如何在保留行的同时用NaN替换每个组的重复项?
我需要保留行而不删除它,或者将第一个原始值保留在它首先显示的位置。
import pandas as pd
from datetime import timedelta
df = pd.DataFrame({
'date': ['2019-01-01 00:00:00','2019-01-01 01:00:00','2019-01-01 02:00:00', '2019-01-01 03:00:00',
'2019-09-01 02:00:00','2019-09-01 03:00:00','2019-09-01 04:00:00', '2019-09-01 05:00:00'],
'value': [10,10,10,10,12,12,12,12],
'ID': ['Jackie','Jackie','Jackie','Jackie','Zoop','Zoop','Zoop','Zoop',]
})
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)
date value ID
0 2019-01-01 00:00:00 10 Jackie
1 2019-01-01 01:00:00 10 Jackie
2 2019-01-01 02:00:00 10 Jackie
3 2019-01-01 03:00:00 10 Jackie
4 2019-09-01 02:00:00 12 Zoop
5 2019-09-01 03:00:00 12 Zoop
6 2019-09-01 04:00:00 12 Zoop
7 2019-09-01 05:00:00 12 Zoop
所需数据框:
date value ID
0 2019-01-01 00:00:00 10 Jackie
1 2019-01-01 01:00:00 NaN Jackie
2 2019-01-01 02:00:00 NaN Jackie
3 2019-01-01 03:00:00 NaN Jackie
4 2019-09-01 02:00:00 12 Zoop
5 2019-09-01 03:00:00 NaN Zoop
6 2019-09-01 04:00:00 NaN Zoop
7 2019-09-01 05:00:00 NaN Zoop
编辑:
重复的值只能在与频率无关的同一天删除。因此,如果值10在1月1日出现两次,在1月2日出现3次,则值10应该只在1月1日出现一次,而在1月2日出现一次。
答案 0 :(得分:3)
我假设您检查value
和ID
列的重复项,并进一步检查date
列的date
df.loc[df.assign(d=df.date.dt.date).duplicated(['value','ID', 'd']), 'value'] = np.nan
Out[269]:
date value ID
0 2019-01-01 00:00:00 10.0 Jackie
1 2019-01-01 01:00:00 NaN Jackie
2 2019-01-01 02:00:00 NaN Jackie
3 2019-01-01 03:00:00 NaN Jackie
4 2019-09-01 02:00:00 12.0 Zoop
5 2019-09-01 03:00:00 NaN Zoop
6 2019-09-01 04:00:00 NaN Zoop
7 2019-09-01 05:00:00 NaN Zoop
正如@Trenton建议的那样,您可以使用pd.NA
来避免导入numpy
(注意:如@rafaelc建议:这是链接,说明pd.NA
和np.nan
https://pandas.pydata.org/pandas-docs/stable/whatsnew/v1.0.0.html#experimental-na-scalar-to-denote-missing-values之间的详细区别)
df.loc[df.assign(d=df.date.dt.date).duplicated(['value','ID', 'd']), 'value'] = pd.NA
Out[273]:
date value ID
0 2019-01-01 00:00:00 10 Jackie
1 2019-01-01 01:00:00 <NA> Jackie
2 2019-01-01 02:00:00 <NA> Jackie
3 2019-01-01 03:00:00 <NA> Jackie
4 2019-09-01 02:00:00 12 Zoop
5 2019-09-01 03:00:00 <NA> Zoop
6 2019-09-01 04:00:00 <NA> Zoop
7 2019-09-01 05:00:00 <NA> Zoop
答案 1 :(得分:1)
如果对数据框进行排序,则可以正常工作-如您的示例:
import numpy as np # to be used for np.nan
df['duplicate'] = df['value'].shift(1) # create a duplicate column
df['value'] = df.apply(lambda x: np.nan if x['value'] == x['duplicate'] \
else x['value'], axis=1) # conditional replace
df = df.drop('duplicate', axis=1) # drop helper column
答案 2 :(得分:1)
对日期进行分组,并获取第一个观察值(按时间排序时不一定是第一个),然后将结果合并回原始数据框。
df2 = df.groupby([df['date'].dt.date, 'ID'], as_index=False).first()
>>> df.drop(columns='value').merge(df2, on=['date', 'ID'], how='left')[df.columns]
date value ID
0 2019-01-01 00:00:00 10.0 Jackie
1 2019-01-01 01:00:00 NaN Jackie
2 2019-01-01 02:00:00 NaN Jackie
3 2019-01-01 03:00:00 NaN Jackie
4 2019-09-01 02:00:00 12.0 Zoop
5 2019-09-01 03:00:00 NaN Zoop
6 2019-09-01 04:00:00 NaN Zoop
7 2019-09-01 05:00:00 NaN Zoop