![](/img/trans.png)
[英]How aggregate data by “group by” hour from datetime python DataFrame?
[英]How to extract 1 hour data from python dataframe?
我有一個由長度和時間組成的Z6A8064B5DF4794555555555555557DZ , 請告訴我)
Length,Time
0.0,2019-08-26 14:46:36.040
0.0,2019-08-26 14:46:36.043
0.0,2019-08-26 14:56:40.156
0.0,2019-08-26 14:56:40.160
6033.0,2019-08-26 15:01:22.963
6033.0,2019-08-26 15:01:23.034
0.0,2019-08-26 15:01:32.759
0.0,2019-08-26 15:01:32.763
0.0,2019-08-26 16:05:13.365
0.0,2019-08-26 16:05:13.368
0.0,2019-08-26 16:12:08.760
0.0,2019-08-26 16:12:08.760
2658.0,2019-08-26 16:14:48.129
2658.0,2019-08-26 17:14:48.132
0.0,2019-08-26 17:22:49.358
0.0,2019-08-26 17:22:49.361
0.0,2019-08-26 17:22:50.152
0.0,2019-08-26 17:22:50.156
0.0,2019-08-26 17:23:08.735
0.0,2019-08-26 18:23:08.735
0.0,2019-08-26 18:23:08.738
0.0,2019-08-26 18:23:08.738
謝謝
您可以在boolean indexing
中過濾每個數據的最大小時數:
h = df['Time'].dt.hour
df = df[h.eq(h.max())]
print (df)
Length Time
19 0.0 2019-08-26 18:23:08.735
20 0.0 2019-08-26 18:23:08.738
21 0.0 2019-08-26 18:23:08.738
將Time
設置為datetime
時間索引,然后按小時將 select 行設置為
df = pd.read_csv('d1.csv')
df.Time = pd.to_datetime(df.Time)
df.set_index('Time',inplace=True)
df.loc['2019-08-26 14']
Length
Time
2019-08-26 14:46:36.040 0.0
2019-08-26 14:46:36.043 0.0
2019-08-26 14:56:40.156 0.0
2019-08-26 14:56:40.160 0.0
df.loc['2019-08-26 15']
Length
Time
2019-08-26 15:01:22.963 6033.0
2019-08-26 15:01:23.034 6033.0
2019-08-26 15:01:32.759 0.0
2019-08-26 15:01:32.763 0.0
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.