如何获得索引，多个发票之间的时差 - 熊猫

Question

我有一个包含来自多个客户的多张发票的pandas.DataFrame。我想找到一种优雅的方法来计算2张发票之间的时间，具体取决于客户。

我的数据框如下所示（索引是发票号，最后一列是我所期待的）：

         CustomerID         InvoiceDate  time between 2 orders
index                                  
536365      17850.0 2010-12-01 08:26:00  0 minutes (or np.nat)
536366      17850.0 2010-12-01 08:28:00  2 minutes
536367      13047.0 2010-12-01 08:34:00  0 minutes (It's a new customer)
536369      13047.0 2010-12-01 08:35:00  1 minute
536371      13748.0 2010-12-01 09:00:00  0 minute  (new customer)
536372      17850.0 2010-12-01 09:01:00  33 minutes (see line #2)
536373      17850.0 2010-12-01 09:02:00  1 minute
536374      15100.0 2010-12-01 09:09:00  0 minute

这是我到目前为止所发现的（但显然它不起作用！）

df = df.sort_values(['CustomerID', 'InvoiceDate']) #To order first according
df = df.set_index('index', drop = True)
for CustomerID in df['CustomerID'].unique():
    index = df.set_index('CustomerID').index.get_loc(CustomerID)
    df['Ordersep'].iloc[index] = df['InvoiceDate'].iloc[index].diff()

有什么好主意帮我吗？

Answer 1

你可以使用groupby()和diff()：

df.InvoiceDate = pd.to_datetime(df.InvoiceDate)
df["timedelta"] = df.groupby(["CustomerID"]).InvoiceDate.apply(lambda x: x.diff())

df
    index  CustomerID         InvoiceDate    timedelta
0  536365     17850.0 2010-12-01 08:26:00          NaT
1  536366     17850.0 2010-12-01 08:28:00     00:02:00
2  536367     13047.0 2010-12-01 08:34:00          NaT
3  536369     13047.0 2010-12-01 08:35:00     00:01:00
4  536371     13748.0 2010-12-01 09:00:00          NaT
5  536372     17850.0 2010-12-01 09:01:00     00:33:00
6  536373     17850.0 2010-12-01 09:02:00     00:01:00
7  536374     15100.0 2010-12-01 09:09:00          NaT

Answer 2

这应该有效，假设您已经在此之前对客户ID和发票日期进行了排序（可能稍微调整一下）

for customer_id in df.CustomerId.unique():
    matching_customer_mask = df.CustomerId == customer_id
    customer_df = df[matching_customer_mask]

    order_times = customer_df.InvoiceDate
    prev_order_times = customer_df.InvoiceDate.shift(1)

    df.loc[matching_customer_mask, 'Ordersep'] = order_times - prev_order_times

这样做是将发票日期列向下移动一步，然后计算您想要的差异。