Given a dict like this:

d={'paris':['a','b'],
  'brussels':['b','c'],
  'mallorca':['a','d']}

#when doing:
df = pd.DataFrame(d)
df.T

I dont get the expected result. What I would like to get is a one_hot_encoding DF, in which the columns are the capitals and the value 1 or 0 corresponds to every of the letters that every city includes being paris, mallorca ect

The desired result is:

df = pd.DataFrame([[1,1,0,0],[0,1,1,0],[1,0,0,1]], index=['paris','brussels','mallorca'], columns=list('abcd'))
df.T

Any clever way to do this without having to multiloop over the first dict to transform it into another one?

CodePudding user response：

Solution 1:

• Combine df.apply with series.value_counts and append df.fillna to fill NaN values with zeros.

out = df.apply(pd.Series.value_counts).fillna(0)
print(out)

   paris  brussels  mallorca
a    1.0       0.0       1.0
b    1.0       1.0       0.0
c    0.0       1.0       0.0
d    0.0       0.0       1.0

Solution 1:

• Transform your df using df.melt and then use the result inside pd.crosstab.
• Again use df.fillna to change NaN values to zeros. Finally, reorder the columns based on the order in the original df.

out = df.melt(value_name='index')
out = pd.crosstab(index=out['index'], columns=out['variable'])\
    .fillna(0).loc[:, df.columns]
print(out)

       paris  brussels  mallorca
index                           
a          1         0         1
b          1         1         0
c          0         1         0
d          0         0         1

CodePudding user response：

I don't know how 'clever' my solution is but it works and it is pretty consise and readable.

import pandas as pd

d = {'paris': ['a', 'b'],
     'brussels': ['b', 'c'],
     'mallorca': ['a', 'd']}

df = pd.DataFrame(d).T
df.columns = ['0', '1']
df = pd.concat([df['0'], df['1']])
df = pd.crosstab(df, columns=df.index)

print(df)

Yields:

       brussels  mallorca  paris                           
a             0         1      1
b             1         0      1
c             1         0      0
d             0         1      0