使用字典中的值过滤 pandas dataframe

Question

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')]

But I want to do something on the lines

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?

EDIT: an example:

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

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

but the expected result was

    A   B   C   D
3   1   0   right   3

only the last one should be selected.

Answer 1

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

---

This works by making a Series to compare against:

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

Selecting the corresponding part of 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:

>>> 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 all match:

>>> (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:

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

Answer 2

Here is a way to do it:

df.loc[df[filter_v.keys()].isin(filter_v.values()).all(axis=1), :]

UPDATE:

With values being the same across columns you could then do something like this:

# 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:

   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:

pd.DataFrame.filter_dict_ = filter_dict

And then use this filter like this:

df1.filter_dict_(filter_v)

Which would yield the same result.

BUT , it is not the right way to do it, clearly. I would use DSM's approach.

Answer 3

For python2, that's OK in @primer's answer. But, you should be careful in Python3 because of dict_keys . For instance,

>> df.loc[df[filter_v.keys()].isin(filter_v.values()).all(axis=1), :]
>> TypeError: unhashable type: 'dict_keys'

The correct way to Python3:

df.loc[df[list(filter_v.keys())].isin(list(filter_v.values())).all(axis=1), :]

Answer 4

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:

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

Answer 5

Here's another way:

filterSeries = pd.Series(np.ones(df.shape[0],dtype=bool))
for column, value in filter_v.items():
    filterSeries = ((df[column] == value) & filterSeries)

This gives:

>>> df[filterSeries]
   A  B      C  D
3  1  0  right  3 

Answer 6

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)]

Answer 7

You can also create a query

query_string = ' and '.join(
    [f'({key} == "{val}")' if type(val) == str else f'({key} == {val})' for key, val in filter_v.items()]
)

df1.query(query_string)

Answer 8

Combining previous answers, here's a function you can feed to df1.loc . Allows for AND/OR (using how='all' / 'any' ), plus it allows comparisons other than == using the op keyword, if desired.

import operator

def quick_mask(df, filters, how='all', op=operator.eq) -> pd.Series:
    if how == 'all':
        comb = pd.Series.all
    elif how == 'any':
        comb = pd.Series.any
    return comb(op(df[[*filters]], pd.Series(filters)), axis=1)

# Usage
df1.loc[quick_mask(df1, filter_v)]

Answer 9

I had an issue due to my dictionary having multiple values for the same key.

I was able to change DSM's query to:

df1.loc[df1[list(filter_v)].isin(filter_v).all(axis=1), :]