我有这个dataframe
:
date value source
0 2020-02-14 0.438767 L8-SR
1 2020-02-15 0.422867 S2A-SR
2 2020-03-01 0.657453 L8-SR
3 2020-03-01 0.603989 S2B-SR
4 2020-03-11 0.717264 S2B-SR
5 2020-04-02 0.737118 L8-SR
我想在groupby
列旁边date
,在其中我根据从source
列中选择的排名/重要性来保留行。例如,我的排名是L8-SR> S2B-SR> GP6_r,这意味着对于所有具有相同日期的行,请将行保留在source==L8-SR
处,如果其中没有包含L8-SR,则将行保留在{{ 1}}等。如何在source==S2B-SR
输出应如下所示:
pandas groupby
答案 0 :(得分:1)
让我们尝试category
dtype和drop_duplicates
:
orders = ['L8-SR','S2B-SR','GP6_r']
df.source = df.source.astype('category')
df.source.cat.set_categories(orders, ordered=True)
df.sort_values(['date','source']).drop_duplicates(['date'])
输出:
date value source
0 2020-02-14 0.438767 L8-SR
1 2020-02-15 0.422867 S2A-SR
2 2020-03-01 0.657453 L8-SR
4 2020-03-11 0.717264 S2B-SR
5 2020-04-02 0.737118 L8-SR
答案 1 :(得分:0)
请按以下代码尝试按操作分组。要在此操作后订购,您可以执行排序方式:
# Import pandas library
import pandas as pd
# Declare a data dictionary contains the data mention in table
pandasdata_dict = {'date':['2020-02-14', '2020-02-15', '2020-03-01', '2020-03-01', '2020-03-11', '2020-04-02'],
'value':[0.438767, 0.422867, 0.657453, 0.603989, 0.717264, 0.737118],
'source':['L8-SR', 'S2A-SR', 'L8-SR', 'S2B-SR', 'S2B-SR', 'L8-SR']}
# Convert above dictionary data to the data frame
df = pd.DataFrame(pandasdata_dict)
# display data frame
df
# Convert date field to datetime
df["date"] = pd.to_datetime(df["date"])
# Once conversion done then do the group by operation on the data frame with date field
df.groupby([df['date'].dt.date])