在带有 groupby 的时间序列列上使用 Pandas .diff()

use*_*044 4 python python-2.7 pandas

我有一个CSV客户购买的文件,没有特定的顺序,我读到了Pandas Dataframe. 我想为每次购买添加一列,并按客户分组显示自上次购买以来已经过去了多长时间。我不确定差异在哪里,但它们太大了(即使在几秒钟内)。

CSV:

Customer Id,Purchase Date
4543,1/1/2015
4543,2/5/2015
4543,3/15/2015
2322,1/1/2015
2322,3/1/2015
2322,2/1/2015
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Python:

import pandas as pd
import time
start = time.time()
data = pd.read_csv('data.csv', low_memory=False)
data = data.sort_values(by=['Customer Id', 'Purchase Date'])
data['Purchase Date'] = pd.to_datetime(data['Purchase Date'])
data['Purchase Difference'] = (data.groupby(['Customer Id'])['Purchase Date']
                         .diff()
                         .fillna('-')
                       )
print data
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输出:

    Customer Id Purchase Date Purchase Difference
3         2322    2015-01-01                   -
5         2322    2015-02-01    2678400000000000
4         2322    2015-03-01    2419200000000000
0         4543    2015-01-01                   -
1         4543    2015-02-05    3024000000000000
2         4543    2015-03-15    328320000000000
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期望输出:

   Customer Id Purchase Date  Purchase Difference
3         2322    2015-01-01                  -
5         2322    2015-02-01              31 days
4         2322    2015-03-01              28 days
0         4543    2015-01-01                  -
1         4543    2015-02-05              35 days
2         4543    2015-03-15              38 days
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Ale*_*der 5

一旦它被转换为时间戳,您就可以应用diff到该Purchase Date列。

df['Purchase Date'] = pd.to_datetime(df['Purchase Date'])
df.sort_values(['Customer Id', 'Purchase Date'], inplace=True)    
df['Purchase Difference'] = \
    [str(n.days) + ' day' + 's' if n > pd.Timedelta(days=1) else '' if pd.notnull(n) else "" 
     for n in df.groupby('Customer Id', sort=False)['Purchase Date'].diff()]

>>> df
   Customer Id Purchase Date Purchase Difference
3         2322    2015-01-01                    
5         2322    2015-02-01             31 days
4         2322    2015-03-01             28 days
0         4543    2015-01-01                    
1         4543    2015-02-05             35 days
2         4543    2015-03-15             38 days
6         4543    2015-03-15                    
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