Reverse a get_dummies encoding in pandas

Column names are: ID,1,2,3,4,5,6,7,8,9.

The col values are either 0 or 1

My dataframe looks like this:

 ID     1    2    3    4    5    6   7   8   9 

1002    0    1    0    1    0    0   0   0   0
1003    0    0    0    0    0    0   0   0   0 
1004    1    1    0    0    0    0   0   0   0
1005    0    0    0    0    1    0   0   0   0
1006    0    0    0    0    0    1   0   0   0
1007    1    0    1    0    0    0   0   0   0
1000    0    0    0    0    0    0   0   0   0
1009    0    0    1    0    0    0   1   0   0

I want the column names in front of the ID where the value in a row is 1.

The Dataframe i want should look like this:

 ID      Col2
1002       2    // has 1 at Col(2) and Col(4)
1002       4    
1004       1    // has 1 at col(1) and col(2)
1004       2
1005       5    // has 1 at col(5)
1006       6    // has 1 at col(6)
1007       1    // has 1 at col(1) and col(3)
1007       3
1009       3    // has 1 at col(3) and col(7)
1009       7

Please help me in this, Thanks in advance

Answers:

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Method 1

Pretty one-liner 🙂

new_df = df.idxmax(axis=1)

Method 2

Several great answers for the OP post. However, often get_dummies is used for multiple categorical features. Pandas uses a prefix separator prefix_sep to distinguish different values for a column.

The following function collapses a “dummified” dataframe while keeping the order of columns:

def undummify(df, prefix_sep="_"):
    cols2collapse = {
        item.split(prefix_sep)[0]: (prefix_sep in item) for item in df.columns
    }
    series_list = []
    for col, needs_to_collapse in cols2collapse.items():
        if needs_to_collapse:
            undummified = (
                df.filter(like=col)
                .idxmax(axis=1)
                .apply(lambda x: x.split(prefix_sep, maxsplit=1)[1])
                .rename(col)
            )
            series_list.append(undummified)
        else:
            series_list.append(df[col])
    undummified_df = pd.concat(series_list, axis=1)
    return undummified_df

Example

>>> df
     a    b    c
0  A_1  B_1  C_1
1  A_2  B_2  C_2
>>> df2 = pd.get_dummies(df)
>>> df2
   a_A_1  a_A_2  b_B_1  b_B_2  c_C_1  c_C_2
0      1      0      1      0      1      0
1      0      1      0      1      0      1
>>> df3 = undummify(df2)
>>> df3
     a    b    c
0  A_1  B_1  C_1
1  A_2  B_2  C_2

Method 3

set_index + stack, stack will dropna by default

df.set_index('ID',inplace=True)

df[df==1].stack().reset_index().drop(0,1)
Out[363]: 
     ID level_1
0  1002       2
1  1002       4
2  1004       1
3  1004       2
4  1005       5
5  1006       6
6  1007       1
7  1007       3
8  1009       3
9  1009       7

Method 4

np.argwhere

v = np.argwhere(df.drop('ID', 1).values).T
pd.DataFrame({'ID' : df.loc[v[0], 'ID'], 'Col2' : df.columns[1:][v[1]]})

  Col2    ID
0    2  1002
0    4  1002
2    1  1004
2    2  1004
3    5  1005
4    6  1006
5    1  1007
5    3  1007
7    3  1009
7    7  1009

argwhere gets the i, j indices of all non-zero elements in your DataFrame. Use the first column of indices to index into column ID, and the second column of indices to index into df.columns.

I transpose v before step 2 for cache efficiency, and less typing.

Method 5

Use:

df = (df.melt('ID', var_name='Col2')
       .query('value== 1')
       .sort_values(['ID', 'Col2'])
       .drop('value',1))

Alternative solution:

df = (df.set_index('ID')
        .mask(lambda x: x == 0)
        .stack()
        .reset_index()
        .drop(0,1))

print (df)
      ID Col2
8   1002    2
24  1002    4
2   1004    1
10  1004    2
35  1005    5
44  1006    6
5   1007    1
21  1007    3
23  1009    3
55  1009    7

Explanation:

1. First reshape values by melt or set_index with unstack
2. Filter only 1 by query or convert 0 to NaNs by mask
3. sort_values for first solution
4. create columns from MultiIndex by reset_index
5. Last remove unnecessary columns by drop

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