Reversing ‘one-hot’ encoding in Pandas

I want to go from this data frame which is basically one hot encoded.

 In [2]: pd.DataFrame({"monkey":[0,1,0],"rabbit":[1,0,0],"fox":[0,0,1]})

    Out[2]:
       fox  monkey  rabbit
    0    0       0       1
    1    0       1       0
    2    1       0       0
    3    0       0       0
    4    0       0       0

To this one which is ‘reverse’ one-hot encoded.

    In [3]: pd.DataFrame({"animal":["monkey","rabbit","fox"]})
    Out[3]:
       animal
    0  monkey
    1  rabbit
    2     fox

I imagine there’s some sort of clever use of apply or zip to do thins but I’m not sure how… Can anyone help?

I’ve not had much success using indexing etc to try to solve this problem.

Answers:

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Method 1

UPDATE: i think ayhan is right and it should be:

df.idxmax(axis=1)

Demo:

In [40]: s = pd.Series(['dog', 'cat', 'dog', 'bird', 'fox', 'dog'])

In [41]: s
Out[41]:
0     dog
1     cat
2     dog
3    bird
4     fox
5     dog
dtype: object

In [42]: pd.get_dummies(s)
Out[42]:
   bird  cat  dog  fox
0   0.0  0.0  1.0  0.0
1   0.0  1.0  0.0  0.0
2   0.0  0.0  1.0  0.0
3   1.0  0.0  0.0  0.0
4   0.0  0.0  0.0  1.0
5   0.0  0.0  1.0  0.0

In [43]: pd.get_dummies(s).idxmax(1)
Out[43]:
0     dog
1     cat
2     dog
3    bird
4     fox
5     dog
dtype: object

OLD answer: (most probably, incorrect answer)

try this:

In [504]: df.idxmax().reset_index().rename(columns={'index':'animal', 0:'idx'})
Out[504]:
   animal  idx
0     fox    2
1  monkey    1
2  rabbit    0

data:

In [505]: df
Out[505]:
   fox  monkey  rabbit
0    0       0       1
1    0       1       0
2    1       0       0
3    0       0       0
4    0       0       0

Method 2

I would use apply to decode the columns:

In [2]: animals = pd.DataFrame({"monkey":[0,1,0,0,0],"rabbit":[1,0,0,0,0],"fox":[0,0,1,0,0]})

In [3]: def get_animal(row):
   ...:     for c in animals.columns:
   ...:         if row[c]==1:
   ...:             return c

In [4]: animals.apply(get_animal, axis=1)
Out[4]: 
0    rabbit
1    monkey
2       fox
3      None
4      None
dtype: object

Method 3

This works with both single and multiple labels.

We can use advanced indexing to tackle this problem. Here is the link.

import pandas as pd

df = pd.DataFrame({"monkey":[1,1,0,1,0],"rabbit":[1,1,1,1,0],
    "fox":[1,0,1,0,0], "cat":[0,0,0,0,1]})

df['tags']='' # to create an empty column

for col_name in df.columns:
    df.ix[df[col_name]==1,'tags']= df['tags']+' '+col_name

print df

And the result is:

   cat  fox  monkey  rabbit                tags
0    0    1       1       1   fox monkey rabbit
1    0    0       1       1       monkey rabbit
2    0    1       0       1          fox rabbit
3    0    0       1       1       monkey rabbit
4    1    0       0       0                 cat

Explanation:
We iterate over the columns on the dataframe.

df.ix[selection criteria, columns to write value] = value
df.ix[df[col_name]==1,'tags']= df['tags']+' '+col_name

The above line basically finds you all the places where df[col_name] == 1, selects column ‘tags’ and set it to the RHS value which is df[‘tags’]+’ ‘+ col_name

Note: .ix has been deprecated since Pandas v0.20. You should instead use .loc or .iloc, as appropriate.

Method 4

I’d do:

cols = df.columns.to_series().values
pd.DataFrame(np.repeat(cols[None, :], len(df), 0)[df.astype(bool).values], df.index[df.any(1)])

Timing

MaxU’s method has edge for large dataframes

Small df 5 x 3

Large df 1000000 x 52

Method 5

You could try using melt(). This method also works when you have multiple OHE labels for a row.

# Your OHE dataframe 
df = pd.DataFrame({"monkey":[0,1,0],"rabbit":[1,0,0],"fox":[0,0,1]})

mel = df.melt(var_name=['animal'], value_name='value') # Melting

mel[mel.value == 1].reset_index(drop=True) # this gives you the result 

Method 6

Try this:

df = pd.DataFrame({"monkey":[0,1,0,1,0],"rabbit":[1,0,0,0,0],"fox":[0,0,1,0,0], "cat":[0,0,0,0,1]})
df 

   cat  fox  monkey  rabbit
0    0    0       0       1
1    0    0       1       0
2    0    1       0       0
3    0    0       1       0
4    1    0       0       0

pd.DataFrame([x for x in np.where(df ==1, df.columns,'').flatten().tolist() if len(x) >0],columns= (["animal"]) )

   animal
0  rabbit
1  monkey
2     fox
3  monkey
4     cat

Method 7

It can be achieved with a simple apply on dataframe

# function to get column name with value one for each row in dataframe
def get_animal(row):
    return(row.index[row.apply(lambda x: x==1)][0])

# prepare a animal column
df['animal'] = df.apply(lambda row:get_animal(row), axis=1)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0