I am currently following the introduction guide to neural networks and deep learning from Michael Nielsen (link) and am stuck in chapter 1.
I set up everything and fixed smaller changes from python2 -> python3 and ran the code, however my value in each epoch remains constant. I assume it means that for whatever reason the weights and biases doesnt get updated.
As stated in the guide I followed it and created two files, one mnist_loader, which loads the data from other sources
Any ideas on where to fix?
I am currently running the newest python 3.10.2.
link to image of cmd
what i expect is the value of each epoch raising like given in the chapter
> Epoch 0: 9129 / 10000
> Epoch 1: 9295 / 10000
> Epoch 2: 9348 / 10000
> ...
> Epoch 27: 9528 / 10000
> Epoch 28: 9542 / 10000
> Epoch 29: 9534 / 10000
Edit: added code
"""
mnist_loader
~~~~~~~~~~~~
A library to load the MNIST image data. For details of the data
structures that are returned, see the doc strings for ``load_data``
and ``load_data_wrapper``. In practice, ``load_data_wrapper`` is the
function usually called by our neural network code.
"""
# Libraries
# Standard library
import pickle
import gzip
# Third-party libraries
import numpy as np
def load_data():
"""Return the MNIST data as a tuple containing the training data,
the validation data, and the test data.
The ``training_data`` is returned as a tuple with two entries.
The first entry contains the actual training images. This is a
numpy ndarray with 50,000 entries. Each entry is, in turn, a
numpy ndarray with 784 values, representing the 28 * 28 = 784
pixels in a single MNIST image.
The second entry in the ``training_data`` tuple is a numpy ndarray
containing 50,000 entries. Those entries are just the digit
values (0...9) for the corresponding images contained in the first
entry of the tuple.
The ``validation_data`` and ``test_data`` are similar, except
each contains only 10,000 images.
This is a nice data format, but for use in neural networks it's
helpful to modify the format of the ``training_data`` a little.
That's done in the wrapper function ``load_data_wrapper()``, see
below.
"""
f = gzip.open("C:/Users/Tai/Documents/mnist/neural-networks-and-deep-learning/data/mnist.pkl.gz")
training_data, validation_data, test_data = pickle.load(f, encoding="latin1")
print(training_data)
f.close()
return (training_data, validation_data, test_data)
def load_data_wrapper():
"""Return a tuple containing ``(training_data, validation_data,
test_data)``. Based on ``load_data``, but the format is more
convenient for use in our implementation of neural networks.
In particular, ``training_data`` is a list containing 50,000
2-tuples ``(x, y)``. ``x`` is a 784-dimensional numpy.ndarray
containing the input image. ``y`` is a 10-dimensional
numpy.ndarray representing the unit vector corresponding to the
correct digit for ``x``.
``validation_data`` and ``test_data`` are lists containing 10,000
2-tuples ``(x, y)``. In each case, ``x`` is a 784-dimensional
numpy.ndarry containing the input image, and ``y`` is the
corresponding classification, i.e., the digit values (integers)
corresponding to ``x``.
Obviously, this means we're using slightly different formats for
the training data and the validation / test data. These formats
turn out to be the most convenient for use in our neural network
code."""
tr_d, va_d, te_d = load_data()
training_inputs = [np.reshape(x, (784, 1)) for x in tr_d[0]]
training_results = [vectorized_result(y) for y in tr_d[1]]
training_data = list(zip(training_inputs, training_results))
validation_inputs = [np.reshape(x, (784, 1)) for x in va_d[0]]
validation_data = list(zip(validation_inputs, va_d[1]))
test_inputs = [np.reshape(x, (784, 1)) for x in te_d[0]]
test_data = list(zip(test_inputs, te_d[1]))
return (training_data, validation_data, test_data)
def vectorized_result(j):
"""Return a 10-dimensional unit vector with a 1.0 in the jth
position and zeroes elsewhere. This is used to convert a digit
(0...9) into a corresponding desired output from the neural
network."""
e = np.zeros((10, 1))
e[j] = 1.0
return e
and second code for network
import numpy as np
import random
"""Network is generating layers of nodes, first layer is the input nodes,
last layer is the output nodes
each layer have random number (gaussian distribution with mean 0 and
standard deviation 1)
applied for weights and bias, first layer has no bias
a network of net = Network([2,3,1]) has 2 input, 3 middle and 1 output
nodes"""
class Network(object):
def __init__(self, sizes):
self.num_layers = len(sizes)
self.sizes = sizes
self.biases = [np.random.randn(y, 1) for y in sizes[1:]]
self.weights = [np.random.randn(y, x)
for x, y in zip(sizes[:-1], sizes[1:])]
# returns output of the network for node a
def feedforward(self, a):
for b, w in zip(self.biases, self.weights):
a = sigmoid(np.dot(w, a) + b)
# stochastic gradiant descent
# Train the neural network using mini-batch stochastic gradient descent.
# training data: list of tuples (x, y) with input and desired output
# epochs: number of epoch to train for
# mini_batch_size: number of mini-batches to use when sampling
# eta: learning rate
# (optional) test data: evaluation of network after each epoch of
traning, print partial process
# slows program down considerably
# each epoch random shuffle of training data, partitioning into mini-
batches of specified size
# for each mini_batch single step of gradient descent specified in
# self.update_mini_batch(mini_batch, eta)
# update weights and biases according to a single iteration of gradient
descent
def SGD(self, training_data, epochs, mini_batch_size, eta, test_data=None):
if test_data: n_test = len(test_data)
n = len(training_data)
for j in range(epochs):
random.shuffle(training_data)
mini_batches = [
training_data[k:k + mini_batch_size]
for k in range(0, n, mini_batch_size)]
for mini_batch in mini_batches:
self.update_mini_batch(mini_batch, eta)
if test_data:
print(
"Epoch {0}: {1} / {2}".format(
j, self.evaluate(test_data), n_test))
else:
print(
"Epoch {0} complete".format(j))
# update network's weights and biases by applying gradient descent using
backpropagation
# to a single mini batch.
# most important line
# delta_nabla_b, delta_nabla_w = self.backprop(x, y)
# self.backprop ensures the backpropagation of (x, y)
def update_mini_batch(self, mini_batch, eta):
nabla_b = [np.zeros(b.shape) for b in self.biases]
nabla_w = [np.zeros(w.shape) for w in self.weights]
for x, y in mini_batch:
delta_nabla_b, delta_nabla_w = self.backprop(x, y)
nabla_b = [nb + dnb for nb, dnb in zip(nabla_b, delta_nabla_b)]
nabla_w = [nw + dnw for nw, dnw in zip(nabla_w, delta_nabla_w)]
self.weights = [w - (eta / len(mini_batch)) * nw
for w, nw in zip(self.weights, nabla_w)]
self.biases = [b - (eta / len(mini_batch)) * nb
for b, nb in zip(self.biases, nabla_b)]
def backprop(self, x, y):
"""Return a tuple ``(nabla_b, nabla_w)`` representing the
gradient for the cost function C_x. ``nabla_b`` and
``nabla_w`` are layer-by-layer lists of numpy arrays, similar
to ``self.biases`` and ``self.weights``."""
nabla_b = [np.zeros(b.shape) for b in self.biases]
nabla_w = [np.zeros(w.shape) for w in self.weights]
# feedforward
activation = x
activations = [x] # list to store all the activations, layer by layer
zs = [] # list to store all the z vectors, layer by layer
for b, w in zip(self.biases, self.weights):
z = np.dot(w, activation) + b
zs.append(z)
activation = sigmoid(z)
activations.append(activation)
# backward pass
delta = self.cost_derivative(activations[-1], y) * \
sigmoid_prime(zs[-1])
nabla_b[-1] = delta
nabla_w[-1] = np.dot(delta, activations[-2].transpose())
# Note that the variable l in the loop below is used a little
# differently to the notation in Chapter 2 of the book. Here,
# l = 1 means the last layer of neurons, l = 2 is the
# second-last layer, and so on. It's a renumbering of the
# scheme in the book, used here to take advantage of the fact
# that Python can use negative indices in lists.
for l in range(2, self.num_layers):
z = zs[-l]
sp = sigmoid_prime(z)
delta = np.dot(self.weights[-l + 1].transpose(), delta) * sp
nabla_b[-l] = delta
nabla_w[-l] = np.dot(delta, activations[-l - 1].transpose())
return nabla_b, nabla_w
def evaluate(self, test_data):
"""Return the number of test inputs for which the neural
network outputs the correct result. Note that the neural
network's output is assumed to be the index of whichever
neuron in the final layer has the highest activation."""
test_results = [(np.argmax(self.feedforward(x)), y)
for (x, y) in test_data]
return sum(int(x == y) for (x, y) in test_results)
#staticmethod
def cost_derivative(output_activations, y):
"""Return the vector of partial derivatives \partial C_x /
\partial a for the output activations."""
return output_activations - y
# z is a vector, numpy automatically applies sigmoid onto each element of z
def sigmoid(z):
return 1.0 / (1.0 + np.exp(-z))
# derivative of the sigmoid function
def sigmoid_prime(z):
return sigmoid(z) * (1 - sigmoid(z))
I found the error thanks to the other comments here, i missed a line in the code
in network:
# returns output of the network for node a
def feedforward(self, a):
for b, w in zip(self.biases, self.weights):
a = sigmoid(np.dot(w, a) + b)
return a
thanks everyone for pointing out!
I have a following loop where I am calculating softmax transform for batches of different sizes as below
import numpy as np
def softmax(Z,arr):
"""
:param Z: numpy array of any shape (output from hidden layer)
:param arr: numpy array of any shape (start, end)
:return A: output of multinum_logit(Z,arr), same shape as Z
:return cache: returns Z as well, useful during back propagation
"""
A = np.zeros(Z.shape)
for i in prange(len(arr)):
shiftx = Z[:,arr[i,1]:arr[i,2]+1] - np.max(Z[:,int(arr[i,1]):int(arr[i,2])+1])
A[:,arr[i,1]:arr[i,2]+1] = np.exp(shiftx)/np.exp(shiftx).sum()
cache = Z
return A,cache
Since this for loop is not vectorized it is the bottleneck in my code. What is a possible solution to make it faster. I have tried using #jit of numba which makes it little faster but not enough. I was wondering if there is another way to make it faster or vectorize/parallelize it.
Sample input data for the function
Z = np.random.random([1,10000])
arr = np.zeros([100,3])
arr[:,0] = 1
temp = int(Z.shape[1]/arr.shape[0])
for i in range(arr.shape[0]):
arr[i,1] = i*temp
arr[i,2] = (i+1)*temp-1
arr = arr.astype(int)
EDIT:
I forgot to stress here that my number of class is varying. For example batch 1 has say 10 classes, batch 2 may have 15 classes. Therefore I am passing an array arr which keeps track of the which rows belong to batch1 and so on. These batches are different than the batches in traditional neural network framework
In the above example arr keeps track of starting index and end index of rows. So the denominator in the softmax function will be sum of only those observations whose index lie between the starting and ending index.
Here's a vectorized softmax function. It's the implementation of an assignment from Stanford's cs231n course on conv nets.
The function takes in optimizable parameters, input data, targets, and a regularizer. (You can ignore the regularizer as that references another class exclusive to some cs231n assignments).
It returns a loss and gradients of the parameters.
def softmax_loss_vectorized(W, X, y, reg):
"""
Softmax loss function, vectorized version.
Inputs and outputs are the same as softmax_loss_naive.
"""
# Initialize the loss and gradient to zero.
loss = 0.0
dW = np.zeros_like(W)
num_train = X.shape[0]
scores = X.dot(W)
shift_scores = scores - np.amax(scores,axis=1).reshape(-1,1)
softmax = np.exp(shift_scores)/np.sum(np.exp(shift_scores), axis=1).reshape(-1,1)
loss = -np.sum(np.log(softmax[range(num_train), list(y)]))
loss /= num_train
loss += 0.5* reg * np.sum(W * W)
dSoftmax = softmax.copy()
dSoftmax[range(num_train), list(y)] += -1
dW = (X.T).dot(dSoftmax)
dW = dW/num_train + reg * W
return loss, dW
For comparison's sake, here is a naive (non-vectorized) implementation of the same method.
def softmax_loss_naive(W, X, y, reg):
"""
Softmax loss function, naive implementation (with loops)
Inputs have dimension D, there are C classes, and we operate on minibatches
of N examples.
Inputs:
- W: A numpy array of shape (D, C) containing weights.
- X: A numpy array of shape (N, D) containing a minibatch of data.
- y: A numpy array of shape (N,) containing training labels; y[i] = c means
that X[i] has label c, where 0 <= c < C.
- reg: (float) regularization strength
Returns a tuple of:
- loss as single float
- gradient with respect to weights W; an array of same shape as W
"""
loss = 0.0
dW = np.zeros_like(W)
num_train = X.shape[0]
num_classes = W.shape[1]
for i in xrange(num_train):
scores = X[i].dot(W)
shift_scores = scores - max(scores)
loss_i = -shift_scores[y[i]] + np.log(sum(np.exp(shift_scores)))
loss += loss_i
for j in xrange(num_classes):
softmax = np.exp(shift_scores[j])/sum(np.exp(shift_scores))
if j==y[i]:
dW[:,j] += (-1 + softmax) * X[i]
else:
dW[:,j] += softmax *X[i]
loss /= num_train
loss += 0.5 * reg * np.sum(W * W)
dW /= num_train + reg * W
return loss, dW
Source
I am learning the neural network and I want to write a function cross_entropy in python. Where it is defined as
where N is the number of samples, k is the number of classes, log is the natural logarithm, t_i,j is 1 if sample i is in class j and 0 otherwise, and p_i,j is the predicted probability that sample i is in class j.
To avoid numerical issues with logarithm, clip the predictions to [10^{−12}, 1 − 10^{−12}] range.
According to the above description, I wrote down the codes by clipping the predictions to [epsilon, 1 − epsilon] range, then computing the cross_entropy based on the above formula.
def cross_entropy(predictions, targets, epsilon=1e-12):
"""
Computes cross entropy between targets (encoded as one-hot vectors)
and predictions.
Input: predictions (N, k) ndarray
targets (N, k) ndarray
Returns: scalar
"""
predictions = np.clip(predictions, epsilon, 1. - epsilon)
ce = - np.mean(np.log(predictions) * targets)
return ce
The following code will be used to check if the function cross_entropy are correct.
predictions = np.array([[0.25,0.25,0.25,0.25],
[0.01,0.01,0.01,0.96]])
targets = np.array([[0,0,0,1],
[0,0,0,1]])
ans = 0.71355817782 #Correct answer
x = cross_entropy(predictions, targets)
print(np.isclose(x,ans))
The output of the above codes is False, that to say my codes for defining the function cross_entropy is not correct. Then I print the result of cross_entropy(predictions, targets). It gave 0.178389544455 and the correct result should be ans = 0.71355817782. Could anybody help me to check what is the problem with my codes?
You're not that far off at all, but remember you are taking the average value of N sums, where N = 2 (in this case). So your code could read:
def cross_entropy(predictions, targets, epsilon=1e-12):
"""
Computes cross entropy between targets (encoded as one-hot vectors)
and predictions.
Input: predictions (N, k) ndarray
targets (N, k) ndarray
Returns: scalar
"""
predictions = np.clip(predictions, epsilon, 1. - epsilon)
N = predictions.shape[0]
ce = -np.sum(targets*np.log(predictions+1e-9))/N
return ce
predictions = np.array([[0.25,0.25,0.25,0.25],
[0.01,0.01,0.01,0.96]])
targets = np.array([[0,0,0,1],
[0,0,0,1]])
ans = 0.71355817782 #Correct answer
x = cross_entropy(predictions, targets)
print(np.isclose(x,ans))
Here, I think it's a little clearer if you stick with np.sum(). Also, I added 1e-9 into the np.log() to avoid the possibility of having a log(0) in your computation. Hope this helps!
NOTE: As per #Peter's comment, the offset of 1e-9 is indeed redundant if your epsilon value is greater than 0.
def cross_entropy(x, y):
""" Computes cross entropy between two distributions.
Input: x: iterabale of N non-negative values
y: iterabale of N non-negative values
Returns: scalar
"""
if np.any(x < 0) or np.any(y < 0):
raise ValueError('Negative values exist.')
# Force to proper probability mass function.
x = np.array(x, dtype=np.float)
y = np.array(y, dtype=np.float)
x /= np.sum(x)
y /= np.sum(y)
# Ignore zero 'y' elements.
mask = y > 0
x = x[mask]
y = y[mask]
ce = -np.sum(x * np.log(y))
return ce
def cross_entropy_via_scipy(x, y):
''' SEE: https://en.wikipedia.org/wiki/Cross_entropy'''
return entropy(x) + entropy(x, y)
from scipy.stats import entropy, truncnorm
x = truncnorm.rvs(0.1, 2, size=100)
y = truncnorm.rvs(0.1, 2, size=100)
print np.isclose(cross_entropy(x, y), cross_entropy_via_scipy(x, y))
This is my custom extension of one of Andrew NG's neural network from deep learning course where instead of producing 0 or 1 for binary classification I'm attempting
to classify multiple examples.
Both the inputs and outputs are one hot encoded.
With not much training I receive an accuracy of 'train accuracy: 67.51658067499625 %'
How can I classify a single training example instead of classifying all training examples?
I think a bug exists in my implementation as an issue with this network is training examples (train_set_x) and output values (train_set_y) both need to have same dimensions or an error related to the dimensionality of matrices is received.
For example using :
train_set_x = np.array([
[1,1,1,1],[0,1,1,1],[0,0,1,1]
])
train_set_y = np.array([
[1,1,1],[1,1,0],[1,1,1]
])
returns error :
ValueError Traceback (most recent call last)
<ipython-input-11-0d356e8d66f3> in <module>()
27 print(A)
28
---> 29 np.multiply(train_set_y,A)
30
31 def initialize_with_zeros(numberOfTrainingExamples):
ValueError: operands could not be broadcast together with shapes (3,3) (1,4)
network code :
import numpy as np
import matplotlib.pyplot as plt
import h5py
import scipy
from scipy import ndimage
import pandas as pd
%matplotlib inline
train_set_x = np.array([
[1,1,1,1],[0,1,1,1],[0,0,1,1]
])
train_set_y = np.array([
[1,1,1,0],[1,1,0,0],[1,1,1,1]
])
numberOfFeatures = 4
numberOfTrainingExamples = 3
def sigmoid(z):
s = 1 / (1 + np.exp(-z))
return s
w = np.zeros((numberOfTrainingExamples , 1))
b = 0
A = sigmoid(np.dot(w.T , train_set_x))
print(A)
np.multiply(train_set_y,A)
def initialize_with_zeros(numberOfTrainingExamples):
w = np.zeros((numberOfTrainingExamples , 1))
b = 0
return w, b
def propagate(w, b, X, Y):
m = X.shape[1]
A = sigmoid(np.dot(w.T , X) + b)
cost = -(1/m)*np.sum(np.multiply(Y,np.log(A)) + np.multiply((1-Y),np.log(1-A)), axis=1)
dw = ( 1 / m ) * np.dot( X, ( A - Y ).T ) # consumes ( A - Y )
db = ( 1 / m ) * np.sum( A - Y ) # consumes ( A - Y ) again
# cost = np.squeeze(cost)
grads = {"dw": dw,
"db": db}
return grads, cost
def optimize(w, b, X, Y, num_iterations, learning_rate, print_cost = True):
costs = []
for i in range(num_iterations):
grads, cost = propagate(w, b, X, Y)
dw = grads["dw"]
db = grads["db"]
w = w - (learning_rate * dw)
b = b - (learning_rate * db)
if i % 100 == 0:
costs.append(cost)
if print_cost and i % 10000 == 0:
print(cost)
params = {"w": w,
"b": b}
grads = {"dw": dw,
"db": db}
return params, grads, costs
def model(X_train, Y_train, num_iterations, learning_rate = 0.5, print_cost = False):
w, b = initialize_with_zeros(numberOfTrainingExamples)
parameters, grads, costs = optimize(w, b, X_train, Y_train, num_iterations, learning_rate, print_cost = True)
w = parameters["w"]
b = parameters["b"]
Y_prediction_train = sigmoid(np.dot(w.T , X_train) + b)
print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - Y_train)) * 100))
model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.0001, print_cost = True)
Update: A bug exists in this implementation in that the training example pairs (train_set_x , train_set_y) must contain the same dimensions. Can point in direction of how linear algebra should be modified?
Update 2 :
I modified #Paul Panzer answer so that learning rate is 0.001 and train_set_x , train_set_y pairs are unique :
train_set_x = np.array([
[1,1,1,1,1],[0,1,1,1,1],[0,0,1,1,0],[0,0,1,0,1]
])
train_set_y = np.array([
[1,0,0],[0,0,1],[0,1,0],[1,0,1]
])
grads = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True)
# To classify single training example :
print(sigmoid(dw # [0,0,1,1,0] + db))
This update produces following output :
-2.09657359028
-3.94918577439
[[ 0.74043089 0.32851512 0.14776077 0.77970162]
[ 0.04810012 0.08033521 0.72846174 0.1063849 ]
[ 0.25956911 0.67148488 0.22029838 0.85223923]]
[[1 0 0 1]
[0 0 1 0]
[0 1 0 1]]
train accuracy: 79.84462279013312 %
[[ 0.51309252 0.48853845 0.50945862]
[ 0.5110232 0.48646923 0.50738869]
[ 0.51354109 0.48898712 0.50990734]]
Should print(sigmoid(dw # [0,0,1,1,0] + db)) produce a vector that once rounded matches train_set_y corresponding value : [0,1,0] ?
Modifying to produce a vector with (adding [0,0,1,1,0] to numpy array and taking transpose):
print(sigmoid(dw # np.array([[0,0,1,1,0]]).T + db))
returns :
array([[ 0.51309252],
[ 0.48646923],
[ 0.50990734]])
Again, rounding these values to nearest whole number produces vector [1,0,1] when [0,1,0] is expected.
These are incorrect operations to produce a prediction for single training example ?
Your difficulties come from mismatched dimensions, so let's walk through the problem and try and get them straight.
Your network has a number of inputs, the features, let's call their number N_in (numberOfFeatures in your code). And it has a number of outputs which correspond to different classes let's call their number N_out. Inputs and outputs are connected by the weights w.
Now here is the problem. Connections are all-to-all, so we need a weight for each of the N_out x N_in pairs of outputs and inputs. Therefore in your code the shape of w must be changed to (N_out, N_in). You probably also want an offset b for each output, so b should be a vector of size (N_out,) or rather (N_out, 1) so it plays well with the 2d terms.
I've fixed that in the modified code below and I tried to make it very explicit. I've also thrown a mock data creator into the bargain.
Re the one-hot encoded categorical output, I'm not an expert on neural networks but I think, most people understand it so that classes are mutually exclusive, so each sample in your mock output should have one one and the rest zeros.
Side note:
At one point a competing answer advised you to get rid of the 1-... terms in the cost function. While that looks like an interesting idea to me my gut feeling (Edit Now confirmed using gradient-free minimizer; use activation="hybrid" in code below. Solver will simply maximize all outputs which are active in at least one training example.) is it won't work just like that because the cost will then fail to penalise false positives (see below for detailed explanation). To make it work you'd have to add some kind of regularization. One method that appears to work is using the softmax instead of the sigmoid. The softmax is to one-hot what the sigmoid is to binary. It makes sure the output is "fuzzy one-hot".
Therefore my recommendation is:
If you want to stick with sigmoid and not explicitly enforce one-hot predictions. Keep the 1-... term.
If you want to use the shorter cost function. Enforce one-hot predictions. For example by using softmax instead of sigmoid.
I've added an activation="sigmoid"|"softmax"|"hybrid" parameter to the code that switches between models. I've also made the scipy general purpose minimizer available, which may be useful when the gradient of the cost is not at hand.
Recap on how the cost function works:
The cost is a sum over all classes and all training samples of the term
-y log (y') - (1-y) log (1-y')
where y is the expected response, i.e. the one given by the "y" training sample for the input (the "x" training sample). y' is the prediction, the response the network with its current weights and biases generates. Now, because the expected response is either 0 or 1 the cost for a single category and a single training sample can be written
-log (y') if y = 1
-log(1-y') if y = 0
because in the first case (1-y) is zero, so the second term vanishes and in the secondo case y is zero, so the first term vanishes.
One can now convince oneself that the cost is high if
the expected response y is 1 and the network prediction y' is close to zero
the expected response y is 0 and the network prediction y' is close to one
In other words the cost does its job in punishing wrong predictions. Now, if we drop the second term (1-y) log (1-y') half of this mechanism is gone. If the expected response is 1, a low prediction will still incur a cost, but if the expected response is 0, the cost will be zero, regardless of the prediction, in particular, a high prediction (or false positive) will go unpunished.
Now, because the total cost is a sum over all training samples, there are three possibilities.
all training samples prescribe that the class be zero:
then the cost will be completely independent of the predictions for this class and no learning can take place
some training samples put the class at zero, some at one:
then because "false negatives" or "misses" are still punished but false positives aren't the net will find the easiest way to minimize the cost which is to indiscriminately increase the prediction of the class for all samples
all training samples prescribe that the class be one:
essentially the same as in the second scenario will happen, only here it's no problem, because that is the correct behavior
And finally, why does it work if we use softmax instead of sigmoid? False positives will still be invisible. Now it is easy to see that the sum over all classes of the softmax is one. So I can only increase the prediction for one class if at least one other class is reduced to compensate. In particular, there can be no false positives without a false negative, and the false negative the cost will detect.
On how to get a binary prediction:
For binary expected responses rounding is indeed the appropriate procedure. For one-hot I'd rather find the largest value, set that to one and all others to zero. I've added a convenience function, predict, implementing that.
import numpy as np
from scipy import optimize as opt
from collections import namedtuple
# First, a few structures to keep ourselves organized
Problem_Size = namedtuple('Problem_Size', 'Out In Samples')
Data = namedtuple('Data', 'Out In')
Network = namedtuple('Network', 'w b activation cost gradient most_likely')
def get_dims(Out, In, transpose=False):
"""extract dimensions and ensure everything is 2d
return Data, Dims"""
# gracefully acccept lists etc.
Out, In = np.asanyarray(Out), np.asanyarray(In)
if transpose:
Out, In = Out.T, In.T
# if it's a single sample make sure it's n x 1
Out = Out[:, None] if len(Out.shape) == 1 else Out
In = In[:, None] if len(In.shape) == 1 else In
Dims = Problem_Size(Out.shape[0], *In.shape)
if Dims.Samples != Out.shape[1]:
raise ValueError("number of samples must be the same for Out and In")
return Data(Out, In), Dims
def sigmoid(z):
s = 1 / (1 + np.exp(-z))
return s
def sig_cost(Net, data):
A = process(data.In, Net)
logA = np.log(A)
return -(data.Out * logA + (1-data.Out) * (1-logA)).sum(axis=0).mean()
def sig_grad (Net, Dims, data):
A = process(data.In, Net)
return dict(dw = (A - data.Out) # data.In.T / Dims.Samples,
db = (A - data.Out).mean(axis=1, keepdims=True))
def sig_ml(z):
return np.round(z).astype(int)
def sof_ml(z):
hot = np.argmax(z, axis=0)
z = np.zeros(z.shape, dtype=int)
z[hot, np.arange(len(hot))] = 1
return z
def softmax(z):
z = z - z.max(axis=0, keepdims=True)
z = np.exp(z)
return z / z.sum(axis=0, keepdims=True)
def sof_cost(Net, data):
A = process(data.In, Net)
logA = np.log(A)
return -(data.Out * logA).sum(axis=0).mean()
sof_grad = sig_grad
def get_net(Dims, activation='softmax'):
activation, cost, gradient, ml = {
'sigmoid': (sigmoid, sig_cost, sig_grad, sig_ml),
'softmax': (softmax, sof_cost, sof_grad, sof_ml),
'hybrid': (sigmoid, sof_cost, None, sig_ml)}[activation]
return Network(w=np.zeros((Dims.Out, Dims.In)),
b=np.zeros((Dims.Out, 1)),
activation=activation, cost=cost, gradient=gradient,
most_likely=ml)
def process(In, Net):
return Net.activation(Net.w # In + Net.b)
def propagate(data, Dims, Net):
return Net.gradient(Net, Dims, data), Net.cost(Net, data)
def optimize_no_grad(Net, Dims, data):
def f(x):
Net.w[...] = x[:Net.w.size].reshape(Net.w.shape)
Net.b[...] = x[Net.w.size:].reshape(Net.b.shape)
return Net.cost(Net, data)
x = np.r_[Net.w.ravel(), Net.b.ravel()]
res = opt.minimize(f, x, options=dict(maxiter=10000)).x
Net.w[...] = res[:Net.w.size].reshape(Net.w.shape)
Net.b[...] = res[Net.w.size:].reshape(Net.b.shape)
def optimize(Net, Dims, data, num_iterations, learning_rate, print_cost = True):
w, b = Net.w, Net.b
costs = []
for i in range(num_iterations):
grads, cost = propagate(data, Dims, Net)
dw = grads["dw"]
db = grads["db"]
w -= learning_rate * dw
b -= learning_rate * db
if i % 100 == 0:
costs.append(cost)
if print_cost and i % 10000 == 0:
print(cost)
return grads, costs
def model(X_train, Y_train, num_iterations, learning_rate = 0.5, print_cost = False, activation='sigmoid'):
data, Dims = get_dims(Y_train, X_train, transpose=True)
Net = get_net(Dims, activation)
if Net.gradient is None:
optimize_no_grad(Net, Dims, data)
else:
grads, costs = optimize(Net, Dims, data, num_iterations, learning_rate, print_cost = True)
Y_prediction_train = process(data.In, Net)
print(Y_prediction_train)
print(data.Out)
print(Y_prediction_train.sum(axis=0))
print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - data.Out)) * 100))
return Net
def predict(In, Net, probability=False):
In = np.asanyarray(In)
is1d = In.ndim == 1
if is1d:
In = In.reshape(-1, 1)
Out = process(In, Net)
if not probability:
Out = Net.most_likely(Out)
if is1d:
Out = Out.reshape(-1)
return Out
def create_data(Dims):
Out = np.zeros((Dims.Out, Dims.Samples), dtype=int)
Out[np.random.randint(0, Dims.Out, (Dims.Samples,)), np.arange(Dims.Samples)] = 1
In = np.random.randint(0, 2, (Dims.In, Dims.Samples))
return Data(Out, In)
train_set_x = np.array([
[1,1,1,1,1],[0,1,1,1,1],[0,0,1,1,0],[0,0,1,0,1]
])
train_set_y = np.array([
[1,0,0],[1,0,0],[0,0,1],[0,0,1]
])
Net1 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='sigmoid')
Net2 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='softmax')
Net3 = model(train_set_x, train_set_y, num_iterations = 20000, learning_rate = 0.001, print_cost = True, activation='hybrid')
Dims = Problem_Size(8, 100, 50)
data = create_data(Dims)
model(data.In.T, data.Out.T, num_iterations = 40000, learning_rate = 0.001, print_cost = True, activation='softmax')
model(data.In.T, data.Out.T, num_iterations = 40000, learning_rate = 0.001, print_cost = True, activation='sigmoid')
Both the idea of how to fix the bug and how you can extend the implementation to classify between more classes can be solved with some dimensionality analysis.
I am assuming that you by classifying multiple examples mean multiple classes and not multiple samples, as we need multiple samples to train even for 2 classes.
Where N = number of samples, D = number of features, K = number of categories(with K=2 being a special case where one can reduce this down to one dimension,ie K=1 with y=0 signifying one class and y=1 the other). The data should have the following dimensions:
X: N * D #input
y: N * K #output
W: D * K #weights, also dW has same dimensions
b: 1 * K #bias, also db has same dimensions
#A should have same dimensions as y
The order of the dimensions can be switched around, as long as the dot products are done correctly.
First dealing with your bug: You are initializing W as N * K instead of D * K ie. in the binary case:
w = np.zeros((numberOfTrainingExamples , 1))
#instead of
w = np.zeros((numberOfFeatures , 1))
This means that the only time you are initializing W to correct dimensions is when y and X (coincidentally) have same dimensions.
This will mess with your dot products as well:
np.dot(X, w) # or np.dot(w.T,X.T) if you define y as [K * N] dimensions
#instead of
np.dot(w.T , X)
and
np.dot( X.T, ( A - Y ) ) #np.dot( X.T, ( A - Y ).T ) if y:[K * N]
#instead of
np.dot( X, ( A - Y ).T )
Also make sure that the cost function returns one number (ie. not an array).
Secondly going on to K>2 you need to make some changes. b is no longer a single number, but a vector (1D-array). y and W go from being 1D-array to 2D array. To avoid confusion and hard-to-find bugs it could be good to set K, N and D to different values
I have implemented and trained a neural network with Theano of k binary inputs (0,1), one hidden layer and one unit in the output layer. Once it has been trained I want to obtain inputs that maximizes the output (e.g. x which makes unit of output layer closest to 1). So far I haven't found an implementation of it, so I am trying the following approach:
Train network => obtain trained weights (theta1, theta2)
Define the neural network function with x as input and trained theta1, theta2 as fixed parameters. That is: f(x) = sigmoid( theta1*(sigmoid (theta2*x ))). This function takes x and with given trained weights (theta1, theta2) gives output between 0 and 1.
Apply gradient descent w.r.t. x on the neural network function f(x) and obtain x that maximizes f(x) with theta1 and theta2 given.
For these I have implemented the following code with a toy example (k = 2). Based on the tutorial on http://outlace.com/Beginner-Tutorial-Theano/ but changed vector y, so that there is only one combination of inputs that gives f(x) ~ 1 which is x = [0, 1].
Edit1: As suggested optimizer was set to None and bias unit was fixed to 1.
Step 1: Train neural network. This runs well and with out error.
import os
os.environ["THEANO_FLAGS"] = "optimizer=None"
import theano
import theano.tensor as T
import theano.tensor.nnet as nnet
import numpy as np
x = T.dvector()
y = T.dscalar()
def layer(x, w):
b = np.array([1], dtype=theano.config.floatX)
new_x = T.concatenate([x, b])
m = T.dot(w.T, new_x) #theta1: 3x3 * x: 3x1 = 3x1 ;;; theta2: 1x4 * 4x1
h = nnet.sigmoid(m)
return h
def grad_desc(cost, theta):
alpha = 0.1 #learning rate
return theta - (alpha * T.grad(cost, wrt=theta))
in_units = 2
hid_units = 3
out_units = 1
theta1 = theano.shared(np.array(np.random.rand(in_units + 1, hid_units), dtype=theano.config.floatX)) # randomly initialize
theta2 = theano.shared(np.array(np.random.rand(hid_units + 1, out_units), dtype=theano.config.floatX))
hid1 = layer(x, theta1) #hidden layer
out1 = T.sum(layer(hid1, theta2)) #output layer
fc = (out1 - y)**2 #cost expression
cost = theano.function(inputs=[x, y], outputs=fc, updates=[
(theta1, grad_desc(fc, theta1)),
(theta2, grad_desc(fc, theta2))])
run_forward = theano.function(inputs=[x], outputs=out1)
inputs = np.array([[0,1],[1,0],[1,1],[0,0]]).reshape(4,2) #training data X
exp_y = np.array([1, 0, 0, 0]) #training data Y
cur_cost = 0
for i in range(5000):
for k in range(len(inputs)):
cur_cost = cost(inputs[k], exp_y[k]) #call our Theano-compiled cost function, it will auto update weights
print(run_forward([0,1]))
Output of run forward for [0,1] is: 0.968905860574.
We can also get values of weights with theta1.get_value() and theta2.get_value()
Step 2: Define neural network function f(x). Trained weights (theta1, theta2) are constant parameters of this function.
Things get a little trickier here because of the bias unit, which is part of he vector of inputs x. To do this I concatenate b and x. But the code now runs well.
b = np.array([[1]], dtype=theano.config.floatX)
#b_sh = theano.shared(np.array([[1]], dtype=theano.config.floatX))
rand_init = np.random.rand(in_units, 1)
rand_init[0] = 1
x_sh = theano.shared(np.array(rand_init, dtype=theano.config.floatX))
th1 = T.dmatrix()
th2 = T.dmatrix()
nn_hid = T.nnet.sigmoid( T.dot(th1, T.concatenate([x_sh, b])) )
nn_predict = T.sum( T.nnet.sigmoid( T.dot(th2, T.concatenate([nn_hid, b]))))
Step 3:
Problem is now in gradient descent as is not limited to values between 0 and 1.
fc2 = (nn_predict - 1)**2
cost3 = theano.function(inputs=[th1, th2], outputs=fc2, updates=[
(x_sh, grad_desc(fc2, x_sh))])
run_forward = theano.function(inputs=[th1, th2], outputs=nn_predict)
cur_cost = 0
for i in range(10000):
cur_cost = cost3(theta1.get_value().T, theta2.get_value().T) #call our Theano-compiled cost function, it will auto update weights
if i % 500 == 0: #only print the cost every 500 epochs/iterations (to save space)
print('Cost: %s' % (cur_cost,))
print x_sh.get_value()
The last iteration prints:
Cost: 0.000220317356533
[[-0.11492753]
[ 1.99729555]]
Furthermore input 1 keeps becoming more negative and input 2 increases, while the optimal solution is [0, 1]. How can this be fixed?
You are adding b=[1] via broadcasting rules as opposed to concatenating it. Also, once you concatenate it, your x_sh has one dimension to many which is why the error occurs at nn_predict and not nn_hid