tf.unstack  |  TensorFlow v2.16.1 (original) (raw)

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

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Compat aliases for migration

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tf.compat.v1.unstack

tf.unstack(
    value, num=None, axis=0, name='unstack'
)

Used in the notebooks

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 acrossaxis 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.functiontrace, you will need to use exlicit loops and a tf.TensorArray instead.