AttributeError: 'DataFrame' object has no attribute 'to_sparse'

Question

sdf = df.to_sparse() has been deprecated. What's the updated way to convert to a sparse DataFrame?

Answer 1

These are the updated sparse conversions in pandas 1.0.0+.

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How to convert dense to sparse

Use DataFrame.astype() with the appropriate SparseDtype() (eg, int ):

>>> df = pd.DataFrame({'A': [1, 0, 0, 0, 1, 0]})
>>> df.dtypes
# A    int64
# dtype: object

>>> sdf = df.astype(pd.SparseDtype(int, fill_value=0))
>>> sdf.dtypes
# A    Sparse[int64, 0]
# dtype: object

Or use the string alias for brevity:

>>> sdf = df.astype('Sparse[int64, 0]')

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How to convert sparse to dense

Use DataFrame.sparse.to_dense() :

>>> from scipy import sparse
>>> sdf = pd.DataFrame.sparse.from_spmatrix(sparse.eye(3), columns=list('ABC'))
>>> sdf.dtypes
# A    Sparse[float64, 0]
# B    Sparse[float64, 0]
# C    Sparse[float64, 0]
# dtype: object

>>> df = sdf.sparse.to_dense()
>>> df.dtypes
# A    float64
# B    float64
# C    float64
# dtype: object

---

How to convert sparse to COO

Use DataFrame.sparse.to_coo() :

>>> from scipy import sparse
>>> sdf = pd.DataFrame.sparse.from_spmatrix(sparse.eye(3), columns=list('ABC'))
>>> sdf.dtypes
# A    Sparse[float64, 0]
# B    Sparse[float64, 0]
# C    Sparse[float64, 0]
# dtype: object

>>> df = sdf.sparse.to_coo()
# <3x3 sparse matrix of type '<class 'numpy.float64'>'
#         with 3 stored elements in COOrdinate format>
# (0, 0)    1.0
# (1, 1)    1.0
# (2, 2)    1.0

Answer 2

You can use scipy to create sparse matrix:

scipy.sparse.csr_matrix(df.values)