Converts a sparse representation into a dense tensor. (deprecated)
tf.compat.v1.sparse_to_dense(
    sparse_indices,
    output_shape,
    sparse_values,
    default_value=0,
    validate_indices=True,
    name=None
)
   Builds an array dense with shape output_shape such that
# If sparse_indices is scalar dense[i] = (i == sparse_indices ? sparse_values : default_value) # If sparse_indices is a vector, then for each i dense[sparse_indices[i]] = sparse_values[i] # If sparse_indices is an n by d matrix, then for each i in [0, n) dense[sparse_indices[i][0], ..., sparse_indices[i][d-1]] = sparse_values[i]
All other values in dense are set to default_value. If sparse_values is a scalar, all sparse indices are set to this single value.
Indices should be sorted in lexicographic order, and indices must not contain any repeats. If validate_indices is True, these properties are checked during execution.
| Args | |
|---|---|
| sparse_indices | A 0-D, 1-D, or 2-D Tensorof typeint32orint64.sparse_indices[i]contains the complete index wheresparse_values[i]will be placed. | 
| output_shape | A 1-D Tensorof the same type assparse_indices. Shape of the dense output tensor. | 
| sparse_values | A 0-D or 1-D Tensor. Values corresponding to each row ofsparse_indices, or a scalar value to be used for all sparse indices. | 
| default_value | A 0-D Tensorof the same type assparse_values. Value to set for indices not specified insparse_indices. Defaults to zero. | 
| validate_indices | A boolean value. If True, indices are checked to make sure they are sorted in lexicographic order and that there are no repeats. | 
| name | A name for the operation (optional). | 
| Returns | |
|---|---|
| Dense Tensorof shapeoutput_shape. Has the same type assparse_values. | 
    © 2022 The TensorFlow Authors. All rights reserved.
Licensed under the Creative Commons Attribution License 4.0.
Code samples licensed under the Apache 2.0 License.
    https://www.tensorflow.org/versions/r2.9/api_docs/python/tf/compat/v1/sparse_to_dense