问题描述
给出数据框：
df = pd.DataFrame({'col1': ['A', 'A', 'A','B','B'], 'col2': ['type1', 'type2', 'type1', 'type2', 'type1'] , 'hour': ['18:03:30','18:00:48', '18:13:46', '18:11:29', '18:06:31'] })
col1 col2 hour
A type1 18:03:30 # Drop this row as (A type1) already present
A type2 18:00:48
A type1 18:13:46 # keep this row as (A type1) already present.
B type2 18:11:29
B type1 18:06:31
我想基于col1，col2删除重复项。
例如（行（0）：A类型1，行（2）：A类型1）
eg.(row(0): A type1, row(2): A type1)
保持仅保留具有最近小时的行，例如（18:13:46）。
keeping only the row that has the latest hour eg.(18:13:46).
我尝试使用 groupby 返回基于col1的子集，并尝试使用 drop_duplicates 将重复项删除到col2中。我需要找到一种通过条件（最近时间）的方法
I tried using groupby to return subset based on col1, and drop_duplicates to drop the duplicate in col2. I need to find a way to pass the condition (latest hour)
示例代码：
for key, grp in df.groupby('col1'):
grp.drop_duplicates(subset='col2', keep="LATEST OF HOUR")
预期结果：
col1 col2 hour
A type1 18:03:30
A type2 18:00:48
B type2 18:11:29
B type1 18:06:31
编辑添加上下文
我的原始数据帧较大，该解决方案还需要工作：
EDIT adding context
my original dataframe is larger, the solution needs to work for also:
col1 col2 other hour
A type1 h 18:03:30 # Drop this row as (A type1) already present
A type2 ss 18:00:48
A type1 ll 18:13:46 # keep this row as (A type1) already present
B type2 mm 18:11:29
B type1 jj 18:06:31
它仍然需要删除lu mn基于小时
it would still need to drop the column based on the hour
推荐答案
按照anky_91的评论，我这样解决了它：
Following anky_91's comment I solved it like this:
df.sort_values('hour').drop_duplicates(['col1','col2'] , keep = 'last')
此选项基于小时列进行排序，因此您可以确保keep ='last'获取最后一个元素
This sorts based on the column 'hour' so that you are sure that keep='last' gets the last element
这篇关于 pandas ：根据col [B]的条件，将col [A]中的重复项保留在行中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！