熊猫：在每一行中按其第一次出现和最后一次出现

Question

My data includes invoices and customers. One customer can have multiple invoices. One invoice belongs to always one customer. The invoices are updated daily (Report Date).

My goal is to calculate the age of the customer in days (see column "Age in Days"). In order to achieve this, I take the first occurrence of a customers report date and calculate the difference to the last occurrence of the report date.

eg Customer 1 occurs from 08-14 till 08-15. Therefore he/she is 1 day old.

Report Date  Invoice No   Customer No  Amount  Age in Days
2018-08-14   A            1            50$     1
2018-08-14   B            1            100$    1
2018-08-14   C            2            75$     2

2018-08-15   A            1            20$     1
2018-08-15   B            1            45$     1
2018-08-15   C            2            70$     2

2018-08-16   C            2            40$     1
2018-08-16   D            3            100$    0
2018-08-16   E            3            60$     0

I solved this, but however, very inefficiently and it takes too long. My data contains 26 million rows. Below I calculated the age for one customer only.

# List every customer no
customerNo = df["Customer No"].unique()
customer_age = []

# Testing for one specific customer
testCustomer = df.loc[df["Customer No"] == customerNo[0]]
testCustomer = testCustomer.sort_values(by="Report Date", ascending=True)

first_occur = testCustomer.iloc[0]['Report Date']
last_occur = testCustomer.iloc[-1]['Report Date']
age = (last_occur - first_occur).days

customer_age.extend([age] * len(testCustomer))
testCustomer.loc[:,'Customer Age']=customer_age 

Is there a better way to solve this problem?

Answer 1

If you need one value per customer, indicating its age you can use a group by(very common):

grpd = my_df.groupby('Customer No')['Report Date'].agg([min, max]).reset_index()
grpd['days_diff'] = (grpd['max'] - grpd['min']).dt.days

Answer 2

Use groupby.transform with first and last aggregations:

grps = df.groupby('Customer No')['Report Date']    
df['Age in Days'] = (grps.transform('last') - grps.transform('first')).dt.days

[out]

  Report Date Invoice No  Customer No Amount  Age in Days
0  2018-08-14          A            1    50$            1
1  2018-08-14          B            1   100$            1
2  2018-08-14          C            2    75$            2
3  2018-08-15          A            1    20$            1
4  2018-08-15          B            1    45$            1
5  2018-08-15          C            2    70$            2
6  2018-08-16          C            2    40$            2
7  2018-08-16          D            3   100$            0
8  2018-08-16          E            3    60$            0