[英]How to keep first two duplicates in a pandas dataframe?
I have a question in regards to finding duplicates in a dataframe, and removing duplicates in a dataframe using a specific column. 关于在数据框中查找重复项并使用特定列删除数据框中的重复项,我有一个问题。 Here is what I am trying to accomplish: 这是我要完成的工作:
Is it possible to remove duplicates but keep the first 2? 是否可以删除重复项但保留前两个?
Here is an example of my current dataframe called df and take a look at the bracket notes I have placed below to give you an idea. 这是我当前的数据框df的示例,请看一下下面放置的方括号内的注释,以使您有所了解。
Note: If 'Roll' = 1 then I want to look at the Date column, see if there is a second duplicate Date in that column... keep those two and delete any others. 注意:如果'Roll'= 1,那么我想查看Date列,看看该列中是否还有第二个重复的Date ...保留这两个并删除其他任何日期。
Date Open High Low Close Roll Dupes
1 19780106 236.00 237.50 234.50 235.50 0 NaN
2 19780113 235.50 239.00 235.00 238.25 0 NaN
3 19780120 238.00 239.00 234.50 237.00 0 NaN
4 19780127 237.00 238.50 235.50 236.00 1 NaN (KEEP)
5 19780203 236.00 236.00 232.25 233.50 0 NaN (KEEP)
6 19780127 237.00 238.50 235.50 236.00 0 NaN (KEEP)
7 19780203 236.00 236.00 232.25 233.50 0 NaN (DELETE)
8 19780127 237.00 238.50 235.50 236.00 0 NaN (DELETE)
9 19780203 236.00 236.00 232.25 233.50 0 NaN (DELETE)
This is what is currently removing the dupes BUT it's removing all dupes (obviously) 这是当前正在删除重复对象的东西,但正在删除所有重复对象(显然)
df = df.drop_duplicates('Date')
EDIT: I forgot to mention something, the only duplicate I want to keep is if column 'Roll' = 1 if it does, then keep that row and the next one that matches based on column 'Date' 编辑:我忘了提些什么,我要保留的唯一重复项是,如果列'Roll'= 1,如果保留的话,则保留该行以及根据列'Date'匹配的下一行
Using head
with a groupby keeps the first x entries in each group, which I think accomplishes what you want. 将head
与groupby一起使用可在每个组中保留前x个条目,我认为这可以满足您的要求。
In [52]: df.groupby('Date').head(2)
Out[52]:
Date Open High Low Close Roll
1 19780106 236.0 237.5 234.50 235.50 0
2 19780113 235.5 239.0 235.00 238.25 0
3 19780120 238.0 239.0 234.50 237.00 0
4 19780127 237.0 238.5 235.50 236.00 0
5 19780203 236.0 236.0 232.25 233.50 0
6 19780127 237.0 238.5 235.50 236.00 0
7 19780203 236.0 236.0 232.25 233.50 0
Edit: 编辑:
In [16]: df['dupe_count'] = df.groupby('Date')['Roll'].transform('max') + 1
In [17]: df.groupby('Date', as_index=False).apply(lambda x: x.head(x['dupe_count'].iloc[0]))
Out[17]:
Date Open High Low Close Roll Dupes dupe_count
0 1 19780106 236.0 237.5 234.50 235.50 0 NaN 1
1 2 19780113 235.5 239.0 235.00 238.25 0 NaN 1
2 3 19780120 238.0 239.0 234.50 237.00 0 NaN 1
3 4 19780127 237.0 238.5 235.50 236.00 1 NaN 2
6 19780127 237.0 238.5 235.50 236.00 0 NaN 2
4 5 19780203 236.0 236.0 232.25 233.50 0 NaN 1
Assuming Roll
can only take the values 0 and 1, if you do 假设Roll
只能取值0和1,如果您这样做
df.groupby(['Date', 'Roll'], as_index=False).first()
you will get two rows for dates for which one of the rows had Roll = 1
and only one row for dates which have only Roll = 0
, which I think is what you want. 您将获得两行日期,其中某一行的Roll = 1
,只有一行日期的Roll = 0
,这就是您想要的。
If passed as_index=False
so that the group keys don't end up in the index as discussed in your comment. 如果通过as_index=False
传递,则组密钥不会像您的注释中所讨论的那样最终出现在索引中。
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