add_n() takes at most 2 arguments (4 given) - tensorflow

This is my first code with tensor flow.
If I split the addition it works, but I am sure there's a way to add N tensors in one line.
import tensorflow as tf
# Create a graph.
g = tf.Graph()
# Establish the graph as the "default" graph.
with g.as_default():
# Assemble a graph consisting of the following three operations:
# * Two tf.constant operations to create the operands.
# * One tf.add operation to add the two operands.
x = tf.constant(8, name="x_const")
y = tf.constant(5, name="y_const")
z = tf.constant(4, name="z_const")
my_sum = tf.add_n(x, y, z, name="z_y_x_sum")
# Now create a session.
# The session will run the default graph.
with tf.Session() as sess:
print my_sum.eval()
Can you help me to figure out what's wrong?

You need to put the tensors into a list:
my_sum = tf.add_n([x, y, z], name="z_y_x_sum")

Related

Tabular data: Implementing a custom tensor layer without resorting to iteration

I have an idea for a tensor operation that would not be difficult to implement via iteration, with batch size one. However I would like to parallelize it as much as possible.
I have two tensors with shape (n, 5) called X and Y. X is actually supposed to represent 5 one-dimensional tensors with shape (n, 1): (x_1, ..., x_n). Ditto for Y.
I would like to compute a tensor with shape (n, 25) where each column represents the output of the tensor operation f(x_i, y_j), where f is fixed for all 1 <= i, j <= 5. The operation f has output shape (n, 1), just like x_i and y_i.
I feel it is important to clarify that f is essentially a fully-connected layer from the concatenated [...x_i, ...y_i] tensor with shape (1, 10), to an output layer with shape (1,5).
Again, it is easy to see how to do this manually with iteration and slicing. However this is probably very slow. Performing this operation in batches, where the tensors X, Y now have shape (n, 5, batch_size) is also desirable, particularly for mini-batch gradient descent.
It is difficult to really articulate here why I desire to create this network; I feel it is suited for my domain of 'itemized tabular data' and cuts down significantly on the number of weights per operation, compared to a fully connected network.
Is this possible using tensorflow? Certainly not using just keras.
Below is an example in numpy per AloneTogether's request
import numpy as np
features = 16
batch_size = 256
X_batch = np.random.random((features, 5, batch_size))
Y_batch = np.random.random((features, 5, batch_size))
# one tensor operation to reduce weights in this custom 'layer'
f = np.random.random((features, 2 * features))
for b in range(batch_size):
X = X_batch[:, :, b]
Y = Y_batch[:, :, b]
for i in range(5):
x_i = X[:, i:i+1]
for j in range(5):
y_j = Y[:, j:j+1]
x_i_y_j = np.concatenate([x_i, y_j], axis=0)
# f(x_i, y_j)
# implemented by a fully-connected layer
f_i_j = np.matmul(f, x_i_y_j)
All operations you need (concatenation and matrix multiplication) can be batched.
Difficult part here is, that you want to concatenate features of all items in X with features of all items in Y (all combinations).
My recommended solution is to expand the dimensions of X to [batch, features, 5, 1], expand dimensions of Y to [batch, features, 1, 5]
Than tf.repeat() both tensors so their shapes become [batch, features, 5, 5].
Now you can concatenate X and Y. You will have a tensor of shape [batch, 2*features, 5, 5]. Observe that this way all combinations are built.
Next step is matrix multiplication. tf.matmul() can also do batch matrix multiplication, but I use here tf.einsum() because I want more control over which dimensions are considered as batch.
Full code:
import tensorflow as tf
import numpy as np
batch_size=3
features=6
items=5
x = np.random.uniform(size=[batch_size,features,items])
y = np.random.uniform(size=[batch_size,features,items])
f = np.random.uniform(size=[2*features,features])
x_reps= tf.repeat(x[:,:,:,tf.newaxis], items, axis=3)
y_reps= tf.repeat(y[:,:,tf.newaxis,:], items, axis=2)
xy_conc = tf.concat([x_reps,y_reps], axis=1)
f_i_j = tf.einsum("bfij, fg->bgij", xy_conc,f)
f_i_j = tf.reshape(f_i_j , [batch_size,features,items*items])

tensorflow model trained on keras.preprocessing.timeseries_dataset_from_array yields unexpected output shape of (sequence_length, 1)

I'm trying to train a tensorflow model where my inputs are a lagged timeseries of multiple features and I want to predict a single value.
Somehow the output shape ends up as an array of (lag/sequence_length, 1) when my lagged dataset has more than one feature, but I haven't been able to figure out why exactly that is. Here is a minimal example of what I'm trying to do
import tensorflow as tf
from tensorflow.keras import layers
import numpy as np
import pandas as pd
# generate some dummy data
x0 = np.array(range(300))
x1 = np.array(range(300)) * 2
df = pd.DataFrame({"x0": x0, "x1": x1})
y = np.array(range(100))
# also tried reshaping my y, but no help
# y = np.array(range(100)).reshape(100,1)
# make a dataset with lagged values
ds = tf.keras.preprocessing.timeseries_dataset_from_array(
data=df,
targets=y,
sequence_length=3,
sequence_stride=1,
sampling_rate=1,
batch_size=5
)
# show an example of what we are working with
list(ds.take(1))
# define simple model and train it
model = tf.keras.Sequential(
[
layers.Dense(32),
layers.Dense(1),
]
)
model.compile(loss="mse", optimizer=tf.optimizers.Adam())
model.fit(ds, epochs=4)
# make predictions on dataset
predictions = model.predict(ds)
# show predictions
predictions
print(predictions.shape)
"""
(100, 3, 1)
"""
If I create the dataset with only a single feature as:
ds = tf.keras.preprocessing.timeseries_dataset_from_array(
data=x1,
targets=y,
sequence_length=3,
sequence_stride=1,
sampling_rate=1,
batch_size=5
)
My outputs are of expected shape.
Would appreciate any pointers. I'm guessing something is probably getting broadcast which then results in the output I'm seeing but I haven't been able to figure out what exactly is going on.

Accessing elements of a placeholder in tensorflow [duplicate]

This question already has answers here:
Weighted cost function in tensorflow
(2 answers)
Closed 4 years ago.
I have a neural network with MSE loss function being implemented something like this:
# input x_ph is of size Nx1 and output should also be of size Nx1
def train_neural_network_batch(x_ph, predict=False):
prediction = neural_network_model(x_ph)
# MSE loss function
cost = tf.reduce_mean(tf.square(prediction - y_ph))
optimizer = tf.train.AdamOptimizer(learn_rate).minimize(cost)
# mini-batch optimization here
I'm fairly new to neural networks and Python, but I understand that each iteration, a sample of training points will be fed into the neural network and the loss function evaluated at the points in this sample. However, I would like to be able to modify the loss function so that it weights certain data more heavily. Some pseudocode of what I mean
# manually compute the MSE of the data without the first sampled element
cost = 0.0
for ii in range(1,len(y_ph)):
cost += tf.square(prediction[ii] - y_ph[ii])
cost = cost/(len(y_ph)-1.0)
# weight the first sampled data point more heavily according to some parameter W
cost += W*(prediction[0] - y_ph[0])
I might have more points I wish to weight differently as well, but for now, I'm just wondering how I can implement something like this in tensorflow. I know len(y_ph) is invalid as y_ph is just a placeholder, and I can't just do something like y_ph[i] or prediction[i].
You can do this in multiple ways:
1) If some of your data instances weighting are simply 2 times or 3 times more than normal instance, you may just copy those instance multiple times in your data set. Thus they would occupy more weight in loss, hence satisfy your intention. This is the simplest way.
2) If your weighting is more complex, say a float weighting. You can define a placeholder for weighting, multiply it to loss, and use feed_dict to feed the weighting in session together with x batch and y batch. Just make sure instance_weight is the same size with batch_size
E.g.
import tensorflow as tf
import numpy as np
with tf.variable_scope("test", reuse=tf.AUTO_REUSE):
x = tf.placeholder(tf.float32, [None,1])
y = tf.placeholder(tf.float32, [None,1])
instance_weight = tf.placeholder(tf.float32, [None,1])
w1 = tf.get_variable("w1", shape=[1, 1])
prediction = tf.matmul(x, w1)
cost = tf.square(prediction - y)
loss = tf.reduce_mean(instance_weight * cost)
opt = tf.train.AdamOptimizer(0.5).minimize(loss)
with tf.Session() as sess:
x1 = [[1.],[2.],[3.]]
y1 = [[2.],[4.],[3.]]
instance_weight1 = [[10.0], [10.0], [0.1]]
sess.run(tf.global_variables_initializer())
print (x1)
print (y1)
print (instance_weight1)
for i in range(1000):
_, loss1, prediction1 = sess.run([opt, loss, prediction], feed_dict={instance_weight : instance_weight1, x : x1, y : y1 })
if (i % 100) == 0:
print(loss1)
print(prediction1)
NOTE instance_weight1, you may change instance_weight1 to see the difference (here batch_size is set to 3)
Where x1,y1 and x2,y2 follow the rule y=2*x
Whereas x3,y3 follow the rule y=x
But with different weight as [10,10,0.1], the prediction1 coverage to y1,y2 rule and almost ignored y3, the output are as:
[[1.9823183]
[3.9646366]
[5.9469547]]
PS: in tensorflow graph, it's highly recommended not to use for loops, but use matrix operator instead to parallel the calculation.

Cleaner way to whiten each image in a batch using keras

I would like to whiten each image in a batch. The code I have to do so is this:
def whiten(self, x):
shape = x.shape
x = K.batch_flatten(x)
mn = K.mean(x, 0)
std = K.std(x, 0) + K.epsilon()
r = (x - mn) / std
r = K.reshape(x, (-1,shape[1],shape[2],shape[3]))
return r
#
where x is (?, 320,320,1). I am not keen on the reshape function with a -1 arg. Is there a cleaner way to do this?
Let's see what the -1 does. From the Tensorflow documentation (Because the documentation from Keras is scarce compared to the one from Tensorflow):
If one component of shape is the special value -1, the size of that dimension is computed so that the total size remains constant.
So what this means:
from keras import backend as K
X = tf.constant([1,2,3,4,5])
K.reshape(X, [-1, 5])
# Add one more dimension, the number of columns should be 5, and keep the number of elements to be constant
# [[1 2 3 4 5]]
X = tf.constant([1,2,3,4,5,6])
K.reshape(X, [-1, 3])
# Add one more dimension, the number of columns should be 3
# For the number of elements to be constant the number of rows should be 2
# [[1 2 3]
# [4 5 6]]
I think it is simple enough. So what happens in your code:
# Let's assume we have 5 images, 320x320 with 3 channels
X = tf.ones((5, 320, 320, 3))
shape = X.shape
# Let's flat the tensor so we can perform the rest of the computation
flatten = K.batch_flatten(X)
# What this did is: Turn a nD tensor into a 2D tensor with same 0th dimension. (Taken from the documentation directly, let's see that below)
flatten.shape
# (5, 307200)
# So all the other elements were squeezed in 1 dimension while keeping the batch_size the same
# ...The rest of the stuff in your code is executed here...
# So we did all we wanted and now we want to revert the tensor in the shape it had previously
r = K.reshape(flatten, (-1, shape[1],shape[2],shape[3]))
r.shape
# (5, 320, 320, 3)
Besides, I can't think of a cleaner way to do what you want to do. If you ask me, your code is already clear enough.

Solving XOR with 3 data points using Multi-Layered Perceptron

The XOR problem is known to be solved by the multi-layer perceptron given all 4 boolean inputs and outputs, it trains and memorizes the weights needed to reproduce the I/O. E.g.
import numpy as np
np.random.seed(0)
def sigmoid(x): # Returns values that sums to one.
return 1 / (1 + np.exp(-x))
def sigmoid_derivative(sx):
# See https://math.stackexchange.com/a/1225116
return sx * (1 - sx)
# Cost functions.
def cost(predicted, truth):
return truth - predicted
xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
xor_output = np.array([[0,1,1,0]]).T
X = xor_input
Y = xor_output
# Define the shape of the weight vector.
num_data, input_dim = X.shape
# Lets set the dimensions for the intermediate layer.
hidden_dim = 5
# Initialize weights between the input layers and the hidden layer.
W1 = np.random.random((input_dim, hidden_dim))
# Define the shape of the output vector.
output_dim = len(Y.T)
# Initialize weights between the hidden layers and the output layer.
W2 = np.random.random((hidden_dim, output_dim))
num_epochs = 10000
learning_rate = 1.0
for epoch_n in range(num_epochs):
layer0 = X
# Forward propagation.
# Inside the perceptron, Step 2.
layer1 = sigmoid(np.dot(layer0, W1))
layer2 = sigmoid(np.dot(layer1, W2))
# Back propagation (Y -> layer2)
# How much did we miss in the predictions?
layer2_error = cost(layer2, Y)
# In what direction is the target value?
# Were we really close? If so, don't change too much.
layer2_delta = layer2_error * sigmoid_derivative(layer2)
# Back propagation (layer2 -> layer1)
# How much did each layer1 value contribute to the layer2 error (according to the weights)?
layer1_error = np.dot(layer2_delta, W2.T)
layer1_delta = layer1_error * sigmoid_derivative(layer1)
# update weights
W2 += learning_rate * np.dot(layer1.T, layer2_delta)
W1 += learning_rate * np.dot(layer0.T, layer1_delta)
We see that we've fully trained the network to memorize the outputs for XOR:
# On the training data
[int(prediction > 0.5) for prediction in layer2]
[out]:
[0, 1, 1, 0]
If we re-feed the same inputs, we get the same output:
for x, y in zip(X, Y):
layer1_prediction = sigmoid(np.dot(W1.T, x)) # Feed the unseen input into trained W.
prediction = layer2_prediction = sigmoid(np.dot(W2.T, layer1_prediction)) # Feed the unseen input into trained W.
print(int(prediction > 0.5), y)
[out]:
0 [0]
1 [1]
1 [1]
0 [0]
But if we retrain the parameters (W1 and W2) without one of the data points, i.e.
xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
xor_output = np.array([[0,1,1,0]]).T
Let's drop the last row of data and use that as unseen test.
X = xor_input[:-1]
Y = xor_output[:-1]
And with the rest of the same code, regardless of how I change the hyperparameters, it's un-able to learn the XOR function and reproduce the I/O.
for x, y in zip(xor_input, xor_output):
layer1_prediction = sigmoid(np.dot(W1.T, x)) # Feed the unseen input into trained W.
prediction = layer2_prediction = sigmoid(np.dot(W2.T, layer1_prediction)) # Feed the unseen input into trained W.
print(int(prediction > 0.5), y)
[out]:
0 [0]
1 [1]
1 [1]
1 [0]
Even if we shuffle the in-/output:
# Shuffle the order of the inputs
_temp = list(zip(X, Y))
random.shuffle(_temp)
xor_input_shuff, xor_output_shuff = map(np.array, zip(*_temp))
We can't train the XOR function fully:'
for x, y in zip(xor_input, xor_output):
layer1_prediction = sigmoid(np.dot(W1.T, x)) # Feed the unseen input into trained W.
prediction = layer2_prediction = sigmoid(np.dot(W2.T, layer1_prediction)) # Feed the unseen input into trained W.
print(x, int(prediction > 0.5), y)
[out]:
[0 0] 1 [0]
[0 1] 1 [1]
[1 0] 1 [1]
[1 1] 0 [0]
So when the literature states that the multi-layered perceptron (Aka the basic deep learning) solves XOR, does it mean that it can fully learn and memorize the weights given the fully set of in-/outputs but cannot generalize the XOR problem given that one of data point is missing?
Here's the link of the Kaggle dataset that answerers can test the network for themselves: https://www.kaggle.com/alvations/xor-with-mlp/
I think learning (generalizing) XOR and memorizing XOR are different things.
A two-layer perceptron can memorize XOR as you have seen, that is there exists a combination of weights where the loss is minimum and equal to 0 (absolute minimum).
If the weights are randomly initialized, you might end up with the situation where you have actually learned XOR and not only memorized.
Note that multi-layer perceptrons are non-convex functions so, there could be multiple minima (multiple global minima even). When data is missing one input, there are multiple minima (and all are equal in value) and there exists minima where the missing point would be correctly classified. Hence, MLP can learn an XOR. (though finding that weight combination might be hard with a missing point).
It is quite often argued that Neural Networks are universal function approximator and can approximate non-sense labels even. In that light, you might want to look at this work https://arxiv.org/abs/1611.03530