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tf.unstack Stay organized with collections Save and categorize content based on your preferences.

Unpacks the given dimension of a rank-R tensor into rank-(R-1) tensors.

View aliases

Compat aliases for migration

See Migration guide for more details.

tf.compat.v1.unstack

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.

value A rank R > 0 Tensor to be unstacked.
num An int. The length of the dimension axis. Automatically inferred if None (the default).
axis An int. The axis to unstack along. Defaults to the first dimension. Negative values wrap around, so the valid range is [-R, R).
name A name for the operation (optional).

The list of Tensor objects unstacked from value.

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.

Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. For details, see the Google Developers Site Policies. Java is a registered trademark of Oracle and/or its affiliates. Some content is licensed under the numpy license.

Last updated 2024-04-26 UTC.

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