如何使用python生成ID号？

Question

I have a dataframe. I want to create a unique ID number for each person and create a column based the person and date(weekly).

import pandas as pd
df = pd.DataFrame({ 'name':['one','one','two','two','two','three','four'],
                     'date':['2019-05-01','2019-05-08','2019-05-01','2019-05-08','2019-05-15','2019-05-01','2019-05-15'],
                    "a":range(7)})
df['date'] = pd.to_datetime(df['date'],yearfirst=True)
df = df.sort_values(['name','date'])
print(df)

This is the data:

    name       date  a
6   four 2019-05-15  6
0    one 2019-05-01  0
1    one 2019-05-08  1
5  three 2019-05-01  5
2    two 2019-05-01  2
3    two 2019-05-08  3
4    two 2019-05-15  4

The expected result is

    name       date  a    id    week
6   four 2019-05-15  6     1    3
0    one 2019-05-01  0     2    1
1    one 2019-05-08  1     2    2
5  three 2019-05-01  5     3    1 
2    two 2019-05-01  2     4    1
3    two 2019-05-08  3     4    2
4    two 2019-05-15  4     4    3

How can I get the "id" and "week"? Thank you!

Answer 1

Like @cs95 commented use GroupBy.ngroup with division days by 7 with numpy.ceil :

df["Id"] = df.groupby("name").ngroup() + 1
df['week'] = np.ceil(df.date.dt.day / 7).astype(int)
print (df)

    name       date  a  Id  week
6   four 2019-05-15  6   1     3
0    one 2019-05-01  0   2     1
1    one 2019-05-08  1   2     2
5  three 2019-05-01  5   3     1
2    two 2019-05-01  2   4     1
3    two 2019-05-08  3   4     2
4    two 2019-05-15  4   4     3

Or:

df["Id"] = df.groupby("name").ngroup() + 1
df['week'] =  df.groupby("date").ngroup() + 1
print (df)

    name       date  a  Id  week
6   four 2019-05-15  6   1     3
0    one 2019-05-01  0   2     1
1    one 2019-05-08  1   2     2
5  three 2019-05-01  5   3     1
2    two 2019-05-01  2   4     1
3    two 2019-05-08  3   4     2
4    two 2019-05-15  4   4     3

Answer 2

I use cumsum to get df['id'] and groupby on df.date to get df['week'] :

df['id'] = df.name.ne(df.name.shift()).cumsum()
df['week'] = df.date.groupby(df.date).ngroup() + 1


Out[408]:
    name       date  a  id  week
6   four 2019-05-15  6   1     3
0    one 2019-05-01  0   2     1
1    one 2019-05-08  1   2     2
5  three 2019-05-01  5   3     1
2    two 2019-05-01  2   4     1
3    two 2019-05-08  3   4     2
4    two 2019-05-15  4   4     3