使用字典中的值过滤 pandas 数据框 [英] Filter a pandas dataframe using values from a dict

问题描述

我需要用dict过滤数据帧，该数据帧的键是列名，值是我要过滤的值.

I need to filter a data frame with a dict, constructed with the key being the column name and the value being the value that I want to filter:

filter_v = {'A':1, 'B':0, 'C':'This is right'}
# this would be the normal approach
df[(df['A'] == 1) & (df['B'] ==0)& (df['C'] == 'This is right')]

但是我想做些事情

for column, value in filter_v.items():
    df[df[column] == value]

但是这将多次过滤数据帧，一次过滤一个值，并且不会同时应用所有过滤器.有没有办法通过编程方式做到这一点?

but this will filter the data frame several times, one value at a time, and not apply all filters at the same time. Is there a way to do it programmatically?

一个例子:

df1 = pd.DataFrame({'A':[1,0,1,1, np.nan], 'B':[1,1,1,0,1], 'C':['right','right','wrong','right', 'right'],'D':[1,2,2,3,4]})
filter_v = {'A':1, 'B':0, 'C':'right'}
df1.loc[df1[filter_v.keys()].isin(filter_v.values()).all(axis=1), :]

给予

    A   B   C   D
0   1   1   right   1
1   0   1   right   2
3   1   0   right   3

但预期结果是

    A   B   C   D
3   1   0   right   3

仅应选择最后一个.

推荐答案

IIUC，您应该可以执行以下操作:

IIUC, you should be able to do something like this:

>>> df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)]
   A  B      C  D
3  1  0  right  3

---

这可以通过与以下系列进行比较来实现:

---

This works by making a Series to compare against:

>>> pd.Series(filter_v)
A        1
B        0
C    right
dtype: object

选择df1的相应部分:

>>> df1[list(filter_v)]
    A      C  B
0   1  right  1
1   0  right  1
2   1  wrong  1
3   1  right  0
4 NaN  right  1

找到匹配的地方:

>>> df1[list(filter_v)] == pd.Series(filter_v)
       A      B      C
0   True  False   True
1  False  False   True
2   True  False  False
3   True   True   True
4  False  False   True

查找所有 匹配的位置:

>>> (df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)
0    False
1    False
2    False
3     True
4    False
dtype: bool

最后使用它索引到df1:

And finally using this to index into df1:

>>> df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)]
   A  B      C  D
3  1  0  right  3

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