Extract a DenseVector from a DenseMatrix based on different column indices

[nejc....@gmail.com]

Hello. I'm looking for a more efficient way of getting a DenseVector from a DenseMatrix where column index is different for every row. My current solution (colIndices is a DenseVector):

val result = 0 until matrix.rows map { rowIndex => val colIndex = colIndices(rowIndex) matrix(rowIndex, colIndex) }

[David Hall]

same basic answers as the other one for efficiency.

If you just want economical, you can probably just do: DenseVector.tabulate(matrix.rows) { rowIndex => ... }

Or you can use a slice vector, which isn't a dense vector, but can be made into one.

matrix(colIndices.toIndexedSeq.zipWithIndex.map(_.swap)).toDenseVector

On Mon, Mar 26, 2018 at 2:47 AM, <nejc....@gmail.com> wrote:

Hello. I'm looking for a more efficient way of getting a DenseVector from a DenseMatrix where column index is different for every row. My current solution (colIndices is a DenseVector):

val result = 0 until matrix.rows map { rowIndex => val colIndex = colIndices(rowIndex) matrix(rowIndex, colIndex) }

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[nejc....@gmail.com]

Thank you for both responses.

On Monday, March 26, 2018 at 8:25:28 PM UTC+2, David Hall wrote:

same basic answers as the other one for efficiency.

If you just want economical, you can probably just do: DenseVector.tabulate(matrix.rows) { rowIndex => ... }

Or you can use a slice vector, which isn't a dense vector, but can be made into one.

matrix(colIndices.toIndexedSeq.zipWithIndex.map(_.swap)).toDenseVector

On Mon, Mar 26, 2018 at 2:47 AM, <nejc....@gmail.com> wrote:

Hello. I'm looking for a more efficient way of getting a DenseVector from a DenseMatrix where column index is different for every row. My current solution (colIndices is a DenseVector):

val result = 0 until matrix.rows map { rowIndex => val colIndex = colIndices(rowIndex) matrix(rowIndex, colIndex) }

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