## DogDogFish

### Data Science, amongst other things.

#### Tag: Numpy

Hi everybody,

Hi Dr Nick. But enough of that – today I’m going to be working through a Kaggle problem. For those of you who don’t know Kaggle, I can’t advise in favour of it strongly enough. It’s a great place to have a go at using real data sets to apply various machine learning techniques. There’s a leaderboard, discussions on methods and some non-too shabby prizes. I’ll come clean at this point – I’m not a natural salesperson.

I believe there’s something of a taboo against posting solutions/methods for Kaggle – however, I think I’m good to write about a method of solving this particular problem. The digit recognizer problem seems to be a rolling competition with a bunch of already published results and a few training classes on how to solve it. Let me know if you think this is overstepping the mark.

So, the problem:
Given a big set (42,000) of labelled training data (28 x 28 black and white images) of handwritten images (0-9) are we able to correctly identify other (identically dimensioned) handwritten digits.

There are a whole bunch of ways of doing this and the method I’ve had best success with is Support Vector Machines (using LibSVM). I may post an example of how to run that for this particular example but today I’d like to look at Adaboost (using the MultiBoost package)…

Until fairly recently I was entirely ignorant of Adaboost – I came across it on a different Kaggle problem (the Higgs one). There, a number of ‘out of the box’ methods were showcased – the most successful of which was Adaboost. A bit of reading on Adaboost suggests that it’s a fairly well-regarded, and successful method of performing a range of machine learning tasks. It’s also sometimes cited as being the best ‘out of the box’ (not specifically designed for the task at hand) algorithm in machine learning.

My current intuition on Adaboost is that it’s basically a ‘rule of thumb’ algorithm. It takes a lot of very simple decision boundaries and uses them to create a more complicated decision space. I say rule of thumb because I imagine a car mechanic or a doctor trying to diagnose a fault. The patient presents with symptom x, that makes a whole bunch of things less likely. However, if the patient falls into this age bucket and this ethnic group, some of the previously discounted things become more likely. I don’t know if that sort of explanation helps you but I quite like it. Basically, you create a simple rule that’s more often right than wrong. However, you can then update it with as many exceptions as you’ve got other bits of data. I think that’s a lot how the human decision-making process goes.

Anyway, all this talking isn’t getting us closer to a juicy set of predictions. Mad props to whoever first generated this particular procedure for the Higgs problem – I’ve shamelessly ripped it off, only making changes where necessary for this problem.

#!/usr/bin/python

import random
import csv
import subprocess
import numpy as np

def DataToArff(dataset, labels, header, title, filename):
"""
With this data structure we're able to turn an arbitrary string of data into a .arff file
These files allow us to import the data into Multiboost or Weka (amongst other machine learning libraries
"""
with open(filename + ".arff", 'w') as f:
f.write('@RELATION ' + title + 'nn')
f.write('@ATTRIBUTE ' + feature + ' NUMERICn')
f.write('@ATTRIBUTE class {0,1,2,3,4,5,6,7,8,9}n')
f.write('n@DATAn')
## We could do this using all_data - however, we need the labels for further work
## Additionally, if the labels were numeric variables we'd be able to leave the rest of our work unchanged and handle them here
for datarow, label in zip(dataset, labels):
for value in datarow:
f.write(str(value) + ',')
f.write(str(label) + 'n')

dataset = np.array([map(float, row[1:]) for row in all_data[1:]])
(numpoints, numfeatures) = dataset.shape

# Labels on the first column of the line
labels = np.array([row[0] for row in all_data[1:]])

randomPermutation = random.sample(range(len(dataset)), len(dataset))
## If this breaks halfway through, we'll be glad to be able to load our random permutation
np.savetxt('randomPermutation.csv', randomPermutation, fmt='%d', delimiter=',')

## I'll change the proportion of the train set and see how we get on.
numpointsTrain = int(numpoints*0.75)
numpointsValidatin = numpoints - numpointsTrain

## Because we've got a random permutation there's no problem taking slices of the total set to sort into train and validation
datasetTrain = dataset[randomPermutation[:numpointsTrain]]
datasetValidation = dataset[randomPermutation[numpointsTrain:]]

labelsTrain = labels[randomPermutation[:numpointsTrain]]
labelsValidation = labels[randomPermutation[numpointsTrain:]]

## Our Adaboost parameters are wholly contained in the relevant config files
p1.wait()
p2.wait()

datasetTest = np.array([map(float, row[:]) for row in testText[1:]])
labelsTest = np.repeat('0', len(testText) - 1)

p3.wait()

with open('submission.csv', 'w') as f:
f.write('ImageId,Labeln')
for index, entry in enumerate(testScoresText):
## Take the index of the maximum value for a given row - this is the most likely value
f.write(str(index + 1) + "," + str(np.argmax(entry)) + 'n')

I’d like to think that that bit of code is reasonably transparent and clear on what it’s doing. If I’m wrong, a basic explanation:
1.) Randomly split the data into a training and validation set
2.) Create .arff files for both of these sets
3.) Run Multiboost (our Adaboost implementation) on the training set and validation set
4.) Using the files created from our train/validation Adaboost, get the test set and generate predictions (again using Multiboost)
5.) Generate a submission file in the required format

Easy does it. Now for the configuration files that I’m using:

config.txt

fileformat arff
verbose 2
learnertype TreeLearner
constant
seed 50
weightpolicy balanced
baselearnertype SingleStumpLearner 8
outputinfo results.dta e01w01auc
traintest training.arff validation.arff 5000
shypname shyp.xml

configScoresValidation.txt

posteriors validation.arff shyp.xml scoresValidation.txt 5000
fileformat arff
verbose 2
learnertype TreeLearner
baselearnertype SingleStumpLearner 8

configScoresTest.txt

posteriors test.arff shyp.xml scoresTest.txt 5000
fileformat arff
verbose 2
learnertype TreeLearner
baselearnertype SingleStumpLearner 8

I’ll be honest and say I’ve not really found the ideal set-up for this problem here. I’m able to get a score of around 0.965 using the ones here but if you look at the leader board you’ll see that’s not all that good. Certainly the LibSVM method performed much better (something like 0.99). Not to worry, it’s doing the right thing, generating good results and is another tool in our arsenal.

The World Cup may stymie my blog posts for a bit – then again, supporting England, it might only be 3 games.

Football’s coming home.

Hi all,

Bit of a different blog coming up – in a previous post I used Markov Clustering and said I’d write a follow-up post on what it was and why you might want to use it. Well, here I am. And here you are. So let’s begin:

In the simplest explanation, imagine an island. The island is connected to a whole bunch of other islands by bridges. The bridges are made out of bricks. Nothing nasty so far – apart from the leader of all the islands. They’re a ‘man versus superman’, ‘survival of the fittest’ sort and so one day the issue a proclamation. “Every day a brick will be taken from every bridge connected to your island and the bricks will be reapportioned on your island back to the bridges, in proportion to the remaining number of bricks in the bridge.”

At first, nobody is especially worried – each day, a brick disappears and then reappears on a different bridge on the island. Some of the islands notice some bridges getting three or four bricks back each day. Some hardly ever seem to see a brick added back to their bridge. Can you see where this will lead in 1000 years? In time, some of the bridges (the smallest ones to start off with) fall apart and end up with no bricks at all. If this is the only way between two islands, these islands become cut off entirely from each other.

This is basically Markov clustering.

For a more mathematical explanation:

import numpy as np
transition_matrix = np.matrix([[0,0.97,0.5],[0.2,0,0.5],[0.8,0.03,0]])

$Transition Matrix = begin{matrix} 0 & 0.97 & 0.5 \ 0.2 & 0 & 0.5 \ 0.8 & 0.03 & 0 end{matrix}$

In the above ‘islands’ picture those numbers represent the number of bricks in the bridges between islands A, B and C. In the random-walk interpretation, those are the probabilities that you’ll end up at each node as the number of random walks tends to infinity. In my previous post on house prices, I used a correlation matrix.

First things first – I’m going to stick a one in each of the identity areas. If you’re interested in why that is, have a read around self-loops and, even better, try this out both with and without self-loops. It sort of fits in nicely with the above islands picture but that’s more of a fluke than anything else – there’s always the strongest bridge possible between an island and itself. The land. Anyway…

np.fill_diagonal(transition_matrix, 1)

$Transition Matrix = begin{matrix} 1 & 0.97 & 0.5 \ 0.2 & 1 & 0.5 \ 0.8 & 0.03 & 1 end{matrix}$

Now let’s normalize – make sure each column sums to 1:

transition_matrix = transition_matrix/np.sum(transition_matrix, axis=0)

$Transition Matrix = begin{matrix} 0.5 & 0.485 & 0.25 \ 0.1 & 0.5 & 0.25 \ 0.4 & 0.015 & 0.5 end{matrix}$

Now we perform an expansion step – that is, we raise the matrix to a power (I’ll use two – you can change this parameter – in the ‘random-walk’ picture this can be thought of as varying how far a person can walk from their original island).

transition_matrix = np.linalg.matrix_power(transition_matrix, 2)

$Expanded Matrix = begin{matrix} 0.3985 & 0.48875 & 0.37125 \ 0.2 & 0.30225 & 0.275 \ 0.4015 & 0.209 & 0.35375 end{matrix}$

Then we perform the inflation step – This involves multiplying each element in the matrix by itself (to a power) and then normalizing on column again. Again, I’ll be using two as a power – increasing this leads to a greater number of smaller clusters:

entry[...] = entry ** 2

$Inflated Matrix = begin{matrix} 0.15880225 & 0.23887556 & 0.13782656 \ 0.04 & 0.09135506 & 0.075625 \ 0.16120225 & 0.043681 & 0.12513906 end{matrix}$

Finally (for this iteration) – we’ll normalize by row.

transition_matrix = transition_matrix/np.sum(transition_matrix, axis=0)

$Normalized Matrix = begin{matrix} 0.44111185 & 0.63885664 & 0.40705959 \ 0.11110972 & 0.24432195 & 0.22335232 \ 0.44777843 & 0.11682141 & 0.36958809 end{matrix}$

And it’s basically that simple. Now all we need to do is rinse and repeat the expansion, inflation and normalization until we hit a stable(ish) solution i.e.

$Normalized Matrix_{n+1} - Normalized Matrix_n < epsilon$
for some small epsilon.

Once we’ve done this (with this particular matrix) we should see something like this:

$Final Matrix = begin{matrix} 1 & 1 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 end{matrix}$

Doesn’t look like a brilliant result be we only started with a tiny matrix. In this case we have all three nodes belonging to one cluster. The first node (the first row) is the ‘attractor’ – as it has values in its row it is attracting itself and the second and third row (the columns). If we were to end up with the following result (from a given initial matrix):

$Final Matrix = begin{matrix} 1 & 0 & 1 & 0 & 0\ 0 & 1 & 0 & 1 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 0 & 0 & 1\ end{matrix}$

This basically says that we have three clusters {1,3} (with 1 as the attractor), {2,4} (with 2 as the attractor) and {5} on its lonesome.

Instead of letting you piece all that together here’s the code for Markov Clustering in Python:

import numpy as np
import math
## How far you'd like your random-walkers to go (bigger number -> more walking)
EXPANSION_POWER = 2
## How tightly clustered you'd like your final picture to be (bigger number -> more clusters)
INFLATION_POWER = 2
## If you can manage 100 iterations then do so - otherwise, check you've hit a stable end-point.
ITERATION_COUNT = 100
def normalize(matrix):
return matrix/np.sum(matrix, axis=0)

def expand(matrix, power)
return np.linalg.matrix_power(matrix, power)

def inflate(matrix, power):
entry[...] = math.pow(entry, power)
return matrix

def run(matrix):
np.fill_diagonal(matrix, 1)
matrix = normalize(matrix)
for _ in range(ITERATION_COUNT):
matrix = normalize(inflate(expand(matrix, EXPANSION_POWER), INFLATION_POWER))
return matrix

If you were in the mood to improve it you could write something that’d check for convergence for you and terminate once you’d achieved a stable solution. You could also write a function to perform the cluster interpretation for you.

As you were.

Hi all,

Today I fancy a bit of a play around with stock prices – I recently took the plunge into the world of stocks & shares and have been getting more and more interested in the financial world as I’ve become more and more exposed to it through savings. I’m a bit sceptical as to being able to find anything ‘new’ or any real arbitrage opportunities – mostly because there’s a billion (trillion?) dollar industry built off of the back of stock trading. It attracts some really smart people with some really powerful gear and a whole lot of money to invest. However, there’s no harm in having a look around and seeing what interesting things we can do with the data.

R is well good but I want a bit more freedom with this little project and I’m missing Python. I find that with R, I spend a lot of my time getting data into the right format to be able to use the tools that already exist. With Python, if I’m silly enough to decide on a strange data structure then I can. I shouldn’t, but I can.

Ordinarily I like the Python & SQL combination and tend not to rely too heavily on the ‘Python analysis stack’ of Pandas/iPython/Scipy/Matplotlib, only pulling things in when necessary. I was going to follow the same pure Python & SQL route for this project until I found an awesome little feature of Pandas – in-built Google and Yahoo stock data integration. It’s not that much work to build this sort of thing yourself but why reinvent the wheel? 🙂

So – I guess we should start with some sort of question, shall we see if we can plot some of the big tech players (AMZ, GOOGL, FB e.t.c.) against the whole tech sector. Given the recent headlines in that area it should be interesting and at least give us some ideas about future work.

As a lazy person, I’m not necessarily inclined to manually go through a list of stock symbols and decide if they’re tech or even go through a list of tech stock and type them into a text file. A quick google shows me that there’s nothing (that I could find) in the way of a regularly updated text file of what I’m after but it shouldn’t be too difficult to coax Python into doing this for me – let’s start off with the NASDAQ site.

If you have a look, you’ll see it’s fairly regular in its URL structure and the URLs are easily craftable – there’s an annoying amount of pagination but you can’t have everything. Actually, hold the phone. You can download the list as a CSV – winner winner chicken dinner.

 Symbol Name LastSale MarketCap ADR TSO IPOyear Sector Industry Summary Quote

All I’m really after for now is the sector and symbol – market cap will prove useful but basically I think I can agree we’ve hit the jackpot!

Time for the Python:

import pandas as pd
from datetime import datetime
import numpy as np
mega_frame = [DataReader(company.strip(),  "yahoo", datetime(2014,1,1), datetime.now().date()) for company in company_information[company_information.Sector == 'Technology']['Symbol']]
symbol_list = [symbol for symbol in company_information[company_information.Sector == 'Technology']['Symbol']]

At this point we’ve got all the data since the start of the year on every tech stock listed on the NASDAQ, NYSE and AMEX and it’s taken us 6 lines. Note that the population of mega_frame takes a fairly long time. In retrospect, we should have filtered further

20 minutes later and I’m regretting my decision to get all of them.

Cancelled it and switched to the first 50 – will just prove concept first.

Right, now I’ve got a list containing data frames – one for each of the first 50 tech stocks. Let’s throw in a percentage change column and make sure all our data frames are of the same length to avoid problems at a later date:

mega_frame = [stock for stock in mega_frame if len(stock) == 79]
symbol_list = [symbol_list[index] for index in len(symbol_list) if len(mega_frame[i]) == 79]
for stock_index in range(len(mega_frame)):
mega_frame[stock_index]['perc_change'] = 100*((mega_frame[stock_index]['Close'] - mega_frame[stock_index]['Open'])/mega_frame[stock_index]['Open'])
## The modal value is 79 hence 79
percentage_change_list = [stock['perc_change'] for stock in mega_frame]

Now we’re going to create a correlation matrix out of those lists to see the most strongly correlated tech stocks over that time period (and in our subset). I’m also going to look at the negatively correlated stocks – you wouldn’t expect to see a strong negative correlation for two stocks in the same sector and region but it won’t hurt to look:

correlation_matrix = np.corrcoef(percentage_change_list)
## Correlation with yourself is no big deal
for i in range(np.shape(correlation_matrix)[0]):
for j in range(np.shape(correlation_matrix)[1]):
if i == j:
correlation_matrix[i][j] = 0
maximum_indices = np.argmax(correlation_matrix, axis=1)
minimum_indices = np.argmin(correlation_matrix, axis=1)
for index in range(np.shape(correlation_matrix)[0]):
print "Stock %s is best correlated with stock %s: %.3g" % (my_list[index], my_list[maximum_indices[index]], correlation_matrix[index][maximum_indices[index]])
print "Stock %s is worst correlated with stock %s: %.3g" % (my_list[index], my_list[minimum_indices[index]], correlation_matrix[index][minimum_indices[index]])

So there we have it (I’ll leave it to run over all the tech stocks overnight) – a fairly quick and simple way to find the most correlated tech stocks in America over a given time period.

Now this isn’t a particularly great way of doing this, as I said earlier, there are people who dedicate their lives to this. If the fancy takes me, I’ll have a look at a few of these (maximal spanning trees, stability of eigenvectors of correlation matrices e.t.c) and see what improvements we can make to our very simple model.