问题描述
所以我有一个如下所示的 Pandas DataFrame:
So I have a pandas DataFrame that looks like this:
r vals positions
1.2 1
1.8 2
2.3 1
1.8 1
2.1 3
2.0 3
1.9 1
... ...
我希望按位置过滤掉所有不出现至少 20 次的行.我见过这样的
I would like the filter out all rows by position that do not appear at least 20 times. I have seen something like this
g=df.groupby('positions')
g.filter(lambda x: len(x) > 20)
但这似乎不起作用，我不明白如何从中获取原始数据帧.预先感谢您的帮助.
but this does not seem to work and I do not understand how to get the original dataframe back from this. Thanks in advance for the help.
推荐答案
在您有限的数据集上，以下工作:
On your limited dataset the following works:
In [125]:
df.groupby('positions')['r vals'].filter(lambda x: len(x) >= 3)
Out[125]:
0 1.2
2 2.3
3 1.8
6 1.9
Name: r vals, dtype: float64
您可以分配此过滤器的结果并将其与 isin 来过滤你的原始文件:
You can assign the result of this filter and use this with isin to filter your orig df:
In [129]:
filtered = df.groupby('positions')['r vals'].filter(lambda x: len(x) >= 3)
df[df['r vals'].isin(filtered)]
Out[129]:
r vals positions
0 1.2 1
1 1.8 2
2 2.3 1
3 1.8 1
6 1.9 1
您只需要在您的情况下将 3 更改为 20
You just need to change 3 to 20 in your case
另一种方法是使用 value_counts 来创建聚合系列，然后我们可以使用它来过滤您的 df:
Another approach would be to use value_counts to create an aggregate series, we can then use this to filter your df:
In [136]:
counts = df['positions'].value_counts()
counts
Out[136]:
1 4
3 2
2 1
dtype: int64
In [137]:
counts[counts > 3]
Out[137]:
1 4
dtype: int64
In [135]:
df[df['positions'].isin(counts[counts > 3].index)]
Out[135]:
r vals positions
0 1.2 1
2 2.3 1
3 1.8 1
6 1.9 1
编辑
如果你想过滤数据框上的 groupby 对象而不是一个系列，那么你可以调用 filter 直接在 groupby 对象上:
If you want to filter the groupby object on the dataframe rather than a Series then you can call filter on the groupby object directly:
In [139]:
filtered = df.groupby('positions').filter(lambda x: len(x) >= 3)
filtered
Out[139]:
r vals positions
0 1.2 1
2 2.3 1
3 1.8 1
6 1.9 1
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