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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.
|