如何用python生成id号？
如何用python生成id号？
提问于 2019-05-24 13:25:17
我有一个数据框架。我想为每个人创建一个唯一的ID号，并基于person和date(每周)创建一个列。
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)以下是数据：
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预期结果为
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如何获取"id“和"week"？谢谢!
回答 2
Stack Overflow用户
回答已采纳
发布于 2019-05-24 13:41:14
就像@cs95评论的那样，通过7与numpy.ceil一起使用带有分区天数的GroupBy.ngroup
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或者：
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 3Stack Overflow用户
发布于 2019-05-24 13:45:58
我使用cumsum获取df['id']，在df.date上使用groupby获取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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56286315
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