end = pd.to_datetime(row['Date_End'], dayfirst=True)
    idx = pd.DatetimeIndex(start=start, end=end, freq='D')
    sales = [row['Total_Number_of_sales']/len(idx)] * len(idx)
    id = [row['Store_ID']] * len(idx)
    res = pd.DataFrame({'Store_ID': id, 'Number_Sales':sales}, index=idx)
    to_concat.append(res)

res = pd.concat(to_concat)

有定义更优雅的解决方案，看看这个thread的例子。

Answer 3

考虑使用DataFrame 构造函数迭代主数据帧的每一行来构建数据帧列表。每次迭代都会扩展从 Start_Date 到范围结束的天数序列，并按天数对总销售额进行所需的销售划分：

from io import StringIO
import pandas as pd
from datetime import timedelta

txt = '''StoreID Date_Start    Date_End     Total_Number_of_sales
78       12/04/2015    17/05/2015    79089
80       12/04/2015    17/05/2015    89089'''

df = pd.read_table(StringIO(txt), sep="\s+", parse_dates=[1, 2], dayfirst=True)
df['Diff_Days'] = (df['Date_End'] - df['Date_Start']).dt.days

def calc_days_sales(row): 
    long_df = pd.DataFrame({'StoreID': row['StoreID'],
                            'Date': [row['Date_Start'] + timedelta(days=i) 
                                          for i in range(row['Diff_Days']+1)],
                            'Number_Sales': row['Total_Number_of_sales'] / row['Diff_Days']})    
    return long_df

df_list = [calc_days_sales(row) for i, row in df.iterrows()]

final_df = pd.concat(df_list).reindex(['StoreID', 'Date', 'Number_Sales'], axis='columns')

print(final_df.head(10))
#    StoreID       Date  Number_Sales
# 0       78 2015-04-12   2259.685714
# 1       78 2015-04-13   2259.685714
# 2       78 2015-04-14   2259.685714
# 3       78 2015-04-15   2259.685714
# 4       78 2015-04-16   2259.685714
# 5       78 2015-04-17   2259.685714
# 6       78 2015-04-18   2259.685714
# 7       78 2015-04-19   2259.685714
# 8       78 2015-04-20   2259.685714
# 9       78 2015-04-21   2259.685714

Python 3.6 不需要在末尾重新索引，因为数据框的输入字典将被排序。