[英]pandas groupby date select earliest per day
I have the following dataset: 我有以下数据集:
value timestamp
0 Fire 2017-10-03 14:33:52
1 Water 2017-10-04 14:33:48
2 Fire 2017-10-04 14:33:45
3 Fire 2017-10-05 14:33:30
4 Water 2017-10-03 14:33:40
5 Water 2017-10-05 14:32:13
6 Water 2017-10-04 14:32:01
7 Fire 2017-10-03 14:31:55
I want to group this set by timestamp
per day and then only select the earliest row per day. 我想按每天的timestamp
将此集合分组,然后仅选择每天最早的行。 For the above example the following should be the result: 对于上面的示例,结果应为:
value timestamp
1 Water 2017-10-05 14:32:13
2 Water 2017-10-04 14:32:01
3 Fire 2017-10-03 14:31:55
For example, for the day 2017-10-03
there are 3 entries but I only want the earliest on that day. 例如,对于2017-10-03
3日这一天,有3个条目,但是我只希望最早的那一天。
If you have unique index, you can use idxmin
on timestamp
to find out the indices of the minimum timestamp and extract them with loc
: 如果您有唯一索引,则可以在timestamp
上使用idxmin
来找出最小时间戳的索引,并使用loc
提取它们:
df.timestamp = pd.to_datetime(df.timestamp)
df.loc[df.groupby(df.timestamp.dt.date, as_index=False).timestamp.idxmin()]
# value timestamp
#7 Fire 2017-10-03 14:31:55
#6 Water 2017-10-04 14:32:01
#5 Water 2017-10-05 14:32:13
Just Making Sure 只是确定
df.timestamp = pd.to_datetime(df.timestamp)
Solution 解
d1 = df.sort_values('timestamp')
d1[~d1.timestamp.dt.date.duplicated()]
value timestamp
7 Fire 2017-10-03 14:31:55
6 Water 2017-10-04 14:32:01
5 Water 2017-10-05 14:32:13
Use dt.floor
and head
: 使用dt.floor
和head
:
df.sort_values('timestamp').groupby(df['timestamp'].dt.floor('D')).head(1)
Output: 输出:
value timestamp
7 Fire 2017-10-03 14:31:55
6 Water 2017-10-04 14:32:01
5 Water 2017-10-05 14:32:13
Or 要么
df.groupby(df.timestamp.dt.date).apply(lambda x:x[x.timestamp==min(x.timestamp)])
Out[714]:
value timestamp
timestamp
2017-10-03 7 Fire 2017-10-03 14:31:55
2017-10-04 6 Water 2017-10-04 14:32:01
2017-10-05 5 Water 2017-10-05 14:32:13
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