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Compute set intersection of elements in last dimension of a
and b
.
tf.sets.intersection( a, b, validate_indices=True )
All but the last dimension of a
and b
must match.
Example:
import tensorflow as tf import collections # Represent the following array of sets as a sparse tensor: # a = np.array([[{1, 2}, {3}], [{4}, {5, 6}]]) a = collections.OrderedDict([ ((0, 0, 0), 1), ((0, 0, 1), 2), ((0, 1, 0), 3), ((1, 0, 0), 4), ((1, 1, 0), 5), ((1, 1, 1), 6), ]) a = tf.sparse.SparseTensor(list(a.keys()), list(a.values()), dense_shape=[2,2,2]) # b = np.array([[{1}, {}], [{4}, {5, 6, 7, 8}]]) b = collections.OrderedDict([ ((0, 0, 0), 1), ((1, 0, 0), 4), ((1, 1, 0), 5), ((1, 1, 1), 6), ((1, 1, 2), 7), ((1, 1, 3), 8), ]) b = tf.sparse.SparseTensor(list(b.keys()), list(b.values()), dense_shape=[2, 2, 4]) # `tf.sets.intersection` is applied to each aligned pair of sets. tf.sets.intersection(a, b) # The result will be equivalent to either of: # # np.array([[{1}, {}], [{4}, {5, 6}]]) # # collections.OrderedDict([ # ((0, 0, 0), 1), # ((1, 0, 0), 4), # ((1, 1, 0), 5), # ((1, 1, 1), 6), # ])
Returns | |
---|---|
A SparseTensor whose shape is the same rank as a and b , and all but the last dimension the same. Elements along the last dimension contain the intersections. |