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
Related
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
I want to change the shape and the content of the tensor in a keras model. Tensor is the output of a layer and has
shape1=(batch_size, max_sentences_in_doc, max_tokens_in_doc, embedding_size)
and I want to convert to
shape2=(batch_size, max_documents_length, embedding_size)
suitable as input of the next layer. Here sentences are made of tokens, and are zero-padded so every sentence has length=max_tokens_in_sentence.
In detail:
I wanto to concatenate all the sentences of a batch taking only the nonzero part of the sentences;
then I zero-pad this concatenation to a length=max_document_length.
So passing from shape1 to shape2 is not only a reshape as mathematical operations are involved.
I created the function embedding_to_docs(x) that iterates on the tensor of shape1 to transform it into shape2. I call the function using a Lambda layer in the model, it works in debug with fictious data, but when I try to call it during the build of the model an error is raised:
Tensor objects are only iterable when eager execution is enabled. To iterate over this tensor use tf.map_fn.
def embedding_to_docs(x):
new_output = []
for doc in x:
document = []
for sentence in doc:
non_zero_indexes = np.nonzero(sentence[:, 0])
max_index = max(non_zero_indexes[0])
if max_index > 0:
document.extend(sentence[0:max_index])
if MAX_DOCUMENT_LENGTH-len(document) > 0:
a = np.zeros((MAX_DOCUMENT_LENGTH-len(document), 1024))
document.extend(a)
else:
document = document[0:MAX_DOCUMENT_LENGTH]
new_output.append(document)
return np.asarray(new_output)
...
# in the model:
tensor_of_shape2 = Lambda(embedding_to_docs)(tensor_of_shape1)
How to fix this?
You can use py_function, which allows you to switch from the graph mode (used by Keras) to the eager mode (where it is possible to iterate over tensors like in your function).
def to_docs(x):
return tf.py_function(embedding_to_docs, [x], tf.float32)
tensor_of_shape2 = Lambda(to_docs)(tensor_of_shape1)
Note that the code run within your embedding_to_docs must be written in tensorflow eager instead of numpy. This means that you'd need to replace some of the numpy calls with tensorflow. You'd surely need to replace the return line with:
return tf.convert_to_tensor(new_output)
Using numpy arrays will stop the gradient computation, but you are likely not interested in gradient flowing through the input data anyway.
I have a regular keras model called e and I would like to compare its output for both y_pred and y_true in my custom loss function.
from keras import backend as K
def custom_loss(y_true, y_pred):
return K.mean(K.square(e.predict(y_pred)-e.predict(y_true)), axis=-1)
I am getting the error: AttributeError: 'Tensor' object has no attribute 'ndim'
This is because y_true and y_pred are both tensor object and keras.model.predict expects to be passed a numpy.array.
Any idea how I may succeed in using my keras.model in my custom loss function?
I am open to getting the output of a specified layer if need be or to converting my keras.model to a tf.estimator object (or anything else).
First, let's try to understand the error message you're getting:
AttributeError: 'Tensor' object has no attribute 'ndim'
Let's take a look at the Keras documentation and find the predict method of Keras model. We can see the description of the function parameters:
x: the input data, as a Numpy array.
So, the model is trying to get a ndims property of a numpy array, because it expects an array as input. On other hand, the custom loss function of the Keras framework gets tensors as inputs. So, don't write any python code inside it - it will never be executed during evaluation. This function is just called to construct the computational graph.
Okay, now that we found out the meaning behind that error message, how can we use a Keras model inside custom loss function? Simple! We just need to get the evaluation graph of the model.
Update
The use of global keyword is a bad coding practice. Also, now in 2020 we have better functional API in Keras that makes hacks with layers uneccessary. Better use something like this:
from keras import backend as K
def make_custom_loss(model):
"""Creates a loss function that uses `model` for evaluation
"""
def custom_loss(y_true, y_pred):
return K.mean(K.square(model(y_pred) - model(y_true)), axis=-1)
return custom_loss
custom_loss = make_custom_loss(e)
Deprecated
Try something like this (only for Sequential models and very old API):
def custom_loss(y_true, y_pred):
# Your model exists in global scope
global e
# Get the layers of your model
layers = [l for l in e.layers]
# Construct a graph to evaluate your other model on y_pred
eval_pred = y_pred
for i in range(len(layers)):
eval_pred = layers[i](eval_pred)
# Construct a graph to evaluate your other model on y_true
eval_true = y_true
for i in range(len(layers)):
eval_true = layers[i](eval_true)
# Now do what you wanted to do with outputs.
# Note that we are not returning the values, but a tensor.
return K.mean(K.square(eval_pred - eval_true), axis=-1)
Please note that the code above is not tested. However, the general idea will stay the same regardless of the implementation: you need to construct a graph, in which the y_true and y_pred will flow through it to the final operations.
I'd like to pass the parameters of the trained model (weights and bias for convolution and fully connected layers) to other frameworks or languages including iOS and Torch by parsing the saved file.
I tried tf.train.write_graph(session.graph_def, '', 'graph.pb'), but it seems it only includes the graph architecture without weights and bias. If so, to create checkpoint file (saver.save(session, "model.ckpt")) is the best way? Is it easy to parse ckpt file type in Swift or other languages?
Please let me know if you have any suggestions.
Instead of parsing a .ckpt file, you can just try evaluating the tensor (in your case the weights of a convolutional layer) and getting the values as a numpy array. Here is a quick toy example (tested on r0.10 - there might some small API changes in newer versions):
import tensorflow as tf
import numpy as np
x = tf.placeholder(np.float32, [2,1])
w = tf.Variable(tf.truncated_normal([2,2], stddev=0.1))
b = tf.Variable(tf.constant(1.0, shape=[2,1]))
z = tf.matmul(w, x) + b
with tf.Session() as sess:
sess.run(tf.initialize_all_variables())
w_val, z_val = sess.run([w, z], feed_dict={x: np.arange(2).reshape(2,1)})
print(w_val)
print(z_val)
Output:
[[-0.02913031 0.13549708]
[ 0.13807134 0.03763327]]
[[ 1.13549709]
[ 1.0376333 ]]
If you have trouble getting a reference to your tensor (say it is in nested into a higher-level "layer" operation), try finding by name. More info here: Tensorflow: How to get a tensor by name?
If you want to see the how the weights change during training, you can also try to save all the values you are interested into tf.Summary objects and parse them later: Parsing `summary_str` byte string evaluated on tensorflow summary object
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.