### Question :

I am looking for for a pythonic way to handle the following problem.

The `pandas.get_dummies()`

method is great to create dummies from a categorical column of a dataframe. For example, if the column has values in `['A', 'B']`

, `get_dummies()`

creates 2 dummy variables and assigns 0 or 1 accordingly.

Now, I need to handle this situation. A single column, let’s call it ‘label’, has values like `['A', 'B', 'C', 'D', 'A*C', 'C*D']`

. `get_dummies()`

creates 6 dummies, but I only want 4 of them, so that a row could have multiple 1s.

Is there a way to handle this in a pythonic way? I could only think of some step-by-step algorithm to get it, but that would not include get_dummies().

Thanks

Edited, hope it is more clear!

##
Answer #1:

I know it’s been a while since this question was asked, but there is (at least *now* there is) a one-liner that is supported by the documentation:

```
In [4]: df
Out[4]:
label
0 (a, c, e)
1 (a, d)
2 (b,)
3 (d, e)
In [5]: df['label'].str.join(sep='*').str.get_dummies(sep='*')
Out[5]:
a b c d e
0 1 0 1 0 1
1 1 0 0 1 0
2 0 1 0 0 0
3 0 0 0 1 1
```

##
Answer #2:

I have a somewhat cleaner solution. Assume we want to transform the following dataframe

```
pageid category
0 0 a
1 0 b
2 1 a
3 1 c
```

into

```
a b c
pageid
0 1 1 0
1 1 0 1
```

One way to do it is to make use of scikit-learn’s DictVectorizer. I would, however, be interested in learning about other methods.

```
df = pd.DataFrame(dict(pageid=[0, 0, 1, 1], category=['a', 'b', 'a', 'c']))
grouped = df.groupby('pageid').category.apply(lambda lst: tuple((k, 1) for k in lst))
category_dicts = [dict(tuples) for tuples in grouped]
v = sklearn.feature_extraction.DictVectorizer(sparse=False)
X = v.fit_transform(category_dicts)
pd.DataFrame(X, columns=v.get_feature_names(), index=grouped.index)
```

##
Answer #3:

You can generate the dummies dataframe with your raw data, isolate the columns that contains a given atom, and then store the result matches back to the atom column.

```
df
Out[28]:
label
0 A
1 B
2 C
3 D
4 A*C
5 C*D
dummies = pd.get_dummies(df['label'])
atom_col = [c for c in dummies.columns if '*' not in c]
for col in atom_col:
...: df[col] = dummies[[c for c in dummies.columns if col in c]].sum(axis=1)
...:
df
Out[32]:
label A B C D
0 A 1 0 0 0
1 B 0 1 0 0
2 C 0 0 1 0
3 D 0 0 0 1
4 A*C 1 0 1 0
5 C*D 0 0 1 1
```

##
Answer #4:

I believe this question needs an updated answer after coming across the MultiLabelBinarizer from sklearn.

The usage of this is as simple as…

```
# Instantiate the binarizer
mlb = MultiLabelBinarizer()
# Using OP's original data frame
df = pd.DataFrame(data=['A', 'B', 'C', 'D', 'A*C', 'C*D'], columns=["label"])
print(df)
label
0 A
1 B
2 C
3 D
4 A*C
5 C*D
# Convert to a list of labels
df = df.apply(lambda x: x["label"].split("*"), axis=1)
print(df)
0 [A]
1 [B]
2 [C]
3 [D]
4 [A, C]
5 [C, D]
dtype: object
# Transform to a binary array
array_out = mlb.fit_transform(df)
print(array_out)
[[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
[1 0 1 0]
[0 0 1 1]]
# Convert back to a dataframe (unnecessary step in many cases)
df_out = pd.DataFrame(data=array_out, columns=mlb.classes_)
print(df_out)
A B C D
0 1 0 0 0
1 0 1 0 0
2 0 0 1 0
3 0 0 0 1
4 1 0 1 0
5 0 0 1 1
```

This is also very fast, took virtually no time (.03 seconds) across 1000 rows and 50K classes.