Question

我有一个CSV客户购买文件，我没有按特定顺序阅读Pandas Dataframe。我想为每次购买添加一列，并显示自上次购买以来已经过了多少时间，按客户分组。我不知道它在哪里得到差异，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

的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

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

期望的输出：

   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

Answer 1

我认为您可以添加read_csv参数parse_dates来解析datetime，sort_values以及使用groupby解析diff：

import pandas as pd
import io

temp=u"""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"""
#after testing replace io.StringIO(temp) to filename
data = pd.read_csv(io.StringIO(temp), parse_dates=['Purchase Date'])

data.sort_values(by=['Customer Id', 'Purchase Date'], inplace=True)

data['Purchase Difference'] = data.groupby(['Customer Id'])['Purchase Date'].diff()
print data
   Customer Id Purchase Date  Purchase Difference
3         2322    2015-01-01                  NaT
5         2322    2015-02-01              31 days
4         2322    2015-03-01              28 days
0         4543    2015-01-01                  NaT
1         4543    2015-02-05              35 days
2         4543    2015-03-15              38 days

Answer 2

一旦转换为时间戳，您就可以将diff应用于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

在具有groupby

2 个答案: