pandas 使用字典中的值过滤熊猫数据框
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Filter a pandas dataframe using values from a dict
提问by Ivan
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:
我需要用字典过滤数据框，键是列名，值是我要过滤的值：
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')]
But I want to do something on the lines
但我想在线上做点什么
for column, value in filter_v.items():
df[df[column] == value]
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?
但这会多次过滤数据框，一次一个值，而不是同时应用所有过滤器。有没有办法以编程方式做到这一点？
EDIT: an example:
编辑：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), :]
gives
给
A B C D
0 1 1 right 1
1 0 1 right 2
3 1 0 right 3
but the expected result was
但预期的结果是
A B C D
3 1 0 right 3
only the last one should be selected.
只应选择最后一个。
回答by DSM
IIUC, you should be able to do something like this:
IIUC，你应该能够做这样的事情：
>>> df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)]
A B C D
3 1 0 right 3
This works by making a Series to compare against:
这通过制作一个系列来比较：
>>> pd.Series(filter_v)
A 1
B 0
C right
dtype: object
Selecting the corresponding part of df1:
选择 的对应部分df1：
>>> 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
Finding where they match:
找到他们匹配的地方：
>>> 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
Finding where they allmatch:
找到它们都匹配的地方：
>>> (df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)
0 False
1 False
2 False
3 True
4 False
dtype: bool
And finally using this to index into df1:
最后使用它来索引 df1：
>>> df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).all(axis=1)]
A B C D
3 1 0 right 3
回答by Primer
Here is a way to do it:
这是一种方法：
df.loc[df[filter_v.keys()].isin(filter_v.values()).all(axis=1), :]
UPDATE:
更新：
With values being the same across columns you could then do something like this:
随着列之间的值相同，您可以执行以下操作：
# Create your filtering function:
def filter_dict(df, dic):
return df[df[dic.keys()].apply(
lambda x: x.equals(pd.Series(dic.values(), index=x.index, name=x.name)), asix=1)]
# Use it on your DataFrame:
filter_dict(df1, filter_v)
Which yields:
其中产生：
A B C D
3 1 0 right 3
If it something that you do frequently you could go as far as to patch DataFrame for an easy access to this filter:
如果您经常这样做，您可以尽可能修补 DataFrame 以便轻松访问此过滤器：
pd.DataFrame.filter_dict_ = filter_dict
And then use this filter like this:
然后像这样使用这个过滤器：
df1.filter_dict_(filter_v)
Which would yield the same result.
这将产生相同的结果。
BUT, it is not the rightway to do it, clearly. I would use DSM's approach.
但是，这显然不是正确的方法。我会使用 DSM 的方法。
回答by E. Zeytinci
For python2, that's OK in @primer's answer. But, you should be careful in Python3 because of dict_keys. For instance,
对于python2，@primer 的回答没问题。但是，由于dict_keys，您应该在 Python3 中小心。例如，
>> df.loc[df[filter_v.keys()].isin(filter_v.values()).all(axis=1), :]
>> TypeError: unhashable type: 'dict_keys'
The correct way to Python3:
Python3的正确方法：
df.loc[df[list(filter_v.keys())].isin(list(filter_v.values())).all(axis=1), :]
回答by efajardo
Here's another way:
这是另一种方式：
filterSeries = pd.Series(np.ones(df.shape[0],dtype=bool))
for column, value in filter_v.items():
filterSeries = ((df[column] == value) & filterSeries)
This gives:
这给出：
>>> df[filterSeries]
A B C D
3 1 0 right 3
回答by Ben Saunders
Abstraction of the above for case of passing array of filter values rather than single value (analogous to pandas.core.series.Series.isin()). Using the same example:
对于传递过滤器值数组而不是单个值的情况的上述抽象（类似于 pandas.core.series.Series.isin()）。使用相同的示例：
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':[1,0], 'C':['right']}
##Start with array of all True
ind = [True] * len(df1)
##Loop through filters, updating index
for col, vals in filter_v.items():
ind = ind & (df1[col].isin(vals))
##Return filtered dataframe
df1[ind]
##Returns
A B C D
0 1.0 1 right 1
3 1.0 0 right 3
回答by Harunobu
To follow up on DSM's answer, you can also use any()to turn your query into an OR operation (instead of AND):
要跟进 DSM 的回答，您还可以使用any()将您的查询转换为 OR 操作（而不是 AND）：
df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).any(axis=1)]
df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).any(axis=1)]
