Reputation: 20101
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.
Upvotes: 74
Views: 50866
Reputation: 10302
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)), axis=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.
Upvotes: 4
Reputation: 417
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)]
Upvotes: 0
Reputation: 41
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)
Upvotes: 3
Reputation: 1
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), :]
Upvotes: -2
Reputation: 1069
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
Upvotes: 5
Reputation: 61
To follow up on DSM's answer, you can also use any()
to turn your query into an OR operation (instead of AND):
df1.loc[(df1[list(filter_v)] == pd.Series(filter_v)).any(axis=1)]
Upvotes: 1
Reputation: 2643
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), :]
Upvotes: 4
Reputation: 797
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
Upvotes: 2
Reputation: 353059
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
Upvotes: 107