从唯一 ID 创建具有第一个和最后一个日期的列

Question

I have the dataset below and I would like to extract the first and last appearance date from that specific ID. The last and first appearance is what the result should be:

id            date           last_apparence     first_apparence
653777        2021-02-19     2021-02-19         2021-02-19
1873547       2021-02-19     2021-02-19         2021-02-19
657443        2021-02-19     2021-02-19         2021-02-19
653777        2021-02-20     2021-02-20         2021-02-19

So, for example, the ID 653777 shows up on the 19th and 20th, in this case, the first appearance would be 19th and the last appearance would be 20th. I tried to use

I tried to use this code below but get the same value for the entire column.

df['latest_apparence'] = df['date'][df.index[-1]]

My last approach was to have a groupby but even trying a bunch of different groups, the closest thing I got was something similar to the countif formula from excel but then I don`t get the first/last date, only how many times the id shows in the dataset

df.groupby(['id'])[['date']].count()

Does anybody have any idea what is the best way to get this result?

thanks

Answer 1

Let us try groupby with transform

df['latest_apparence'] = df.groupby('id')['date'].transform('max')
df['first_apparence'] = df.groupby('id')['date'].transform('min')

Answer 2

Depending on how you want your data, you can do

dt = df.groupby('id')['date'].agg(['min','max']).rename(columns={'min':'first_date','max':'last_date'})

print(dt)

# or broadcasting as Beny suggested with 
# AGG is aggregation of min and max
# df.groupby('id')['date'].transform(AGG)