假设我有这样的pandas DataFrame:
Suppose I have pandas DataFrame like this:
>>> df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4],'value':[1,2,3,1,2,3,4,1,1]}) >>> df id value 0 1 1 1 1 2 2 1 3 3 2 1 4 2 2 5 2 3 6 2 4 7 3 1 8 4 1我想获得一个新的DataFrame,其中每个ID的前2个记录都是这样,
I want to get a new DataFrame with top 2 records for each id, like this:
id value 0 1 1 1 1 2 3 2 1 4 2 2 7 3 1 8 4 1我可以对分组依据中的一组记录进行编号:
I can do it with numbering records within group after group by:
>>> dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index() >>> dfN id level_1 index value 0 1 0 0 1 1 1 1 1 2 2 1 2 2 3 3 2 0 3 1 4 2 1 4 2 5 2 2 5 3 6 2 3 6 4 7 3 0 7 1 8 4 0 8 1 >>> dfN[dfN['level_1'] <= 1][['id', 'value']] id value 0 1 1 1 1 2 3 2 1 4 2 2 7 3 1 8 4 1但是,有没有更有效/更优雅的方法来做到这一点?还有一种更优雅的方法来对每个组中的数字进行记录(例如SQL窗口函数 row_number ()).
But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).
推荐答案您尝试过df.groupby('id').head(2)
生成的输出:
>>> df.groupby('id').head(2) id value id 1 0 1 1 1 1 2 2 3 2 1 4 2 2 3 7 3 1 4 8 4 1(请记住,根据数据的不同,您可能需要先进行订购/排序)
(Keep in mind that you might need to order/sort before, depending on your data)
如发问者所述,使用df.groupby('id').head(2).reset_index(drop=True)删除多义词并展平结果.
As mentioned by the questioner, use df.groupby('id').head(2).reset_index(drop=True) to remove the multindex and flatten the results.
>>> df.groupby('id').head(2).reset_index(drop=True) id value 0 1 1 1 1 2 2 2 1 3 2 2 4 3 1 5 4 1更多推荐
pandas 在每个组中获得最高的n条记录
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