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The Introduction to gradients and automatic differentiation guide includes everything required to calculate gradients in TensorFlow. This guide focuses on deeper, less common features of the tf.GradientTape API.
Setup
import tensorflow as tf
import matplotlib as mpl
import matplotlib.pyplot as plt
mpl.rcParams['figure.figsize'] = (8, 6)
2024-07-19 02:50:43.916438: E external/local_xla/xla/stream_executor/cuda/cuda_fft.cc:485] Unable to register cuFFT factory: Attempting to register factory for plugin cuFFT when one has already been registered 2024-07-19 02:50:43.937690: E external/local_xla/xla/stream_executor/cuda/cuda_dnn.cc:8454] Unable to register cuDNN factory: Attempting to register factory for plugin cuDNN when one has already been registered 2024-07-19 02:50:43.944003: E external/local_xla/xla/stream_executor/cuda/cuda_blas.cc:1452] Unable to register cuBLAS factory: Attempting to register factory for plugin cuBLAS when one has already been registered
Controlling gradient recording
In the automatic differentiation guide you saw how to control which variables and tensors are watched by the tape while building the gradient calculation.
The tape also has methods to manipulate the recording.
Stop recording
If you wish to stop recording gradients, you can use tf.GradientTape.stop_recording to temporarily suspend recording.
This may be useful to reduce overhead if you do not wish to differentiate a complicated operation in the middle of your model. This could include calculating a metric or an intermediate result:
x = tf.Variable(2.0)
y = tf.Variable(3.0)
with tf.GradientTape() as t:
x_sq = x * x
with t.stop_recording():
y_sq = y * y
z = x_sq + y_sq
grad = t.gradient(z, {'x': x, 'y': y})
print('dz/dx:', grad['x']) # 2*x => 4
print('dz/dy:', grad['y'])
dz/dx: tf.Tensor(4.0, shape=(), dtype=float32) dz/dy: None WARNING: All log messages before absl::InitializeLog() is called are written to STDERR I0000 00:00:1721357446.860587 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.864384 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.868144 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.871707 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.884190 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.887603 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.891098 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.894451 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.897962 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.901403 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.904883 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357446.908301 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.171935 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.174025 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.175971 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.177986 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.180036 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.181937 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.183763 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.185685 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.187626 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.189545 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.191377 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.193295 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.231363 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.233482 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.235953 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.237931 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.239900 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.241810 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.243663 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.245603 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.247582 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.249942 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.252134 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355 I0000 00:00:1721357448.254535 136363 cuda_executor.cc:1015] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero. See more at https://github.com/torvalds/linux/blob/v6.0/Documentation/ABI/testing/sysfs-bus-pci#L344-L355
Reset/start recording from scratch
If you wish to start over entirely, use tf.GradientTape.reset. Simply exiting the gradient tape block and restarting is usually easier to read, but you can use the reset method when exiting the tape block is difficult or impossible.
x = tf.Variable(2.0)
y = tf.Variable(3.0)
reset = True
with tf.GradientTape() as t:
y_sq = y * y
if reset:
# Throw out all the tape recorded so far.
t.reset()
z = x * x + y_sq
grad = t.gradient(z, {'x': x, 'y': y})
print('dz/dx:', grad['x']) # 2*x => 4
print('dz/dy:', grad['y'])
dz/dx: tf.Tensor(4.0, shape=(), dtype=float32) dz/dy: None
Stop gradient flow with precision
In contrast to the global tape controls above, the tf.stop_gradient function is much more precise. It can be used to stop gradients from flowing along a particular path, without needing access to the tape itself:
x = tf.Variable(2.0)
y = tf.Variable(3.0)
with tf.GradientTape() as t:
y_sq = y**2
z = x**2 + tf.stop_gradient(y_sq)
grad = t.gradient(z, {'x': x, 'y': y})
print('dz/dx:', grad['x']) # 2*x => 4
print('dz/dy:', grad['y'])
dz/dx: tf.Tensor(4.0, shape=(), dtype=float32) dz/dy: None
Custom gradients
In some cases, you may want to control exactly how gradients are calculated rather than using the default. These situations include:
- There is no defined gradient for a new op you are writing.
- The default calculations are numerically unstable.
- You wish to cache an expensive computation from the forward pass.
- You want to modify a value (for example, using
tf.clip_by_valueortf.math.round) without modifying the gradient.
For the first case, to write a new op you can use tf.RegisterGradient to set up your own (refer to the API docs for details). (Note that the gradient registry is global, so change it with caution.)
For the latter three cases, you can use tf.custom_gradient.
Here is an example that applies tf.clip_by_norm to the intermediate gradient:
# Establish an identity operation, but clip during the gradient pass.
@tf.custom_gradient
def clip_gradients(y):
def backward(dy):
return tf.clip_by_norm(dy, 0.5)
return y, backward
v = tf.Variable(2.0)
with tf.GradientTape() as t:
output = clip_gradients(v * v)
print(t.gradient(output, v)) # calls "backward", which clips 4 to 2
tf.Tensor(2.0, shape=(), dtype=float32)
Refer to the tf.custom_gradient decorator API docs for more details.
Custom gradients in SavedModel
Custom gradients can be saved to SavedModel by using the option tf.saved_model.SaveOptions(experimental_custom_gradients=True).
To be saved into the SavedModel, the gradient function must be traceable (to learn more, check out the Better performance with tf.function guide).
class MyModule(tf.Module):
@tf.function(input_signature=[tf.TensorSpec(None)])
def call_custom_grad(self, x):
return clip_gradients(x)
model = MyModule()
tf.saved_model.save(
model,
'saved_model',
options=tf.saved_model.SaveOptions(experimental_custom_gradients=True))
# The loaded gradients will be the same as the above example.
v = tf.Variable(2.0)
loaded = tf.saved_model.load('saved_model')
with tf.GradientTape() as t:
output = loaded.call_custom_grad(v * v)
print(t.gradient(output, v))
INFO:tensorflow:Assets written to: saved_model/assets INFO:tensorflow:Assets written to: saved_model/assets tf.Tensor(2.0, shape=(), dtype=float32)
A note about the above example: If you try replacing the above code with tf.saved_model.SaveOptions(experimental_custom_gradients=False), the gradient will still produce the same result on loading. The reason is that the gradient registry still contains the custom gradient used in the function call_custom_op. However, if you restart the runtime after saving without custom gradients, running the loaded model under the tf.GradientTape will throw the error: LookupError: No gradient defined for operation 'IdentityN' (op type: IdentityN).
Multiple tapes
Multiple tapes interact seamlessly.
For example, here each tape watches a different set of tensors:
x0 = tf.constant(0.0)
x1 = tf.constant(0.0)
with tf.GradientTape() as tape0, tf.GradientTape() as tape1:
tape0.watch(x0)
tape1.watch(x1)
y0 = tf.math.sin(x0)
y1 = tf.nn.sigmoid(x1)
y = y0 + y1
ys = tf.reduce_sum(y)
tape0.gradient(ys, x0).numpy() # cos(x) => 1.0
1.0
tape1.gradient(ys, x1).numpy() # sigmoid(x1)*(1-sigmoid(x1)) => 0.25
0.25
Higher-order gradients
Operations inside of the tf.GradientTape context manager are recorded for automatic differentiation. If gradients are computed in that context, then the gradient computation is recorded as well. As a result, the exact same API works for higher-order gradients as well.
For example:
x = tf.Variable(1.0) # Create a Tensorflow variable initialized to 1.0
with tf.GradientTape() as t2:
with tf.GradientTape() as t1:
y = x * x * x
# Compute the gradient inside the outer `t2` context manager
# which means the gradient computation is differentiable as well.
dy_dx = t1.gradient(y, x)
d2y_dx2 = t2.gradient(dy_dx, x)
print('dy_dx:', dy_dx.numpy()) # 3 * x**2 => 3.0
print('d2y_dx2:', d2y_dx2.numpy()) # 6 * x => 6.0
dy_dx: 3.0 d2y_dx2: 6.0
While that does give you the second derivative of a scalar function, this pattern does not generalize to produce a Hessian matrix, since tf.GradientTape.gradient only computes the gradient of a scalar. To construct a Hessian matrix, go to the Hessian example under the Jacobian section.
"Nested calls to tf.GradientTape.gradient" is a good pattern when you are calculating a scalar from a gradient, and then the resulting scalar acts as a source for a second gradient calculation, as in the following example.
Example: Input gradient regularization
Many models are susceptible to "adversarial examples". This collection of techniques modifies the model's input to confuse the model's output. The simplest implementation—such as the Adversarial example using the Fast Gradient Signed Method attack—takes a single step along the gradient of the output with respect to the input; the "input gradient".
One technique to increase robustness to adversarial examples is input gradient regularization (Finlay & Oberman, 2019), which attempts to minimize the magnitude of the input gradient. If the input gradient is small, then the change in the output should be small too.
Below is a naive implementation of input gradient regularization. The implementation is:
- Calculate the gradient of the output with respect to the input using an inner tape.
- Calculate the magnitude of that input gradient.
- Calculate the gradient of that magnitude with respect to the model.
x = tf.random.normal([7, 5])
layer = tf.keras.layers.Dense(10, activation=tf.nn.relu)
with tf.GradientTape() as t2:
# The inner tape only takes the gradient with respect to the input,
# not the variables.
with tf.GradientTape(watch_accessed_variables=False) as t1:
t1.watch(x)
y = layer(x)
out = tf.reduce_sum(layer(x)**2)
# 1. Calculate the input gradient.
g1 = t1.gradient(out, x)
# 2. Calculate the magnitude of the input gradient.
g1_mag = tf.norm(g1)
# 3. Calculate the gradient of the magnitude with respect to the model.
dg1_mag = t2.gradient(g1_mag, layer.trainable_variables)
[var.shape for var in dg1_mag]
[TensorShape([5, 10]), TensorShape([10])]
Jacobians
All the previous examples took the gradients of a scalar target with respect to some source tensor(s).
The Jacobian matrix represents the gradients of a vector valued function. Each row contains the gradient of one of the vector's elements.
The tf.GradientTape.jacobian method allows you to efficiently calculate a Jacobian matrix.
Note that:
- Like
gradient: Thesourcesargument can be a tensor or a container of tensors. - Unlike
gradient: Thetargettensor must be a single tensor.
Scalar source
As a first example, here is the Jacobian of a vector-target with respect to a scalar-source.
x = tf.linspace(-10.0, 10.0, 200+1)
delta = tf.Variable(0.0)
with tf.GradientTape() as tape:
y = tf.nn.sigmoid(x+delta)
dy_dx = tape.jacobian(y, delta)
When you take the Jacobian with respect to a scalar the result has the shape of the target, and gives the gradient of the each element with respect to the source:
print(y.shape)
print(dy_dx.shape)
(201,) (201,)
plt.plot(x.numpy(), y, label='y')
plt.plot(x.numpy(), dy_dx, label='dy/dx')
plt.legend()
_ = plt.xlabel('x')
Tensor source
Whether the input is scalar or tensor, tf.GradientTape.jacobian efficiently calculates the gradient of each element of the source with respect to each element of the target(s).
For example, the output of this layer has a shape of (10, 7):
x = tf.random.normal([7, 5])
layer = tf.keras.layers.Dense(10, activation=tf.nn.relu)
with tf.GradientTape(persistent=True) as tape:
y = layer(x)
y.shape
TensorShape([7, 10])
And the layer's kernel's shape is (5, 10):
layer.kernel.shape
TensorShape([5, 10])
The shape of the Jacobian of the output with respect to the kernel is those two shapes concatenated together:
j = tape.jacobian(y, layer.kernel)
j.shape
TensorShape([7, 10, 5, 10])
If you sum over the target's dimensions, you're left with the gradient of the sum that would have been calculated by tf.GradientTape.gradient:
g = tape.gradient(y, layer.kernel)
print('g.shape:', g.shape)
j_sum = tf.reduce_sum(j, axis=[0, 1])
delta = tf.reduce_max(abs(g - j_sum)).numpy()
assert delta < 1e-3
print('delta:', delta)
g.shape: (5, 10) delta: 2.3841858e-07
Example: Hessian
While tf.GradientTape doesn't give an explicit method for constructing a Hessian matrix it's possible to build one using the tf.GradientTape.jacobian method.
x = tf.random.normal([7, 5])
layer1 = tf.keras.layers.Dense(8, activation=tf.nn.relu)
layer2 = tf.keras.layers.Dense(6, activation=tf.nn.relu)
with tf.GradientTape() as t2:
with tf.GradientTape() as t1:
x = layer1(x)
x = layer2(x)
loss = tf.reduce_mean(x**2)
g = t1.gradient(loss, layer1.kernel)
h = t2.jacobian(g, layer1.kernel)
print(f'layer.kernel.shape: {layer1.kernel.shape}')
print(f'h.shape: {h.shape}')
layer.kernel.shape: (5, 8) h.shape: (5, 8, 5, 8)
To use this Hessian for a Newton's method step, you would first flatten out its axes into a matrix, and flatten out the gradient into a vector:
n_params = tf.reduce_prod(layer1.kernel.shape)
g_vec = tf.reshape(g, [n_params, 1])
h_mat = tf.reshape(h, [n_params, n_params])
The Hessian matrix should be symmetric:
def imshow_zero_center(image, **kwargs):
lim = tf.reduce_max(abs(image))
plt.imshow(image, vmin=-lim, vmax=lim, cmap='seismic', **kwargs)
plt.colorbar()
imshow_zero_center(h_mat)
The Newton's method update step is shown below:
eps = 1e-3
eye_eps = tf.eye(h_mat.shape[0])*eps
# X(k+1) = X(k) - (∇²f(X(k)))^-1 @ ∇f(X(k))
# h_mat = ∇²f(X(k))
# g_vec = ∇f(X(k))
update = tf.linalg.solve(h_mat + eye_eps, g_vec)
# Reshape the update and apply it to the variable.
_ = layer1.kernel.assign_sub(tf.reshape(update, layer1.kernel.shape))
While this is relatively simple for a single tf.Variable, applying this to a non-trivial model would require careful concatenation and slicing to produce a full Hessian across multiple variables.
Batch Jacobian
In some cases, you want to take the Jacobian of each of a stack of targets with respect to a stack of sources, where the Jacobians for each target-source pair are independent.
For example, here the input x is shaped (batch, ins) and the output y is shaped (batch, outs):
x = tf.random.normal([7, 5])
layer1 = tf.keras.layers.Dense(8, activation=tf.nn.elu)
layer2 = tf.keras.layers.Dense(6, activation=tf.nn.elu)
with tf.GradientTape(persistent=True, watch_accessed_variables=False) as tape:
tape.watch(x)
y = layer1(x)
y = layer2(y)
y.shape
TensorShape([7, 6])
The full Jacobian of y with respect to x has a shape of (batch, ins, batch, outs), even if you only want (batch, ins, outs):
j = tape.jacobian(y, x)
j.shape
TensorShape([7, 6, 7, 5])
If the gradients of each item in the stack are independent, then every (batch, batch) slice of this tensor is a diagonal matrix:
imshow_zero_center(j[:, 0, :, 0])
_ = plt.title('A (batch, batch) slice')
def plot_as_patches(j):
# Reorder axes so the diagonals will each form a contiguous patch.
j = tf.transpose(j, [1, 0, 3, 2])
# Pad in between each patch.
lim = tf.reduce_max(abs(j))
j = tf.pad(j, [[0, 0], [1, 1], [0, 0], [1, 1]],
constant_values=-lim)
# Reshape to form a single image.
s = j.shape
j = tf.reshape(j, [s[0]*s[1], s[2]*s[3]])
imshow_zero_center(j, extent=[-0.5, s[2]-0.5, s[0]-0.5, -0.5])
plot_as_patches(j)
_ = plt.title('All (batch, batch) slices are diagonal')
To get the desired result, you can sum over the duplicate batch dimension, or else select the diagonals using tf.einsum:
j_sum = tf.reduce_sum(j, axis=2)
print(j_sum.shape)
j_select = tf.einsum('bxby->bxy', j)
print(j_select.shape)
(7, 6, 5) (7, 6, 5)
It would be much more efficient to do the calculation without the extra dimension in the first place. The tf.GradientTape.batch_jacobian method does exactly that:
jb = tape.batch_jacobian(y, x)
jb.shape
WARNING:tensorflow:5 out of the last 5 calls to <function pfor.<locals>.f at 0x7fb3886ecca0> triggered tf.function retracing. Tracing is expensive and the excessive number of tracings could be due to (1) creating @tf.function repeatedly in a loop, (2) passing tensors with different shapes, (3) passing Python objects instead of tensors. For (1), please define your @tf.function outside of the loop. For (2), @tf.function has reduce_retracing=True option that can avoid unnecessary retracing. For (3), please refer to https://www.tensorflow.org/guide/function#controlling_retracing and https://www.tensorflow.org/api_docs/python/tf/function for more details. WARNING:tensorflow:5 out of the last 5 calls to <function pfor.<locals>.f at 0x7fb3886ecca0> triggered tf.function retracing. Tracing is expensive and the excessive number of tracings could be due to (1) creating @tf.function repeatedly in a loop, (2) passing tensors with different shapes, (3) passing Python objects instead of tensors. For (1), please define your @tf.function outside of the loop. For (2), @tf.function has reduce_retracing=True option that can avoid unnecessary retracing. For (3), please refer to https://www.tensorflow.org/guide/function#controlling_retracing and https://www.tensorflow.org/api_docs/python/tf/function for more details. TensorShape([7, 6, 5])
error = tf.reduce_max(abs(jb - j_sum))
assert error < 1e-3
print(error.numpy())
0.0
x = tf.random.normal([7, 5])
layer1 = tf.keras.layers.Dense(8, activation=tf.nn.elu)
bn = tf.keras.layers.BatchNormalization()
layer2 = tf.keras.layers.Dense(6, activation=tf.nn.elu)
with tf.GradientTape(persistent=True, watch_accessed_variables=False) as tape:
tape.watch(x)
y = layer1(x)
y = bn(y, training=True)
y = layer2(y)
j = tape.jacobian(y, x)
print(f'j.shape: {j.shape}')
WARNING:tensorflow:6 out of the last 6 calls to <function pfor.<locals>.f at 0x7fb3886ec9d0> triggered tf.function retracing. Tracing is expensive and the excessive number of tracings could be due to (1) creating @tf.function repeatedly in a loop, (2) passing tensors with different shapes, (3) passing Python objects instead of tensors. For (1), please define your @tf.function outside of the loop. For (2), @tf.function has reduce_retracing=True option that can avoid unnecessary retracing. For (3), please refer to https://www.tensorflow.org/guide/function#controlling_retracing and https://www.tensorflow.org/api_docs/python/tf/function for more details. WARNING:tensorflow:6 out of the last 6 calls to <function pfor.<locals>.f at 0x7fb3886ec9d0> triggered tf.function retracing. Tracing is expensive and the excessive number of tracings could be due to (1) creating @tf.function repeatedly in a loop, (2) passing tensors with different shapes, (3) passing Python objects instead of tensors. For (1), please define your @tf.function outside of the loop. For (2), @tf.function has reduce_retracing=True option that can avoid unnecessary retracing. For (3), please refer to https://www.tensorflow.org/guide/function#controlling_retracing and https://www.tensorflow.org/api_docs/python/tf/function for more details. j.shape: (7, 6, 7, 5)
plot_as_patches(j)
_ = plt.title('These slices are not diagonal')
_ = plt.xlabel("Don't use `batch_jacobian`")
In this case, batch_jacobian still runs and returns something with the expected shape, but its contents have an unclear meaning:
jb = tape.batch_jacobian(y, x)
print(f'jb.shape: {jb.shape}')
jb.shape: (7, 6, 5)