Python Pandas groupby：根据值的条件过滤
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Python Pandas groupby: filter according to condition on values
提问by Adam
Consider a dataframe like the following.
考虑如下所示的数据框。
import pandas as pd
# Initialize dataframe
df1 = pd.DataFrame(columns=['bar', 'foo'])
df1['bar'] = ['001', '001', '001', '001', '002', '002', '003', '003', '003']
df1['foo'] = [-1, 0, 2, 3, -8, 1, 0, 1, 2]
>>> print df1
bar foo
0 001 -1
1 001 0
2 001 2
3 001 3
4 002 -8
5 002 1
6 003 0
7 003 1
8 003 2
# Lower and upper bound for desired range
lower_bound = -5
upper_bound = 5
I would like to use groupby in Pandas to return a dataframe that filters out rows with an barthat meets a condition. In particular, I would like to filter out rows with an barif one of the values of foofor this baris not between lower_boundand upper_bound.
我想在 Pandas 中使用 groupby 返回一个数据框，该数据框过滤掉bar满足条件的行。我特别想用过滤掉行bar，如果值的一个foo本bar不是之间lower_bound和upper_bound。
In the above example, rows with bar = 002should be filtered out since not all of the rows with bar = 002contain a value of foobetween -5and 5(namely, row index 4contains foo = -8). The desired output for this example is the following.
在上面的示例中，bar = 002应过滤掉 的行，因为并非所有的 行bar = 002都包含foo介于-5和之间的值5（即，行索引4包含foo = -8）。此示例所需的输出如下。
# Desired output
bar foo
0 001 -1
1 001 0
2 001 2
3 001 3
6 003 0
7 003 1
8 003 2
I have tried the following approach.
我尝试了以下方法。
# Attempted solution
grouped = df1.groupby('bar')['foo']
grouped.filter(lambda x: x < lower_bound or x > upper_bound)
However, this yields a TypeError: the filter must return a boolean result. Furthermore, this approach might return a groupby object, when I want the result to return a dataframe object.
但是，这会产生一个TypeError: the filter must return a boolean result. 此外，当我希望结果返回一个数据帧对象时，这种方法可能会返回一个 groupby 对象。
采纳答案by Psidom
Most likely you will not use andand orbut vectorized &and |with pandas, and for your case, then apply all()function in the filter to construct the boolean condition, this keeps barwhere all corresponding foovalues are between lower_boundand upper_bound:
很可能您不会使用andandor而是使用矢量化&和|with pandas，并且对于您的情况，然后all()在过滤器中应用函数来构造布尔条件，这将保持bar所有相应foo值在lower_bound和upper_bound之间的位置：
df1.groupby('bar').filter(lambda x: ((x.foo >= lower_bound) & (x.foo <= upper_bound)).all())
# bar foo
#0 001 -1
#1 001 0
#2 001 2
#3 001 3
#6 003 0
#7 003 1
#8 003 2
回答by mrPuzzle
Psidom's answer works fine, but can be slow on large datasets. Mine is somewhat of a workaround, but it is fast.
Psidom 的答案工作正常，但在大型数据集上可能会很慢。我的有点解决方法，但速度很快。
df1['conditions_apply'] = (df1.foo >= lower_bound) & (df1.foo <= upper_bound)
selection = df1.groupby('bar')['conditions_apply'].min() # any False will return False
selection = selection[selection].index.tolist() # get all bars with Trues
df1 = df1[df1.bar.isin(selection)] # make selection
df1.drop(columns=['conditions_apply'], inplace=True) # drop newly made column
