Pandas sparse dataFrame to sparse matrix, without generating a dense matrix in memory

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Question :

Pandas sparse dataFrame to sparse matrix, without generating a dense matrix in memory

Is there a way to convert from a pandas.SparseDataFrame to scipy.sparse.csr_matrix, without generating a dense matrix in memory?

scipy.sparse.csr_matrix(df.values)

doesn’t work as it generates a dense matrix which is cast to the csr_matrix.

Thanks in advance!

Asked By: Jake0x32

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Answer #1:

Pandas docs talks about an experimental conversion to scipy sparse, SparseSeries.to_coo:

http://pandas-docs.github.io/pandas-docs-travis/sparse.html#interaction-with-scipy-sparse

================

edit – this is a special function from a multiindex, not a data frame. See the other answers for that. Note the difference in dates.

============

As of 0.20.0, there is a sdf.to_coo() and a multiindex ss.to_coo(). Since a sparse matrix is inherently 2d, it makes sense to require multiindex for the (effectively) 1d dataseries. While the dataframe can represent a table or 2d array.

When I first responded to this question this sparse dataframe/series feature was experimental (june 2015).

Answered By: hpaulj

Answer #2:

Pandas 0.20.0+:

As of pandas version 0.20.0, released May 5, 2017, there is a one-liner for this:

from scipy import sparse


def sparse_df_to_csr(df):
    return sparse.csr_matrix(df.to_coo())

This uses the new to_coo() method.

Earlier Versions:

Building on Victor May’s answer, here’s a slightly faster implementation, but it only works if the entire SparseDataFrame is sparse with all BlockIndex (note: if it was created with get_dummies, this will be the case).

Edit: I modified this so it will work with a non-zero fill value. CSR has no native non-zero fill value, so you will have to record it externally.

import numpy as np
import pandas as pd
from scipy import sparse

def sparse_BlockIndex_df_to_csr(df):
    columns = df.columns
    zipped_data = zip(*[(df[col].sp_values - df[col].fill_value,
                         df[col].sp_index.to_int_index().indices)
                        for col in columns])
    data, rows = map(list, zipped_data)
    cols = [np.ones_like(a)*i for (i,a) in enumerate(data)]
    data_f = np.concatenate(data)
    rows_f = np.concatenate(rows)
    cols_f = np.concatenate(cols)
    arr = sparse.coo_matrix((data_f, (rows_f, cols_f)),
                            df.shape, dtype=np.float64)
    return arr.tocsr()
Answered By: T.C. Proctor

Answer #3:

The answer by @Marigold does the trick, but it is slow due to accessing all elements in each column, including the zeros. Building on it, I wrote the following quick n’ dirty code, which runs about 50x faster on a 1000×1000 matrix with a density of about 1%. My code also handles dense columns appropriately.

def sparse_df_to_array(df):
    num_rows = df.shape[0]   

    data = []
    row = []
    col = []

    for i, col_name in enumerate(df.columns):
        if isinstance(df[col_name], pd.SparseSeries):
            column_index = df[col_name].sp_index
            if isinstance(column_index, BlockIndex):
                column_index = column_index.to_int_index()

            ix = column_index.indices
            data.append(df[col_name].sp_values)
            row.append(ix)
            col.append(len(df[col_name].sp_values) * [i])
        else:
            data.append(df[col_name].values)
            row.append(np.array(range(0, num_rows)))
            col.append(np.array(num_rows * [i]))

    data_f = np.concatenate(data)
    row_f = np.concatenate(row)
    col_f = np.concatenate(col)

    arr = coo_matrix((data_f, (row_f, col_f)), df.shape, dtype=np.float64)
    return arr.tocsr()
Answered By: nojka_kruva

Answer #4:

As of Pandas version 0.25 SparseSeries and SparseDataFrame are deprecated. DataFrames now support Sparse Dtypes for columns with sparse data. Sparse methods are available through sparse accessor, so conversion one-liner now looks like this:

sparse_matrix = scipy.sparse.csr_matrix(df.sparse.to_coo())
Answered By: Claygirl

Answer #5:

Here’s a solution that fills the sparse matrix column by column (assumes you can fit at least one column to memory).

import pandas as pd
import numpy as np
from scipy.sparse import lil_matrix

def sparse_df_to_array(df):
    """ Convert sparse dataframe to sparse array csr_matrix used by
    scikit learn. """
    arr = lil_matrix(df.shape, dtype=np.float32)
    for i, col in enumerate(df.columns):
        ix = df[col] != 0
        arr[np.where(ix), i] = df.ix[ix, col]

    return arr.tocsr()
Answered By: Marigold

Answer #6:

EDIT: This method is actually having a dense representation at some stage, so it doesn’t solve the question.

You should be able to use the experimental .to_coo() method in pandas [1] in the following way:

df, idx_rows, idx_cols = df.stack().to_sparse().to_coo()
df = df.tocsr()

This method, instead of taking a DataFrame (rows / columns) it takes a Series with rows and columns in a MultiIndex (this is why you need the .stack() method). This Series with the MultiIndex needs to be a SparseSeries, and even if your input is a SparseDataFrame, .stack() returns a regular Series. So, you need to use the .to_sparse() method before calling .to_coo().

The Series returned by .stack(), even if it’s not a SparseSeries only contains the elements that are not null, so it shouldn’t take more memory than the sparse version (at least with np.nan when the type is np.float).

  1. http://pandas.pydata.org/pandas-docs/stable/sparse.html#interaction-with-scipy-sparse
Answered By: Marc Garcia

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