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Using tensorflow.stack what does it mean to have axis=-1 ?
I'm using tensorflow==1.14
Using axis=-1 simply means to stack the tensors along the last axis (as per the python list indexing syntax).
Let's take a look at how this looks like using these tensors of shape (2, 2):
>>> x = tf.constant([[1, 2], [3, 4]])
>>> y = tf.constant([[5, 6], [7, 8]])
>>> z = tf.constant([[9, 10], [11, 12]])
The default behavior for tf.stack as described in the documentation is to stack the tensors along the first axis (index 0) resulting in a tensor of shape (3, 2, 2)
>>> tf.stack([x, y, z], axis=0)
<tf.Tensor: shape=(3, 2, 2), dtype=int32, numpy=
array([[[ 1, 2],
[ 3, 4]],
[[ 5, 6],
[ 7, 8]],
[[ 9, 10],
[11, 12]]], dtype=int32)>
Using axis=-1, the three tensors are stacked along the last axis instead, resulting in a tensor of shape (2, 2, 3)
>>> tf.stack([x, y, z], axis=-1)
<tf.Tensor: shape=(2, 2, 3), dtype=int32, numpy=
array([[[ 1, 5, 9],
[ 2, 6, 10]],
[[ 3, 7, 11],
[ 4, 8, 12]]], dtype=int32)>
I want to apply the operation tf.tile, e.g. tf.tile(A, [1, 1, b]) where A has shape [5,4,3]. How to generate [1, 1, 1] according to A? Then I set the [1, 1, 1]'s third element to b, where b is a placeholder.
This is my code, but it doesn't work, how to fix it?
d = tf.shape(A)
for i in range(tf.rank(A)): #wrong, tf.rank(A) as a tensor can't be here
d[i] = 1
d[2] = b
result = tf.tile(A, d)
The easiest solution is probably to use tf.one_hot to build your multiples tensor directly.
>>> b = 2
>>> tf.one_hot(indices=[b], depth=tf.rank(A), on_value=b, off_value=1)
<tf.Tensor: shape=(3,), dtype=int32, numpy=array([1, 1, 2], dtype=int32)>
Alternatively, you can use tf.ones_like to generate a tensor of 1 with the same shape as the Tensor passed as an argument.
>>> A = tf.random.uniform((5,4,3))
>>> tf.shape(A)
<tf.Tensor: shape=(3,), dtype=int32, numpy=array([5, 4, 3], dtype=int32)>
>>> tf.ones_like(tf.shape(A))
<tf.Tensor: shape=(3,), dtype=int32, numpy=array([1, 1, 1], dtype=int32)>
Note that in tensorflow, you can't do item assignment on a tensor (so d[2] = b won't work for example). To generate your tensor [1,1,b] you can use tf.concat:
>>> b = 2
>>> tf.concat([tf.ones_like(tf.shape(A)[:-1]),[b]],axis=0)
<tf.Tensor: shape=(3,), dtype=int32, numpy=array([1, 1, 2], dtype=int32)>
Question
Please confirm if the below is as designed and expected, or an issue of tf. slice, or a mistake in the usage of tf. slice. If a mistake, kindly suggest how to correct it.
Background
Introduction to tensor slicing - Extract tensor slices says Numpy-like slice syntax is an alternative of tf. slice.
Perform NumPy-like tensor slicing using tf. slice.
t1 = tf.constant([0, 1, 2, 3, 4, 5, 6, 7])
print(tf.slice(t1,
begin=[1],
size=[3]))
Alternatively, you can use a more Pythonic syntax. Note that tensor slices are evenly spaced over a start-stop range.
print(t1[1:4])
Problem
To Update the dark orange region.
TYPE = tf.int32
N = 4
D = 5
shape = (N,D)
# Target to update
Y = tf.Variable(
initial_value=tf.reshape(tf.range(N*D,dtype=TYPE), shape=shape),
trainable=True
)
print(f"Target Y: \n{Y}\n")
---
Target Y:
<tf.Variable 'Variable:0' shape=(4, 5) dtype=int32, numpy=
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]], dtype=int32)>
tf. slice does not work.
# --------------------------------------------------------------------------------
# Slice region in the target to be updated
# --------------------------------------------------------------------------------
S = tf.slice( # Error "EagerTensor' object has no attribute 'assign'"
Y,
begin=[0,1], # Coordinate (n,d) as the start point
size=[3,2] # Shape (3,2) -> (n+3, n+2) as the end point
)
print(f"Slice to update S: \n{S}\n")
# Values to set
V = tf.ones(shape=tf.shape(S), dtype=TYPE)
print(f"Values to set V: \n{V}\n")
# Assing V to S region of T
S.assign(V)
---
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-17-e5692b1750c8> in <module>
24
25 # Assing V to S region of T
---> 26 S.assign(V)
AttributeError: 'tensorflow.python.framework.ops.EagerTensor' object has no attribute 'assign'
However, slice syntax works.
S = Y[
0:3, # From coordinate (n=0,d), slice rows (0,1,2) or 'size'=3 -> shape (3,?)
1:3 # From coordinate (n=0,d=1), slice columns (1,2) or 'size'=2 -> shape (3,2)
]
print(f"Slice to update S: \n{S}\n")
# Values to set
V = tf.ones(shape=tf.shape(S), dtype=TYPE)
print(f"Values to set V: \n{V}\n")
# Assing V to S region of T
S.assign(V)
---
<tf.Variable 'UnreadVariable' shape=(4, 5) dtype=int32, numpy=
array([[ 0, 1, 1, 3, 4],
[ 5, 1, 1, 8, 9],
[10, 1, 1, 13, 14],
[15, 16, 17, 18, 19]], dtype=int32)>
In my understanding, the above behavior is expected or not a bug at least. As the error said, there is no attribute called assign in tf. Tensor (EagerTensor for eager execution) but there is in tf. Variable. And generally, tf. slice returns a tensor as its output and thus it doesn't possess assign attribute.
AttributeError: 'tensorflow.python.framework.ops.EagerTensor' object has no attribute 'assign'
But when we do np like slicing and use it to modify the original tf. Variable, it seamlessly works.
Possible Solution
A workaround is to use tf.strided_slice instead of tf.slice. If we follow the source code of it, we will see, it takes the var argument which is a variable corresponding to input_
#tf_export("strided_slice")
#dispatch.add_dispatch_support
def strided_slice(input_,
begin,
end,
..........
var=None,
name=None):
And when we pass a parameter for var that basically corresponding to the input_, it then calls assign function that is defined within it
def assign(val, name=None):
"""Closure that holds all the arguments to create an assignment."""
if var is None:
raise ValueError("Sliced assignment is only supported for variables")
else:
if name is None:
name = parent_name + "_assign"
return var._strided_slice_assign(
begin=begin,
end=end,
strides=strides,
value=val,
name=name,
begin_mask=begin_mask,
end_mask=end_mask,
ellipsis_mask=ellipsis_mask,
new_axis_mask=new_axis_mask,
shrink_axis_mask=shrink_axis_mask)
So, when we pass var in the tf.strided_slice, it will return an assignable object.
Code
Here is the full working code for reference.
import tensorflow as tf
print(tf.__version__)
TYPE = tf.int32
N = 4
D = 5
shape = (N,D)
# Target to update
Y = tf.Variable(
initial_value=tf.reshape(tf.range(N*D,dtype=TYPE), shape=shape),
trainable=True
)
Y
2.4.1
<tf.Variable 'Variable:0' shape=(4, 5) dtype=int32, numpy=
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]], dtype=int32)>
Now, we use tf.stried_slice instead of tf.slice.
S = tf.strided_slice(
Y,
begin = [0, 1],
end = [3, 3],
var = Y,
name ='slice_op'
)
S
<tf.Tensor: shape=(3, 2), dtype=int32, numpy=
array([[ 1, 2],
[ 6, 7],
[11, 12]], dtype=int32)>
Update the variables with no attribution error.
# Values to set
V = tf.ones(shape=tf.shape(S), dtype=TYPE)
print(V)
print()
# Assing V to S region of T
S.assign(V)
tf.Tensor(
[[1 1]
[1 1]
[1 1]], shape=(3, 2), dtype=int32)
<tf.Variable 'UnreadVariable' shape=(4, 5) dtype=int32, numpy=
array([[ 0, 1, 1, 3, 4],
[ 5, 1, 1, 8, 9],
[10, 1, 1, 13, 14],
[15, 16, 17, 18, 19]], dtype=int32)>
Using np like slicing.
# slicing
S = Y[
0:3,
1:3
]
S
<tf.Tensor: shape=(3, 2), dtype=int32, numpy=
array([[ 1, 2],
[ 6, 7],
[11, 12]], dtype=int32)>
# Values to set
V = tf.ones(shape=tf.shape(S), dtype=TYPE)
print(V)
# Assing V to S region of T
S.assign(V)
tf.Tensor(
[[1 1]
[1 1]
[1 1]], shape=(3, 2), dtype=int32)
<tf.Variable 'UnreadVariable' shape=(4, 5) dtype=int32, numpy=
array([[ 0, 1, 1, 3, 4],
[ 5, 1, 1, 8, 9],
[10, 1, 1, 13, 14],
[15, 16, 17, 18, 19]], dtype=int32)>
Materials
tf.Variable - tf.Tensor.
Is there a way to directly update the elements in tf.Variable X at indices without creating a new tensor having the same shape as X?
tf.tensor_scatter_nd_update create a new tensor hence it appears not updateing the original tf.Variable.
This operation creates a new tensor by applying sparse updates to the input tensor.
tf.Variable assign apparently needs a new tensor value which has the same shape of X to update the tf.Variable X.
assign(
value, use_locking=False, name=None, read_value=True
)
value A Tensor. The new value for this variable.
About the tf.tensor_scatter_nd_update, you're right that it returns a new tf.tensor (and not tf.Variable). But about the assign which is an attribute of tf.Variable, I think you somewhat misread the document; the value is just the new item that you want to assign in particular indices of your old variable.
AFAIK, in tensorflow all tensors are immutable like python numbers and strings; you can never update the contents of a tensor, only create a new one, source. And directly updating or manipulating of tf.tensor or tf.Variable such as numpy like item assignment is still not supported. Check the following Github issues to follow up the discussions: #33131, #14132.
In numpy, we can do an in-place item assignment that you showed in the comment box.
import numpy as np
a = np.array([1,2,3])
print(a) # [1 2 3]
a[1] = 0
print(a) # [1 0 3]
A similar result can be achieved in tf.Variable with assign attribute.
import tensorflow as tf
b = tf.Variable([1,2,3])
b.numpy() # array([1, 2, 3], dtype=int32)
b[1].assign(0)
b.numpy() # array([1, 0, 3], dtype=int32)
Later, we can convert it to tf. tensor as follows.
b_ten = tf.convert_to_tensor(b)
b_ten.numpy() # array([1, 0, 3], dtype=int32)
We can do such item assignment in tf.tensor too but we need to convert it to tf.Variable first, (I know, not very intuitive).
tensor = [[1, 1], [1, 1], [1, 1]] # tf.rank(tensor) == 2
indices = [[0, 1], [2, 0]] # num_updates == 2, index_depth == 2
updates = [5, 10] # num_updates == 2
x = tf.tensor_scatter_nd_update(tensor, indices, updates)
x
<tf.Tensor: shape=(3, 2), dtype=int32, numpy=
array([[ 1, 5],
[ 1, 1],
[10, 1]], dtype=int32)>
x = tf.Variable(x)
x
<tf.Variable 'Variable:0' shape=(3, 2) dtype=int32, numpy=
array([[ 1, 5],
[ 1, 1],
[10, 1]], dtype=int32)>
x[0].assign([5,1])
x
<tf.Variable 'Variable:0' shape=(3, 2) dtype=int32, numpy=
array([[ 5, 1],
[ 1, 1],
[10, 1]], dtype=int32)>
x = tf.convert_to_tensor(x)
x
<tf.Tensor: shape=(3, 2), dtype=int32, numpy=
array([[ 5, 1],
[ 1, 1],
[10, 1]], dtype=int32)>
I have a sobel filter
sobel_x = tf.constant([[-1, 0, 1], [-2, 0, 2], [-1, 0, 1]], tf.float32)
I want to get a depth of 64. The shape is momentarily [3,3,1], but it should result in [3,3,64].
How do to that? With the following line, I get shape errors.
tf.tile(sobel_x, [1, 1, 64])
ValueError: Shape must be rank 2 but is rank 3 for 'Tile' (op: 'Tile') with input shapes: [3,3], [3].
The reason you cannot broadcast is that the third dimension does not exist, and so you actually have a rank 2 tensor.
>>> sess.run(tf.shape(sobel_x))
array([3, 3], dtype=int32)
We can solve this problem by reshaping the tensor first.
>>> sobel_x = tf.reshape(sobel_x, [3, 3, 1])
>>> tf.tile(sobel_x, [1, 1, 64])
<tf.Tensor 'Tile_6:0' shape=(3, 3, 64) dtype=float32>
I think your issue is with sobel_x.
sobel_x.get_shape(): TensorShape([Dimension(3), Dimension(3)])
sobel_x: <tf.tensor 'Const:0' shape=(3, 3) dtype=float32
So sobel_x is a two dimension matrix and your passing a rank 3 input to tile hence the error.
Fix: Make sobel_x 3 dimensional such that the shape is shape=(3, 3, 1)
then tf.tile(sobel_x, [1, 1, 64] will output shape=(1, 1, 64)