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I was converting a program from Octave to Python, I used OMPC for this to try and see if it works, and when I get the conversion of the code for Python at the top there is a line that says #mfunction("") which I'm not sure what is for and I haven't found a lot of information about it. Also about elmul which I also don´t know what it is for and haven't found information about it.
This is the code on Octave
function [J, grad] = costFunction(theta, X, y)
%COSTFUNCTION Compute cost and gradient for logistic regression
% J = COSTFUNCTION(theta, X, y) computes the cost of using theta as the
% parameter for logistic regression and the gradient of the cost
% w.r.t. to the parameters.
% Initialize some useful values
m = length(y); % number of training examples
% You need to return the following variables correctly
J = 0;
grad = zeros(size(theta));
% ====================== YOUR CODE HERE ======================
% Instructions: Compute the cost of a particular choice of theta.
% You should set J to the cost.
% Compute the partial derivatives and set grad to the partial
% derivatives of the cost w.r.t. each parameter in theta
%
% Note: grad should have the same dimensions as theta
% Use sigmoid function previously programed
hypothesis = sigmoid(X*theta); % Hypothesis for logistic regression
Weight = 1/m;
J = -Weight*sum( ( y.*log(hypothesis) + (1 - y).*log(1 - hypothesis) ) );
for i = 1 : m
grad = grad + (hypothesis(i) - y(i)) * X(i,:)'; % X must be transposed
end
grad = Weight*grad;
% =============================================================
end
This is the code produced by OMPC:
#mfunction("J, grad")
def costFunction(theta=None, X=None, y=None):
#COSTFUNCTION Compute cost and gradient for logistic regression
# J = COSTFUNCTION(theta, X, y) computes the cost of using theta as the
# parameter for logistic regression and the gradient of the cost
# w.r.t. to the parameters.
# Initialize some useful values
m = length(y) # number of training examples
# You need to return the following variables correctly
J = 0
grad = zeros(size(theta))
# ====================== YOUR CODE HERE ======================
# Instructions: Compute the cost of a particular choice of theta.
# You should set J to the cost.
# Compute the partial derivatives and set grad to the partial
# derivatives of the cost w.r.t. each parameter in theta
#
# Note: grad should have the same dimensions as theta
# Use sigmoid function previously programed
hypothesis = sigmoid(X * theta)# Hypothesis for logistic regression
Weight = 1 / m
J = -Weight * sum((y *elmul* log(hypothesis) + (1 - y) *elmul* log(1 - hypothesis)))
for i in mslice[1:m]:
grad = grad + (hypothesis(i) - y(i)) * X(i, mslice[:]).cT # X must be transposed
end
grad = Weight * grad
# =============================================================
end
Both #mfunction and elmul are OMPC-defined functions. The first one is to correctly deal with the special marray type defined by OMPC, the second one is for doing element-wise multiplication on the aforementioned marrays. They are intended to leverage the semantic differences between NumPy and MATLAB.
However, these differences are difficult to translate in full generality, especially so because often MATLAB code relies on these semantic details.
That is why MATLAB-to-Python converters cannot be used reliably in production code.
I would translate the Octave code above into something like (assuming NumPy arrays inputs):
def cost_function(theta, x, y):
m = np.size(y)
grad = np.zeros(theta.shape)
hypothesis = sigmoid(x # theta)
j = -np.sum((y * np.log(hypothesis) + (1 - y) * np.log(1 - hypothesis))) / m
for i in range(m):
grad = grad + (hypothesis[i] - y[i]) # x[i, :].transpose().conjugate()
grad = grad / m
return j, grad
Please carefully check that x[i, :].transpose().conjugate() is actually needed (it depends on what x and hypothesis actually are).
I am trying to implement the optimization algorithm from Scipy. It works fine when I implement it without inputting the Jacobian gradient function. I believe the issue that I am getting when I input the gradient is because the minimize function itself is changing the shape of the initial guess x0. You can see this from the output of the code below.
Input:
import numpy as np
from costFunction import *
import scipy.optimize as op
def sigmoid(z):
epsilon = np.finfo(z.dtype).eps
g = 1/(1+np.exp(-z))
g = np.clip(g,epsilon,1-epsilon)
return g
def costFunction(theta,X,y):
m = y.size
h = sigmoid(X#theta)
J = 1/(m)*(-y.T#np.log(h)-(1-y).T#np.log(1-h))
grad = 1/m*X.T#(h-y)
print ('Shape of theta is',np.shape(theta),'\n')
print ('Shape of gradient is',np.shape(grad),'\n')
return J, grad
X = np.array([[1, 3],[5,7]])
y = np.array([[1],[0]])
m,n = np.shape(X)
one_vec = np.ones((m,1))
X = np.hstack((one_vec,X))
initial_theta = np.zeros((n+1,1))
print ('Running costFunction before executing minimize function...\n')
cost, grad = costFunction(initial_theta,X,y) #To test the shape of gradient before calling minimize
print ('Executing minimize function...\n')
Result = op.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
Output:
Running costFunction before executing minimize function...
Shape of theta is (3, 1)
Traceback (most recent call last):
Shape of gradient is (3, 1)
Executing minimize function...
Shape of theta is (3,)
File "C:/Users/#####/minimizeshapechange.py", line 34, in <module>
Shape of gradient is (3, 2)
Result = op.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
File "C:\Users\#####\anaconda3\lib\site-packages\scipy\optimize\_minimize.py", line 453, in minimize
**options)
File "C:\Users\#####\anaconda3\lib\site-packages\scipy\optimize\tnc.py", line 409, in _minimize_tnc
xtol, pgtol, rescale, callback)
ValueError: tnc: invalid gradient vector from minimized function.
Process finished with exit code 1
I will not analyze your exact computations, but some remarks:
(1) Your gradient is broken!
scipy expects a partial derivative resulting in an array of shape equal to your x0.
your gradient is of shape (3,2), while (n+1, 1) is expected
compare with the example given in the tutorial which uses scipy.optimize.rosen_der (der = derivative)
(2) It seems your scipy-version is a bit older, because mine (0.19.0) is telling me:
ValueError: tnc: invalid gradient vector from minimized function.
Some supporting source-code from scipy:
if (PyArray_SIZE(arr_grad) != py_state->n)
{
PyErr_SetString(PyExc_ValueError,
"tnc: invalid gradient vector from minimized function.");
goto failure;
Remark: This code above was changed / touched / introduced 5 years ago. If you really don't get this error while using your code listed (with removal of the import of costFunction), it seems you are using scipy < v0.13.0b1, which i do no recommend! I assume you are using some deprecated windows-based inofficial distribution with outdated scipy. You should change that!
I had the same problem with Scipy trying to do the same thing as you. I don't understand exactly why this solves the problem but playing with array shapes until it worked gave me the following:
Gradient function defined as follows
def Gradient(theta,X,y):
#Initializing variables
m = len(y)
theta = theta[:,np.newaxis] #<---- THIS IS THE TRICK
grad = np.zeros(theta.shape)
#Vectorized computations
z = X # theta
h = sigmoid(z)
grad = (1/m)*(X.T # ( h - y));
return grad #< --- also works with grad.ravel()
Initial_theta initialized as
initial_theta = np.zeros((n+1))
initial_theta.shape
(3,)
i.e. a simple numpy array rather than a column vector.
Gradient function returns
Gradient(initial_theta,X,y).shape
(3,1) or (3,) depending on whether the function returns grad or grad.ravel
scipy.optimize called as
import scipy.optimize as opt
model = opt.minimize(fun = CostFunc, x0 = initial_theta, args = (X, y), method = 'TNC', jac = Gradient)
What does not work with Scipy
initial_theta of shape (3,1) using initial_theta = np.zeros((n+1))[:,np.newaxis] crashes the scipy.minimize function call.
ValueError: tnc: invalid gradient vector from minimized function.
If someone could clarify these points that would be great ! Thanks
your code of costFunctuion is wrong,maybe you should look that
def costFunction(theta,X,y):
h_theta = sigmoid(X#theta)
J = (-y) * np.log(h_theta) - (1 - y) * np.log(1 - h_theta)
return np.mean(J)
please copy and past in jpuiter in1 and so on in separte cell
In 1
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
filepath =('C:/Pythontry/MachineLearning/dataset/couresra/ex2data1.txt')
data =pd.read_csv(filepath,sep=',',header=None)
#print(data)
X = data.values[:,:2] #(100,2)
y = data.values[:,2:3] #(100,1)
#print(np.shape(y))
#In 2
#%% ==================== Part 1: Plotting ====================
postive_value = data.loc[data[2] == 1]
#print(postive_value.values[:,2:3])
negative_value = data.loc[data[2] == 0]
#print(len(postive_value))
#print(len(negative_value))
ax1 = postive_value.plot(kind='scatter',x=0,y=1,s=50,color='b',marker="+",label="Admitted") # S is line width #https://matplotlib.org/api/_as_gen/matplotlib.axes.Axes.scatter.html#matplotlib.axes.Axes.scatter
ax2 = negative_value.plot(kind='scatter',x=0,y=1,s=50,color='y',ax=ax1,label="Not Admitted")
ax1.set_xlabel("Exam 1 score")
ax2.set_ylabel("Exam 2 score")
plt.show()
#print(ax1 == ax2)
#print(np.shape(X))
# In 3
#============ Part 2: Compute Cost and Gradient ===========
[m,n] = np.shape(X) #(100,2)
print(m,n)
additional_coulmn = np.ones((m,1))
X = np.append(additional_coulmn,X,axis=1)
initial_theta = np.zeros((n+1), dtype=int)
print(initial_theta)
# In4
#Sigmoid and cost function
def sigmoid(z):
g = np.zeros(np.shape(z));
g = 1/(1+np.exp(-z));
return g
def costFunction(theta, X, y):
J = 0;
#print(theta)
receive_theta = np.array(theta)[np.newaxis] ##This command is used to create the 1D array
#print(receive_theta)
theta = np.transpose(receive_theta)
#print(np.shape(theta))
#grad = np.zeros(np.shape(theta))
z = np.dot(X,theta) # where z = theta*X
#print(z)
h = sigmoid(z) #formula h(x) = g(z) whether g = 1/1+e(-z) #(100,1)
#print(np.shape(h))
#J = np.sum(((-y)*np.log(h)-(1-y)*np.log(1-h))/m);
J = np.sum(np.dot((-y.T),np.log(h))-np.dot((1-y).T,np.log(1-h)))/m
#J = (-y * np.log(h) - (1 - y) * np.log(1 - h)).mean()
#error = h-y
#print(np.shape(error))
#print(np.shape(X))
grad =np.dot(X.T,(h-y))/m
#print(grad)
return J,grad
#In5
[cost, grad] = costFunction(initial_theta, X, y)
print('Cost at initial theta (zeros):', cost)
print('Expected cost (approx): 0.693\n')
print('Gradient at initial theta (zeros): \n',grad)
print('Expected gradients (approx):\n -0.1000\n -12.0092\n -11.2628\n')
In6 # Compute and display cost and gradient with non-zero theta
test_theta = [-24, 0.2, 0.2]
#test_theta_value = np.array([-24, 0.2, 0.2])[np.newaxis] #This command is used to create the 1D row array
#test_theta = np.transpose(test_theta_value) # Transpose
#test_theta = test_theta_value.transpose()
[cost, grad] = costFunction(test_theta, X, y)
print('\nCost at test theta: \n', cost)
print('Expected cost (approx): 0.218\n')
print('Gradient at test theta: \n',grad);
print('Expected gradients (approx):\n 0.043\n 2.566\n 2.647\n')
#IN6
# ============= Part 3: Optimizing using range =============
import scipy.optimize as opt
#initial_theta_initialize = np.array([0, 0, 0])[np.newaxis]
#initial_theta = np.transpose(initial_theta_initialize)
print ('Executing minimize function...\n')
# Working models
#result = opt.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
result = opt.fmin_tnc(func=costFunction, x0=initial_theta, args=(X, y))
# Not working model
#costFunction(initial_theta,X,y)
#model = opt.minimize(fun = costFunction, x0 = initial_theta, args = (X, y), method = 'TNC',jac = costFunction)
print('Thetas found by fmin_tnc function: ', result);
print('Cost at theta found : \n', cost);
print('Expected cost (approx): 0.203\n');
print('theta: \n',result[0]);
print('Expected theta (approx):\n');
print(' -25.161\n 0.206\n 0.201\n');
Result:
Executing minimize function...
Thetas found by fmin_tnc function: (array([-25.16131854, 0.20623159, 0.20147149]), 36, 0)
Cost at theta found :
0.218330193827
Expected cost (approx): 0.203
theta:
[-25.16131854 0.20623159 0.20147149]
Expected theta (approx):
-25.161
0.206
0.201
scipy’s fmin_tnc doesn’t work well with column or row vector. It expects the parameters to be in an array format.
Python Implementation of Andrew Ng’s Machine Learning Course (Part 2.1)
opt.fmin_tnc(func = costFunction, x0 = theta.flatten(),fprime = gradient, args = (X, y.flatten()))
What worked for me is to reshape y as a vector (1-D) rather than a matrix (2-D array). I simply used the following code and then reran the SciPy's minimize function and it worked.
y = np.reshape(y,100) #e.g., if your y variable has 100 data points.
Little bit late but I also started anderw assignment to implement with Python and put a lot of effort to resolve the mentioned issue. Finally is works for me.
This blog help me but with one changes in fmin_tnc function calling, refer below :-
result = op.fmin_tnc(func=costFunction, x0=initial_theta, fprime=None, approx_grad=True, args=(X, y)) Got this info from here
I'm working through my Matlab code for the Andrew NG Coursera course and turning it into python. I am working on non-regularized logistic regression and after writing my gradient and cost functions I needed something similar to fminunc and after some googling, I found a couple options. They are both returning the same results, but they do not match what is in Andrew NG's expected results code. Others seem to be getting this to work correctly, but I'm wondering why my specific code does not seem to return the desired result when using scipy.optimize functions, but does for the cost and gradient pieces earlier in the code.
The data I'm using can be found at the link below;
ex2data1
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as op
#Machine Learning Online Class - Exercise 2: Logistic Regression
#Load Data
#The first two columns contains the exam scores and the third column contains the label.
data = pd.read_csv('ex2data1.txt', header = None)
X = np.array(data.iloc[:, 0:2]) #100 x 3
y = np.array(data.iloc[:,2]) #100 x 1
y.shape = (len(y), 1)
#Creating sub-dataframes for plotting
pos_plot = data[data[2] == 1]
neg_plot = data[data[2] == 0]
#==================== Part 1: Plotting ====================
#We start the exercise by first plotting the data to understand the
#the problem we are working with.
print('Plotting data with + indicating (y = 1) examples and o indicating (y = 0) examples.')
plt.plot(pos_plot[0], pos_plot[1], "+", label = "Admitted")
plt.plot(neg_plot[0], neg_plot[1], "o", label = "Not Admitted")
plt.xlabel('Exam 1 score')
plt.ylabel('Exam 2 score')
plt.legend()
plt.show()
def sigmoid(z):
'''
SIGMOID Compute sigmoid function
g = SIGMOID(z) computes the sigmoid of z.
Instructions: Compute the sigmoid of each value of z (z can be a matrix,
vector or scalar).
'''
g = 1 / (1 + np.exp(-z))
return g
def costFunction(theta, X, y):
'''
COSTFUNCTION Compute cost and gradient for logistic regression
J = COSTFUNCTION(theta, X, y) computes the cost of using theta as the
parameter for logistic regression and the gradient of the cost
w.r.t. to the parameters.
'''
m = len(y) #number of training examples
h = sigmoid(X.dot(theta)) #logisitic regression hypothesis
J = (1/m) * np.sum((-y*np.log(h)) - ((1-y)*np.log(1-h)))
#h is 100x1, y is %100x1, these end up as 2 vector we subtract from each other
#then we sum the values by rows
#cost function for logisitic regression
return J
def gradient(theta, X, y):
m = len(y)
grad = np.zeros((theta.shape))
h = sigmoid(X.dot(theta))
for i in range(len(theta)): #number of rows in theta
XT = X[:,i]
XT.shape = (len(X),1)
grad[i] = (1/m) * np.sum((h-y)*XT) #updating each row of the gradient
return grad
#============ Part 2: Compute Cost and Gradient ============
#In this part of the exercise, you will implement the cost and gradient
#for logistic regression. You neeed to complete the code in costFunction.m
#Add intercept term to x and X_test
Bias = np.ones((len(X), 1))
X = np.column_stack((Bias, X))
#Initialize fitting parameters
initial_theta = np.zeros((len(X[0]), 1))
#Compute and display initial cost and gradient
(cost, grad) = costFunction(initial_theta, X, y), gradient(initial_theta, X, y)
print('Cost at initial theta (zeros): %f' % cost)
print('Expected cost (approx): 0.693\n')
print('Gradient at initial theta (zeros):')
print(grad)
print('Expected gradients (approx):\n -0.1000\n -12.0092\n -11.2628')
#Compute and display cost and gradient with non-zero theta
test_theta = np.array([[-24], [0.2], [0.2]]);
(cost, grad) = costFunction(test_theta, X, y), gradient(test_theta, X, y)
print('\nCost at test theta: %f' % cost)
print('Expected cost (approx): 0.218\n')
print('Gradient at test theta:')
print(grad)
print('Expected gradients (approx):\n 0.043\n 2.566\n 2.647\n')
result = op.fmin_tnc(func = costFunction, x0 = initial_theta, fprime = gradient, args = (X,y))
result[1]
Result = op.minimize(fun = costFunction,
x0 = initial_theta,
args = (X, y),
method = 'TNC',
jac = gradient, options={'gtol': 1e-3, 'disp': True, 'maxiter': 1000})
theta = Result.x
theta
test = np.array([[1, 45, 85]])
prob = sigmoid(test.dot(theta))
print('For a student with scores 45 and 85, we predict an admission probability of %f,' % prob)
print('Expected value: 0.775 +/- 0.002\n')
This was a very difficult problem to debug, and illustrates a poorly documented aspect of the scipy.optimize interface. The documentation vaguely indicates that theta will be passed around as a vector:
Minimization of scalar function of one or more variables.
In general, the optimization problems are of the form:
minimize f(x) subject to
g_i(x) >= 0, i = 1,...,m
h_j(x) = 0, j = 1,...,p
where x is a vector of one or more variables.
What's important is that they really mean vector in the most primitive sense, a 1-dimensional array. So you have to expect that whenever theta is passed into one of your callbacks, it will be passed in as a 1-d array. But in numpy, 1-d arrays sometimes behave differently from 2-d row arrays (and, obviously, from 2-d column arrays).
I don't know exactly why it's causing a problem in your case, but it's easily fixed regardless. You just have to add the following at the top of both your cost function and your gradient function:
theta = theta.reshape(-1, 1)
This guarantees that theta will be a 2-d column array, as expected. Once you've done this, the results are correct.
I have had similar issues with Scipy dealing with the same problem as you. As senderle points out the interface is not the easiest to deal with, especially combined with the numpy array interface... Here is my implementation which works as expected.
Defining the cost and gradient functions
Note that initial_theta is passed as a simple array of shape (3,) and converted to a column vector of shape (3,1) within the function. The gradient function then returns the grad.ravel() which has shape (3,) again. This is important as doing otherwise caused an error message with various optimization methods in Scipy.optimize.
Note that different methods have different behaviours but returning .ravel() seems to fix most issues...
import pandas as pd
import numpy as np
import scipy.optimize as opt
def sigmoid(x):
return 1 / (1 + np.exp(-x))
def CostFunc(theta,X,y):
#Initializing variables
m = len(y)
J = 0
grad = np.zeros(theta.shape)
#Vectorized computations
z = X # theta
h = sigmoid(z)
J = (1/m) * ( (-y.T # np.log(h)) - (1 - y).T # np.log(1-h));
return J
def Gradient(theta,X,y):
#Initializing variables
m = len(y)
theta = theta[:,np.newaxis]
grad = np.zeros(theta.shape)
#Vectorized computations
z = X # theta
h = sigmoid(z)
grad = (1/m)*(X.T # ( h - y));
return grad.ravel() #<-- This is the trick
Initializing variables and parameters
Note that initial_theta.shape returns (3,)
X = data1.iloc[:,0:2].values
m,n = X.shape
X = np.concatenate((np.ones(m)[:,np.newaxis],X),1)
y = data1.iloc[:,-1].values[:,np.newaxis]
initial_theta = np.zeros((n+1))
Calling Scipy.optimize
model = opt.minimize(fun = CostFunc, x0 = initial_theta, args = (X, y), method = 'TNC', jac = Gradient)
Any comments from more knowledgeable people are welcome, this Scipy interface is a mystery to me, thanks
I have create a theano.Op that returns distance between each pair of the two collections of inputs, converting the scipy cdist:
class Cdist(theano.Op):
__props__ = ()
def __init__(self):
#self.fn = scipy_cdist2
super(Cdist, self).__init__()
def make_node(self, x, w):
#print('make_node')
return gof.Apply(self, [x, w], [x.type()])
def perform(self, node, inputs, output_storage):
#print('perform')
x, w = inputs[0], inputs[1]
z = output_storage[0]
z[0] = distance.cdist(x, w, 'euclidean')
It works, but now want to add the grad method. I have read the guide and the documentation about the grad method. But i still dont't understand how it works. For example in the guide to get the gradient of a method that return a*x + b, they use:
def grad(self, inputs, output_grads):
return [a * output_grads[0] + b]
why? I'm going to quote what is written in the documentation about the grad:
If the output list of the op is [f_1, ... f_n], then the list
output_gradients is [grad_{f_1}(C), grad_{f_2}(C), ... ,
grad_{f_n}(C)]. If inputs consists of the list [x_1, ..., x_m], then
Op.grad should return the list [grad_{x_1}(C), grad_{x_2}(C), ...,
grad_{x_m}(C)], where (grad_{y}(Z))_i = \frac{\partial Z}{\partial
y_i} (and i can stand for multiple dimensions).
They are told me that i have to write the gradient? But in the example the make a combination of output_grads and integger values. Really i'm not understanding.
There's nothing wrong about docs. In grad method you should write a symbolic expression, as opposed to perform method where you write a numerical expression.
grad method is called from theano.grad, while perform is called inside the compiled function.
For example, assuming euclidean distance:
def grad(self, inputs, out_grads):
x, y = inputs # matrices of shape [mA, n] and [mB, n]]
g, = out_grads # matrix of shape [mA, mB]
diff = x.dimshuffle(0, 'x', 1) - y.dimshuffle('x', 0, 1) # [mA, mB, n] tensor
z = T.sqrt(T.sum(T.sqr(diff), axis=2, keepdims=True))
diff = g * diff / z
return [T.sum(diff, axis=1), -T.sum(diff, axis=0)]
For this particular case, I'd suggest writing a L_op instead of grad. L_op additionally reuses output in the forward Op.
def L_op(self, inputs, outputs, out_grads):
x, y = inputs # matrices of shape [mA, n] and [mB, n]
z, = outputs # matrix of shape [mA, mB]
g, = out_grads # idem
diff = x.dimshuffle(0, 'x', 1) - y.dimshuffle('x', 0, 1) # [mA, mB, n] tensor
diff = g.dimshuffle(0, 1, 'x') * diff / z.dimshuffle(0, 1, 'x')
return [T.sum(diff, axis=1), -T.sum(diff, axis=0)]
Well, the grad expressions are probably wrong but you get the idea.
As you can see, we are calling symbolic functions such as dimshuffle. However there are cases where you want to write a class for grad Op. Either because the symbolic graph is too inefficient or you want a custom gradient.
For example:
class CDistGrad(theano.Op):
def __init__(...):
# <...>
pass
def c_code(...):
# implement this in case you want more performance
pass
def perform(...):
# <...>
pass
def make_node(...):
# <...>
pass
class CDist(theano.Op):
# <...>
def grad(self, inputs, output_grads):
return CDistGrad()(*inputs, *output_grads)
Still, symbolic expression is used in grad method. Just a custom Op replaced vanilla Theano expression.
Darn thing just won't learn. Sometimes weights seem to become nan.
I haven't played with different numbers of hidden layers/inputs/outputs but the bug appears consistent across different sizes of hidden layer.
from __future__ import division
import numpy
import matplotlib
import random
class Net:
def __init__(self, *sizes):
sizes = list(sizes)
sizes[0] += 1
self.sizes = sizes
self.weights = [numpy.random.uniform(-1, 1, (sizes[i+1],sizes[i])) for i in range(len(sizes)-1)]
#staticmethod
def activate(x):
return 1/(1+numpy.exp(-x))
def y(self, x_):
x = numpy.concatenate(([1], numpy.atleast_1d(x_.copy())))
o = [x] #o[i] is the (activated) output of hidden layer i, "hidden layer 0" is inputs
for weight in self.weights[:-1]:
x = weight.dot(x)
x = Net.activate(x)
o.append(x)
o.append(self.weights[-1].dot(x))
return o
def __call__(self, x):
return self.y(x)[-1]
def delta(self, x, t):
o = self.y(x)
delta = [(o[-1]-t) * o[-1] * (1-o[-1])]
for i, weight in enumerate(reversed(self.weights)):
delta.append(weight.T.dot(delta[-1]) * o[-i-2] * (1-o[-i-2]))
delta.reverse()
return o, delta
def train(self, inputs, outputs, epochs=100, rate=.1):
for epoch in range(epochs):
pairs = zip(inputs, outputs)
random.shuffle(pairs)
for x, t in pairs: #shuffle? subset?
o, d = self.delta(x, t)
for layer in range(len(self.sizes)-1):
self.weights[layer] -= rate * numpy.outer(o[layer+1], d[layer])
n = Net(1, 4, 1)
x = numpy.linspace(0, 2*3.14, 10)
t = numpy.sin(x)
matplotlib.pyplot.plot(x, t, 'g')
matplotlib.pyplot.plot(x, map(n, x), 'r')
n.train(x, t)
print n.weights
matplotlib.pyplot.plot(x, map(n, x), 'b')
matplotlib.pyplot.show()
I haven't looked for a particular bug in your code, but can you please try the following things to narrow down your problem further? Otherwise it is very tedious to find the needle in the haystack.
1) Please try to use a real dataset to have an idea what to expect, e.g., MNIST, and/or standardize your data, because your weights may become NaN if they become too small.
2) Try different learning rates and plot the cost function vs. epochs to check if you are converging. It should look somewhat like this (note that I used minibatch learning and averaged the minibatch chunks for each epoch).
3) I see that you are using a sigmoid activation, your implementation is correct, but to make it numerically more stable, replace 1.0 / (1.0 + np.exp(-z)) by expit(z) from scipy.special (same function but more efficient).
4) Implement gradient checking. Here, you compare the analytical solution to a numerically approximated gradient
Or an even better approach that yields a more accurate approximation of the gradient is to compute the symmetric (or centered) difference quotient given by the two-point formula
PS: If you are interested and find it useful, I have a working vanilla NumPy neural net implemented here.
I fixed it! Thanks for all the suggestions. I worked out numeric partials and found that my o and deltas were correct, but I was multiplying the wrong ones. That's why I now take numpy.outer(d[layer+1], o[layer]) instead of numpy.outer(d[layer], o[layer+1]).
I was also skipping the update on one layer. That's why I changed for layer in range(self.hidden_layers) to for layer in range(self.hidden_layers+1).
I'll add that I caught a bug just before posting originally. My output layer delta was incorrect because my net (intentionally) doesn't activate the final outputs, but my delta was computed as though it did.
Debugged primarily with a one hidden layer, one hidden unit net, then moved to a 2 input, 3 hidden layers of 2 neurons each, 2 output model.
from __future__ import division
import numpy
import scipy
import scipy.special
import matplotlib
#from pylab import *
#numpy.random.seed(23)
def nmap(f, x):
return numpy.array(map(f, x))
class Net:
def __init__(self, *sizes):
self.hidden_layers = len(sizes)-2
self.weights = [numpy.random.uniform(-1, 1, (sizes[i+1],sizes[i])) for i in range(self.hidden_layers+1)]
#staticmethod
def activate(x):
return scipy.special.expit(x)
#return 1/(1+numpy.exp(-x))
#staticmethod
def activate_(x):
s = scipy.special.expit(x)
return s*(1-s)
def y(self, x):
o = [numpy.array(x)] #o[i] is the (activated) output of hidden layer i, "hidden layer 0" is inputs and not activated
for weight in self.weights[:-1]:
o.append(Net.activate(weight.dot(o[-1])))
o.append(self.weights[-1].dot(o[-1]))
# for weight in self.weights:
# o.append(Net.activate(weight.dot(o[-1])))
return o
def __call__(self, x):
return self.y(x)[-1]
def delta(self, x, t):
x = numpy.array(x)
t = numpy.array(t)
o = self.y(x)
#delta = [(o[-1]-t) * o[-1] * (1-o[-1])]
delta = [o[-1]-t]
for i, weight in enumerate(reversed(self.weights)):
delta.append(weight.T.dot(delta[-1]) * o[-i-2] * (1-o[-i-2]))
delta.reverse() #surely i need this
return o, delta
def train(self, inputs, outputs, epochs=1000, rate=.1):
errors = []
for epoch in range(epochs):
for x, t in zip(inputs, outputs): #shuffle? subset?
o, d = self.delta(x, t)
for layer in range(self.hidden_layers+1):
grad = numpy.outer(d[layer+1], o[layer])
self.weights[layer] -= rate * grad
return errors
def rmse(self, inputs, outputs):
return ((outputs - nmap(self, inputs))**2).sum()**.5/len(inputs)
n = Net(1, 8, 1)
X = numpy.linspace(0, 2*3.1415, 10)
T = numpy.sin(X)
Y = map(n, X)
Y = numpy.array([y[0,0] for y in Y])
matplotlib.pyplot.plot(X, T, 'g')
matplotlib.pyplot.plot(X, Y, 'r')
print 'output successful'
print n.rmse(X, T)
errors = n.train(X, T)
print 'tried to train successfully'
print n.rmse(X, T)
Y = map(n, X)
Y = numpy.array([y[0,0] for y in Y])
matplotlib.pyplot.plot(x, Y, 'b')
matplotlib.pyplot.show()