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Unpacks the given dimension of a rank-R
tensor into rank-(R-1)
tensors.
tf.unstack(
value, num=None, axis=0, name='unstack'
)
Used in the notebooks
Used in the tutorials |
---|
Unpacks tensors from value
by chipping it along the axis
dimension.
x = tf.reshape(tf.range(12), (3,4))
p, q, r = tf.unstack(x)
p.shape.as_list()
[4]
i, j, k, l = tf.unstack(x, axis=1)
i.shape.as_list()
[3]
This is the opposite of stack.
x = tf.stack([i, j, k, l], axis=1)
More generally if you have a tensor of shape (A, B, C, D)
:
A, B, C, D = [2, 3, 4, 5]
t = tf.random.normal(shape=[A, B, C, D])
The number of tensor returned is equal to the length of the target axis
:
axis = 2
items = tf.unstack(t, axis=axis)
len(items) == t.shape[axis]
True
The shape of each result tensor is equal to the shape of the input tensor,
with the target axis
removed.
items[0].shape.as_list() # [A, B, D]
[2, 3, 5]
The value of each tensor items[i]
is equal to the slice of input
across
axis
at index i
:
for i in range(len(items)):
slice = t[:,:,i,:]
assert tf.reduce_all(slice == items[i])
Python iterable unpacking
With eager execution you can unstack the 0th axis of a tensor using python's iterable unpacking:
t = tf.constant([1,2,3])
a,b,c = t
unstack
is still necessary because Iterable unpacking doesn't work in
a @tf.function
: Symbolic tensors are not iterable.
You need to use tf.unstack
here:
@tf.function
def bad(t):
a,b,c = t
return a
bad(t)
Traceback (most recent call last):
OperatorNotAllowedInGraphError: ...
@tf.function
def good(t):
a,b,c = tf.unstack(t)
return a
good(t).numpy()
1
Unknown shapes
Eager tensors have concrete values, so their shape is always known.
Inside a tf.function
the symbolic tensors may have unknown shapes.
If the length of axis
is unknown tf.unstack
will fail because it cannot
handle an unknown number of tensors:
@tf.function(input_signature=[tf.TensorSpec([None], tf.float32)])
def bad(t):
tensors = tf.unstack(t)
return tensors[0]
bad(tf.constant([1.0, 2.0, 3.0]))
Traceback (most recent call last):
ValueError: Cannot infer argument `num` from shape (None,)
If you know the axis
length you can pass it as the num
argument. But this
must be a constant value.
If you actually need a variable number of tensors in a single tf.function
trace, you will need to use exlicit loops and a tf.TensorArray
instead.
Returns | |
---|---|
The list of Tensor objects unstacked from value .
|
Raises | |
---|---|
ValueError
|
If axis is out of the range [-R, R) .
|
ValueError
|
If num is unspecified and cannot be inferred.
|
InvalidArgumentError
|
If num does not match the shape of value .
|