Question

我有一个带有两列x和y的df。列y是x值的计数。 x值具有不同的计数。如何在不迭代行的情况下获得每个x的前两个y计数的结果。

示例df：

df = pd.DataFrame({"x": [101, 101, 101, 101, 201, 201, 201, 405, 405], "y": [1, 2, 3, 4, 1, 2, 3, 1, 2]})

     x  y
0  101  1
1  101  2
2  101  3
3  101  4
4  201  1
5  201  2
6  201  3
7  405  1
8  405  2

期望的结果：

Answer 1

你可以这样做：

def clean_dob(value):
  pass

class MyModel(ndb.Model):
  # ...

因此In [35]: df.loc[df.groupby(['x'])['y'].apply(lambda x: x.iloc[-2:]).index.get_level_values(1)] Out[35]: x y 2 101 3 3 101 4 5 201 2 6 201 3 7 405 1 8 405 2在＆＃39; x＆＃39;列并返回最后2个值，假设df已经按照您的显示排序。这会生成带有多索引的df，并且可以使用groupby

将第二级值用于索引回原始df

修改

要回答您的评论，您可以get_level_values再次使用groupby与transform一起将值重置为rank和1：

2

Answer 2

如果您的数据框没有排序，这是一个解决方案：

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不幸的是，In [1]: df.groupby('x')['y'].nlargest(2) Out[1]: x 101 3 4 2 3 201 6 3 5 2 405 8 2 7 1 dtype: int64无法应用于分组数据框，因此需要进行一些重新格式化。

更有效的方法来查找pandas dataframe列中的顶级值

2 个答案: