CODEWITHSUNDEEP
pythonpandas
jim jarnac
jim jarnac

Reputation: 5152

Pandas - get row and column name for each element during applymap

I am trying to compare one list of strings for similarity and get the results in a pandas dataframe for inspection; so I use one list as index and the other as column list. I then want to compute the "Levenshtein similarity" on them (a function that compares the similarity between two words).

I am trying to do that using applymap on every cell, and compare the cell index to the cell column. How could I do that? Or simpler alternatives?

things = ['car', 'bike', 'sidewalk', 'eatery']
action = ['walking', 'caring', 'biking', 'eating']
matrix = pd.DataFrame(index = things, columns = action)

def lev(x):
    x = Levenshtein.distance(x.index, x.column)  
matrix.applymap(lev)

so far I resorted to use the following (below) but I find it clumsy and slow

matrix = pd.DataFrame(data = [action for i in things], index = things, columns = action)
for i, values in matrix.iterrows():
    for j, value in enumerate(values):
        matrix.ix[i,j] = Levenshtein.distance(i, value)

Upvotes: 14

Views: 14308

Answers (3)

Stefan
Stefan

Reputation: 12350

Here is a combination of apply and comprehension:

def mapping_function(value, index, column_name):
    # this is called for each cell
    mapping_result = column_name + '|' + str(index) + '|' + str(value)
    return mapping_result


def _column_mapping_function(column_series):
    column_name = column_series.name
    new_series_data = [mapping_function(value, index, column_name) for index, value in column_series.items()]
    new_series = pd.Series(data=new_series_data, index=column_series.index)
    return new_series
    
result = indexed_data_frame.apply(_column_mapping_function)

Upvotes: 0

jezrael
jezrael

Reputation: 863196

I think you can use apply on the dataframe, and to access columns' values use .name:

def lev(x):
    #replace your function
    return x.index + x.name
a = matrix.apply(lev)
print (a)
                  walking          caring          biking          eating
car            carwalking       carcaring       carbiking       careating
bike          bikewalking      bikecaring      bikebiking      bikeeating
sidewalk  sidewalkwalking  sidewalkcaring  sidewalkbiking  sidewalkeating
eatery      eaterywalking    eaterycaring    eaterybiking    eateryeating

EDIT:

If need some arithemtic operation use broadcasting:

a = pd.DataFrame(matrix.index.values + matrix.columns.values[:,None], 
                 index=matrix.index, 
                 columns=matrix.columns)
print (a)
             walking       caring           biking         eating
car       carwalking  bikewalking  sidewalkwalking  eaterywalking
bike       carcaring   bikecaring   sidewalkcaring   eaterycaring
sidewalk   carbiking   bikebiking   sidewalkbiking   eaterybiking
eatery     careating   bikeeating   sidewalkeating   eateryeating

Or:

a = pd.DataFrame(matrix.index.values + matrix.columns.values[:, np.newaxis], 
                 index=matrix.index, 
                 columns=matrix.columns)
print (a)
             walking       caring           biking         eating
car       carwalking  bikewalking  sidewalkwalking  eaterywalking
bike       carcaring   bikecaring   sidewalkcaring   eaterycaring
sidewalk   carbiking   bikebiking   sidewalkbiking   eaterybiking
eatery     careating   bikeeating   sidewalkeating   eateryeating

Upvotes: 17

chaonan99
chaonan99

Reputation: 91

You can do that by "nested apply" as follows:

things = ['car', 'bike', 'sidewalk', 'eatery']
action = ['walking', 'caring', 'biking', 'eating']
matrix = pd.DataFrame(index=things, columns=action)
matrix.apply(lambda x: pd.DataFrame(x).apply(lambda y: LD(x.name, y.name), axis=1))

Output:

          walking  caring  biking  eating
car             6       3       6       5
bike            6       5       3       5
sidewalk        7       8       7       8
eatery          6       5       6       3

The call pd.DataFrame(x) here is because x is a Series object and the Series.apply is similar to applymap, which does not carry index or columns information.

Upvotes: 9

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