3
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I am a machine learning noob attempting to implement regularized logistic regression via Newton's method.

The example data have two features which are to be expanded to 28 through finding all monomial terms of (u,v) up to degree 6.

My code converges to the correct solution of norm(theta)=0.9384 after around 500 or so iterations when it should only take around 15 for lambda = 10, though the exercise is based on Matlab instead of Python. Each cycle of the parameter update is also very slow with my code and I am not sure exactly why. If anyone could explain why my code takes so many iterations to converge and why each iteration is painfully slow I would be very grateful! Please be brutally honest about my code habits as well.

The data are taken from Andrew Ng's open course exercise 5. The problem information and data can be found here, although I posted the data and my code below.

X values, 2 features

0.051267,0.69956
-0.092742,0.68494
-0.21371,0.69225
-0.375,0.50219
-0.51325,0.46564
-0.52477,0.2098
-0.39804,0.034357
-0.30588,-0.19225
0.016705,-0.40424
0.13191,-0.51389
0.38537,-0.56506
0.52938,-0.5212
0.63882,-0.24342
0.73675,-0.18494
0.54666,0.48757
0.322,0.5826
0.16647,0.53874
-0.046659,0.81652
-0.17339,0.69956
-0.47869,0.63377
-0.60541,0.59722
-0.62846,0.33406
-0.59389,0.005117
-0.42108,-0.27266
-0.11578,-0.39693
0.20104,-0.60161
0.46601,-0.53582
0.67339,-0.53582
-0.13882,0.54605
-0.29435,0.77997
-0.26555,0.96272
-0.16187,0.8019
-0.17339,0.64839
-0.28283,0.47295
-0.36348,0.31213
-0.30012,0.027047
-0.23675,-0.21418
-0.06394,-0.18494
0.062788,-0.16301
0.22984,-0.41155
0.2932,-0.2288
0.48329,-0.18494
0.64459,-0.14108
0.46025,0.012427
0.6273,0.15863
0.57546,0.26827
0.72523,0.44371
0.22408,0.52412
0.44297,0.67032
0.322,0.69225
0.13767,0.57529
-0.0063364,0.39985
-0.092742,0.55336
-0.20795,0.35599
-0.20795,0.17325
-0.43836,0.21711
-0.21947,-0.016813
-0.13882,-0.27266
0.18376,0.93348
0.22408,0.77997
0.29896,0.61915
0.50634,0.75804
0.61578,0.7288
0.60426,0.59722
0.76555,0.50219
0.92684,0.3633
0.82316,0.27558
0.96141,0.085526
0.93836,0.012427
0.86348,-0.082602
0.89804,-0.20687
0.85196,-0.36769
0.82892,-0.5212
0.79435,-0.55775
0.59274,-0.7405
0.51786,-0.5943
0.46601,-0.41886
0.35081,-0.57968
0.28744,-0.76974
0.085829,-0.75512
0.14919,-0.57968
-0.13306,-0.4481
-0.40956,-0.41155
-0.39228,-0.25804
-0.74366,-0.25804
-0.69758,0.041667
-0.75518,0.2902
-0.69758,0.68494
-0.4038,0.70687
-0.38076,0.91886
-0.50749,0.90424
-0.54781,0.70687
0.10311,0.77997
0.057028,0.91886
-0.10426,0.99196
-0.081221,1.1089
0.28744,1.087
0.39689,0.82383
0.63882,0.88962
0.82316,0.66301
0.67339,0.64108
1.0709,0.10015
-0.046659,-0.57968
-0.23675,-0.63816
-0.15035,-0.36769
-0.49021,-0.3019
-0.46717,-0.13377
-0.28859,-0.060673
-0.61118,-0.067982
-0.66302,-0.21418
-0.59965,-0.41886
-0.72638,-0.082602
-0.83007,0.31213
-0.72062,0.53874
-0.59389,0.49488
-0.48445,0.99927
-0.0063364,0.99927

Y values

1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0

My code

import pandas as pd
import numpy as np
import math

def sigmoid(theta, x):

    return 1/(1 + math.exp(-1*theta.T.dot(x)))


def cost_function(X, y, theta):

    s = 0
    for i in range(m):
        loss = -y[i]*np.log(sigmoid(theta, X[i])) - (1-y[i])*np.log(1-sigmoid(theta, X[i]))
        s += loss
    s /= m
    s += (lamb/(2*m))*sum(theta[j]**2 for j in range(1, 28)) 
    return s


def gradient(theta, X, y):

    add_column = theta * (lamb/m)
    add_column[0] = 0
    a = sum((sigmoid(theta, X[i]) - y[i])*X[i] + add_column for i in range(m))/m
    return a


def hessian(theta, X, reg_matrix):

    matrix = []
    for i in range(28):
        row = []
        for j in range(28):
            cell = sum(sigmoid(theta, X[k])*(1-sigmoid(theta, X[k]))*X[k][i]*X[k][j] for k in range(m))
            row.append(cell)
        matrix.append(row)

    H = np.array(matrix)
    H = np.add(H, reg_matrix)
    return H


def newtons_method(theta, iterations):

    for i in range(iterations):
        g = gradient(theta, X, y)
        H = hessian(theta, X, reg_matrix)
        theta = theta - np.linalg.inv(H).dot(g)
        cost = cost_function(X,y,theta)
        print(cost)    
    return theta

def map_feature(u, v): # expand features according to problem instructions

    new_row = [] 
    new_row.append(1)
    new_row.append(u)
    new_row.append(v)
    new_row.append(u**2)
    new_row.append(u*v)
    new_row.append(v**2)
    new_row.append(u**3)
    new_row.append(u**2*v)
    new_row.append(u*v**2)
    new_row.append(v**3)
    new_row.append(u**4)
    new_row.append(u**3*v)
    new_row.append(u*v**3)
    new_row.append(v**4)
    new_row.append(u**2*v**2)
    new_row.append(u**5)
    new_row.append(u**4*v)
    new_row.append(u*v**4)
    new_row.append(v**5)
    new_row.append(u**2*v**3)
    new_row.append(u**3*v**2)
    new_row.append(u**6)
    new_row.append(u**5*v)
    new_row.append(u*v**5)
    new_row.append(v**6)
    new_row.append(u**4*v**2)
    new_row.append(u**2*v**4)
    new_row.append(u**3*v**3)
    return np.array(new_row)

with open('ex5Logx.dat', 'r') as f:
    array = []
    for line in f.readlines():
        array.append(line.strip().split(','))

    for a in array:

        a[0], a[1] = float(a[0]), float(a[1].strip())

    xdata= np.array(array)

with open('ex5Logy.dat', 'r') as f:
    array = []
    for line in f.readlines():
        array.append(line.strip())

    for i in range(len(array)):
        array[i] = float(array[i])
    ydata= np.array(array)


X_df = pd.DataFrame(xdata, columns=['score1', 'score2'])

y_df = pd.DataFrame(ydata, columns=['acceptence'])

m = len(y_df)

iterations = 1500

ones = np.ones((m,1)) # intercept term in first column
X = np.array(X_df)
X = np.append(ones, X, axis=1)
y = np.array(y_df).flatten()

new_X = [] # prepare new array for expanded features
for i in range(m):
    new_row = map_feature(X[i][1], X[i][2])

    new_X.append(new_row)

X = np.array(new_X)

theta = np.array([0 for i in range(28)]) # initialize parameters to 0

lamb = 1 # lambda constant for regularization

reg_matrix = np.zeros((28,28),dtype=int) # n+1*n+1 regularization matrix 
np.fill_diagonal(reg_matrix, 1)
reg_matrix[0] = 0
reg_matrix = (lamb/m)*reg_matrix

theta = newtons_method(theta, iterations)
print(np.linalg.norm(theta))
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1
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A few tips for clarity, before you can even begin to analyze what goes slow

global state

You pass on global state to your functions, but not as arguments

for example in your cost_function: m and lamb are from the global scope

Use pandas csv_reader

Depending on how exactly the datafiles look like

with open('ex5Logx.dat', 'r') as f:
    array = []
    for line in f.readlines():
        array.append(line.strip().split(','))

    for a in array:

        a[0], a[1] = float(a[0]), float(a[1].strip())

    xdata= np.array(array)

with open('ex5Logy.dat', 'r') as f:
    array = []
    for line in f.readlines():
        array.append(line.strip())

    for i in range(len(array)):
        array[i] = float(array[i])
    ydata= np.array(array)


X_df = pd.DataFrame(xdata, columns=['score1', 'score2'])

y_df = pd.DataFrame(ydata, columns=['acceptence'])

Can be replace by something as

X_df = pd.read_csv(x_filename, sep=',', header=None, index_col=None).rename(columns={0: 'score1', 1: 'score2'})
y_df = pd.read_csv(y_filename, sep=',', header=None, index_col=None).rename(columns={0: 'acceptance'})

map_feature

Instead of appending new rows (which is slow) to a list for every item, why not build it in one go, either literal or via the indices?

def map_feature(u, v): # expand features according to problem instructions
    indices = [
    (0, 0),  
    (1, 0),  (0, 1),  
    (2, 0),  (1, 1),  (0, 2),  
    (3, 0),  (2, 1),  (1, 2),  (0, 3),  
    (4, 0),  (3, 1),  (1, 3),  (0, 4),  (2, 2),  
    (5, 0),  (4, 1),  (1, 4),  (0, 5),  (3, 2),  (2, 3),  
    (6, 0),  (5, 1),  (1, 5),  (0, 6),  (4, 2),  (2, 4),  (3, 3),
    ]

    return np.array([u ** i * v ** j for i, j in indices])

If needed, you could generate the indices on the fly with something like this:

def indices(n):
    yield from sorted(set(itertools.chain.from_iterable(((n - i, i), (i, n - i)) for i in range(min(n + 1, 2)))), reverse=True)
    yield from itertools.chain.from_iterable(sorted({(n - i, i), (i, n - i)}, reverse=True) for i in range(2, n//2 + 1))
idx = itertools.chain.from_iterable(indices(i) for i in range(7))

conversions

Why all the conversions between pd.DataFrame and np.array? like here?

y = np.array(y_df).flatten()

The insertion of the column of 1 can be handled like this:

X = X_df.insert('1'=1)

vectorize

The map_feature is done for all rows of the X after adding a column of 1s. Why not do it in 1 go?

new_X = pd.concat((X['score1'] ** i * X['score2'] ** j for i, j in indices), axis=1)

your cost_function can be vectorized too, especially this part:

for i in range(m):
    loss = -y[i]*np.log(sigmoid(theta, X[i])) - (1-y[i])*np.log(1-sigmoid(theta, X[i]))
    s += loss

The Hessian can use some vectorization too

Magical value

where does the 28 come from? is it len(indices)?

use the libraries functions

reg_matrix = np.zeros((28,28),dtype=int) # n+1*n+1 regularization matrix 
np.fill_diagonal(reg_matrix, 1)
reg_matrix[0] = 0
reg_matrix = (lamb/m)*reg_matrix

creates a diagonal matrix with 1/len(y_df) on the diagonal except the first item. numpy had diag_flat

reg_matrix = np.diagflat([0] + [lamb/m] * 27)

There are still a lot of open points, but without further clarification what it does and how the data looks like, that will be hard

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  • \$\begingroup\$ The 28 comes from the expansion of the two given features per training example to 28 per x. It was part of the problem instructions presumably to test the effectiveness of restricting the parameters. The data were not said to represent any real quantities but the objective is to classify an input as either 1 or 0. This answer is very helpful, thank you. I am still unable to upvote but will when possible. \$\endgroup\$ – Cam Mar 4 '18 at 0:24
0
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This is a small improvement to Maarten's answer. Instead of hard-coding indices, you could use

import itertools
indices = list(itertools.product(range(7), repeat=2))

This is clearly shorter, and also cleaner.

Better yet,

def map_feature(u, v): # expand features according to problem instructions
    return np.array([u ** i * v ** j for i, j in itertools.product(range(7), repeat=2)])

This will be even faster, as it removes the need to create the indices list.

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  • \$\begingroup\$ [(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3), (3, 0), (3, 1), (3, 2), (3, 3)] is not the same list as indices. I've looked into creating that list in that order with itertools, bu!t it becomes convoluted quickly \$\endgroup\$ – Maarten Fabré Apr 3 '18 at 7:37
  • \$\begingroup\$ Oh, the order matters. Never mind. \$\endgroup\$ – Oscar Smith Apr 3 '18 at 14:26
  • \$\begingroup\$ I don't know whether the order matters. OP is unclear about that, but even so, not all combinations are generated \$\endgroup\$ – Maarten Fabré Apr 3 '18 at 14:56
  • \$\begingroup\$ oops, 4 should be 7, and then it's just order. \$\endgroup\$ – Oscar Smith Apr 3 '18 at 15:43
  • \$\begingroup\$ are you even paying attention? There is no coefficient (6, 6). If you want to do it without regards to the order, you'd need something like: itertools.chain.from_iterable(set((j, i-j) for j in range(i + 1)) for i in range(7)) \$\endgroup\$ – Maarten Fabré Apr 4 '18 at 8:04

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