writing a custom cost function in tensorflow - numpy

I'm trying to write my own cost function in tensor flow, however apparently I cannot 'slice' the tensor object?
import tensorflow as tf
import numpy as np
# Establish variables
x = tf.placeholder("float", [None, 3])
W = tf.Variable(tf.zeros([3,6]))
b = tf.Variable(tf.zeros([6]))
# Establish model
y = tf.nn.softmax(tf.matmul(x,W) + b)
# Truth
y_ = tf.placeholder("float", [None,6])
def angle(v1, v2):
return np.arccos(np.sum(v1*v2,axis=1))
def normVec(y):
return np.cross(y[:,[0,2,4]],y[:,[1,3,5]])
angle_distance = -tf.reduce_sum(angle(normVec(y_),normVec(y)))
# This is the example code they give for cross entropy
cross_entropy = -tf.reduce_sum(y_*tf.log(y))
I get the following error:
TypeError: Bad slice index [0, 2, 4] of type <type 'list'>

At present, tensorflow can't gather on axes other than the first - it's requested.
But for what you want to do in this specific situation, you can transpose, then gather 0,2,4, and then transpose back. It won't be crazy fast, but it works:
tf.transpose(tf.gather(tf.transpose(y), [0,2,4]))
This is a useful workaround for some of the limitations in the current implementation of gather.
(But it is also correct that you can't use a numpy slice on a tensorflow node - you can run it and slice the output, and also that you need to initialize those variables before you run. :). You're mixing tf and np in a way that doesn't work.
x = tf.Something(...)
is a tensorflow graph object. Numpy has no idea how to cope with such objects.
foo = tf.run(x)
is back to an object python can handle.
You typically want to keep your loss calculation in pure tensorflow, so do the cross and other functions in tf. You'll probably have to do the arccos the long way, as tf doesn't have a function for it.

just realized that the following failed:
cross_entropy = -tf.reduce_sum(y_*np.log(y))
you cant use numpy functions on tf objects, and the indexing my be different too.

I think you can use "Wraps Python function" method in tensorflow. Here's the link to the documentation.
And as for the people who answered "Why don't you just use tensorflow's built in function to construct it?" - sometimes the cost function people are looking for cannot be expressed in tf's functions or extremely difficult.

This is because you have not initialized your variable and because of this it does not have your Tensor there right now (can read more in my answer here)
Just do something like this:
def normVec(y):
print y
return np.cross(y[:,[0,2,4]],y[:,[1,3,5]])
t1 = normVec(y_)
# and comment everything after it.
To see that you do not have a Tensor now and only Tensor("Placeholder_1:0", shape=TensorShape([Dimension(None), Dimension(6)]), dtype=float32).
Try initializing your variables
init = tf.initialize_all_variables()
sess = tf.Session()
sess.run(init)
and evaluate your variable sess.run(y). P.S. you have not fed your placeholders up till now.

Related

K-Means of Tensorflow - Graph disconnected error

I am trying to write a function that runs KMeans on a dataset and outputs the cluster centroids. My aim is to use this in a custom keras layer, so I am using TensorFlow's implementation of KMeans that takes a tensor as the input dataset.
My problem however is that I can't make it work even as a standalone function. The problem comes from the fact that KMeans accepts a generator function that provides mini-batches instead of a plain tensor, but when I am using closure to do that, I get a graph disconnected error:
import tensorflow as tf # version: 2.4.1
from tensorflow.compat.v1.estimator.experimental import KMeans
#tf.function
def KMeansCentroids(inputs, num_clusters, steps, use_mini_batch=False):
# `inputs` is a 2D tensor
def input_fn():
# Each one of the lines below results in the same "Graph Disconnected" error. Tuples don't really needed but just to be consistent with the documentation
return (inputs, None)
return (tf.data.Dataset.from_tensor_slices(inputs), None)
return (tf.convert_to_tensor(inputs), None)
kmeans = KMeans(
num_clusters=num_clusters,
use_mini_batch=use_mini_batch)
kmeans.train(input_fn, steps=steps) # This is where the error happens
return kmeans.cluster_centers()
>>> x = tf.random.uniform((100, 2))
>>> c = KMeansCentroids(x, 5, 10)
The exact error is:
ValueError:
Tensor("strided_slice:0", shape=(), dtype=int32)
must be from the same graph as
Tensor("Equal:0", shape=(), dtype=bool)
(graphs are FuncGraph(name=KMeansCentroids, id=..) and <tensorflow.python.framework.ops.Graph object at ...>).
If I were to use a numpy dataset and convert to tensor inside the function, the code would work just fine.
Also, making input_fn() return directly tf.random.uniform((100, 2)) (ignoring the inputs argument), would again work. That's why I am guessing that tensorflow doesn't support closures since it needs to build the computation graph at the beginning.
But I don't see how to work around that.
Could it be a version error due to KMeans being a compat.v1.experimental module?
Note that the documentation of KMeans states for the input_fn():
The function should construct and return one of the following:
A tf.data.Dataset object: Outputs of Dataset object must be a tuple (features, labels) with same constraints as below.
A tuple (features, labels): Where features is a tf.Tensor or a dictionary of string feature name to Tensor and labels is a Tensor or a dictionary of string label name to Tensor. Both features and labels are consumed by model_fn. They should satisfy the expectation of model_fn from inputs.
The problem you're facing is more about invoking tensor outside the created graph. Basically, when you called the .train function, a new graph will be created and that is with the graph defined in that input_fn and the graph defined in the model_fn.
kmeans.train(input_fn, steps=steps)
And, after that all the tensors those coming outside these functions will be treated as outsiders and won't part of this new graph. That's why you're getting a graph disconnected error for trying to use outsider tensor. To resolve this, you need to create the necessary tensors within these graphs.
import tensorflow as tf
from tensorflow.compat.v1.estimator.experimental import KMeans
#tf.function
def KMeansCentroids(num_clusters, steps, use_mini_batch=False):
def input_fn(batch_size):
pinputs = tf.random.uniform((100, 2))
dataset = tf.data.Dataset.from_tensor_slices((pinputs))
dataset = dataset.shuffle(1000).repeat()
return dataset.batch(batch_size)
kmeans = KMeans(
num_clusters=num_clusters,
use_mini_batch=use_mini_batch)
kmeans.train(input_fn = lambda: input_fn(5),
steps=steps)
return kmeans.cluster_centers()
c = KMeansCentroids(5, 10)
Here is some more info for reading, 1. FYI, I tested your code with a few versions of tf > 2, and I don't think it's related to version error or something.
Re-mentioning here for future readers. An alternative of using KMeans within Keras layers:
tf_kmeans.py
ClusteringLayer

Simple way to convert tensor to numpy array without eager mode in TF 2.2

I can't find a simple way to convert a tensor to a NumPy array without enabling eager mode, which gives a nice .numpy() method, but also slows down my model training.
I'd be super grateful for your suggestions. For context, I'm writing a custom metric for my TensorFlow model that relies on a scikit learn function, which only takes numpy arrays.
I've tried wrapping the tensors with np.array(), which throws a not implemented error. Also gave sessions and .eval() a go, but didn't get it to work either and seemed like too much for this simple job.
My specific error:
NotImplementedError: Cannot convert a symbolic Tensor (model_17/dense_17/Sigmoid:0) to a numpy array.
# Custom metric
def accuracy_ml(y_true, y_pred):
return accuracy_score(y_true, np.round(y_pred)) # ERROR here feeding tensor to sklearn function
# Model
cnn = simple_model(input_shape=(224, 224, 3),
num_classes=10,
base_model = base_ResNet101)
lr = 1e-2
loss_fn = tf.keras.losses.BinaryCrossentropy()
metrics = [accuracy_ml]
cnn.compile(optimizer=tf.keras.optimizers.Adam(learning_rate=lr),
loss=loss_fn,
metrics=metrics)
# Simple baseline eval that fails
validation_steps=17
loss0, accuracy0 = cnn.evaluate(validation_batches, steps = validation_steps)
Wrapping my NumPy metric with tf.numpy_function() solved it. https://www.tensorflow.org/api_docs/python/tf/numpy_function

How do you set a Tensorflow 2 Keras optimizer to a state, before you've applied grads with it?

I'm working at a slightly lower-level of Keras than the Model fit API. I would like to be able to set the state of a newly constructed optimizer to the state of it from previous training.
The get_weights and set_weights methods seem promising; they just return and receive numpy arrays or standard scalar data for the state of the optimizer. However, the problem is you cannot set_weights if the weights have not yet been created, and as far as I can tell, the only public way they get created is on the first call to apply_gradients.
For example, the following fails because opt2 will not have its weights created.
import tensorflow as tf
import numpy as np
opt1 = tf.keras.optimizers.Adam()
opt2 = tf.keras.optimizers.Adam()
layer = tf.keras.layers.Dense(1)
# dummy data
x = np.array([[-1, 1], [1, 1]])
y = np.array([[-1], [1]])
# do one optimization step
with tf.GradientTape() as tape:
loss = (layer(x) - y)**2
grads = tape.gradient(loss, layer.trainable_weights)
opt1.apply_gradients(zip(grads, layer.trainable_weights))
# copy state to optimizer 2
opt2.set_weights(opt1.get_weights()) # this fails!
Lets assume I do have on hand the relevant model weights on which the optimizer operates. What is the right way restore state? Based on the implementation of the apply_gradients method, it seems like this is the path:
_ = opt2.iterations # must be called to make this weight appear
opt2._create_hypers()
opt2._create_slots(layer.trainable_weights)
# now we can safely set weights
opt2.set_weights(opt1.get_weights())
But that feels really hacky to me and prone to fail if implementation details change at a future point. Are there better approaches that I'm missing?

Tensorflow: How to feed a placeholder variable with a tensor?

I have a placeholder variable that expects a batch of input images:
input_placeholder = tf.placeholder(tf.float32, [None] + image_shape, name='input_images')
Now I have 2 sources for the input data:
1) a tensor and
2) some numpy data.
For the numpy input data, I know how to feed data to the placeholder variable:
sess = tf.Session()
mLoss, = sess.run([loss], feed_dict = {input_placeholder: myNumpyData})
How can I feed a tensor to that placeholder variable?
mLoss, = sess.run([loss], feed_dict = {input_placeholder: myInputTensor})
gives me an error:
TypeError: The value of a feed cannot be a tf.Tensor object. Acceptable feed values include Python scalars, strings, lists, or numpy ndarrays.
I don't want to convert the tensor into a numpy array using .eval(), since that would slow my program down, is there any other way?
This has been discussed on GitHub in 2016, and please check here. Here is the key point by concretevitamin:
One key thing to note is that Tensor is simply a symbolic object. The values of your feed_dict are the actual values, e.g. a Numpy ndarry.
The tensor as a symbolic object is flowing in the graph while the actual values are outside of it, then we can only pass the actual values into the graph and the symbolic object can not exist outside the graph.
You can use feed_dict to feed data into non-placeholders. So, first, wire up your dataflow graph directly to your myInputTensor tensor data source (i.e. don't use a placeholder). Then when you want to run with your numpy data you can effectively mask myImportTensor with myNumpyData, like this:
mLoss, = sess.run([loss], feed_dict={myImportTensor: myNumpyData})
[I'm still trying to figure out how to do this with multiple tensor data sources however.]
One way of solving the problem is to actually remove the Placeholder tensor and replace it by your "myInputTensor".
You will use the myInputTensor as the source for the other operations in the graph and when you want to infer the graph with your np array as input data, you will feed a value to this tensor directly.
Here is a quick example:
import tensorflow as tf
import numpy as np
# Input Tensor
myInputTensor = tf.ones(dtype=tf.float32, shape=1) # In your case, this would be the results of some ops
output = myInputTensor * 5.0
with tf.Session() as sess:
print(sess.run(output)) # == 5.0, using the Tensor value
myNumpyData = np.zeros(1)
print(sess.run(output, {myInputTensor: myNumpyData}) # == 0.0 * 5.0 = 0.0, using the np value
This works for me in latest version...maybe you have older version of TF?
a = tf.Variable(1)
sess.run(2*a, feed_dict={a:5}) # prints 10

How to make contrib.learn (SKFLOW) regression predict multiple values (TensorFlow 0.12.1)

This is very similar to the question skflow regression predict multiple values. However, later versions of TensorFlow seem to rendered the answer from this question obsolete.
I would like to be able to have multiple output neurons in a TensorFlow Learn regression neural network (DNNRegressor). I upgraded the code from the referenced question to account for breaking changes in TensorFlow, but still get an error.
import numpy as np
import tensorflow.contrib.learn as skflow
import tensorflow as tf
from sklearn.metrics import mean_squared_error
# Create random dataset.
rng = np.random.RandomState(1)
X = np.sort(200 * rng.rand(100, 1) - 100, axis=0)
y = np.array([np.pi * np.sin(X).ravel(), np.pi * np.cos(X).ravel()]).T
# Fit regression DNN model.
feature_columns = [tf.contrib.layers.real_valued_column("", dimension=X.shape[0])]
regressor = skflow.DNNRegressor(hidden_units=[5, 5],feature_columns=feature_columns)
regressor.fit(X, y)
score = mean_squared_error(regressor.predict(X), y)
print("Mean Squared Error: {0:f}".format(score))
But this results in:
ValueError: Shapes (?, 1) and (?, 2) are incompatible
I don't seen any release notes about breaking changes that indicate that the method for multiple outputs have changed. Is there another way to do this?
As mentioned in the tf.contrib.learn.DNNRegressor docs, you may use the label_dimension parameter, which is exactly what you are looking for.
Your code line with this param will do what you want:
regressor = skflow.DNNRegressor(hidden_units=[5, 5],
feature_columns=feature_columns,
label_dimension=2)
The standard predict() returns an generator object. To get an array, you have to add as_iterable=False:
score = metrics.mean_squared_error(regressor.predict(X, as_iterable=False), y)