If you don't have Python 3.4 or later you'll have to make your own homegrown Enum class, or use something like def enum(.sequential,.named): enums = dict(zip(sequential, range(1, len(sequential) + 1)),.named) return type('Enum', , enums) Reel = enum('CHERRY', 'LEMON', 'ORANGE', 'PLUM', 'BELL', 'BAR', 'SEVEN').
This is the fundamental concept of Upper Confidence Bound algorithm. Now let's see what steps are involved to perform this algorithm on our dataset: Implementing the Multi-Armed Bandit Problem in Python. We will implement the whole algorithm in Python. First of all, we need to import some essential libraries.

SlotMachine

import random

print(''Welcome to the Slot Machine Simulator

You'll start with $50. You'll be asked if you want to play.

Answer with yes/no. you can also use y/n

No case sensitivity in your answer.

For example you can answer with YEs, yEs, Y, nO, N.

To win you must get one of the following combinations:

BARtBARtBARttpayst$250

BELLtBELLtBELL/BARtpayst$20

PLUMtPLUMtPLUM/BARtpayst$14

ORANGEtORANGEtORANGE/BARtpayst$10

CHERRYtCHERRYtCHERRYttpayst$7

CHERRYtCHERRYt -ttpayst$5

CHERRYt -t -ttpayst$2

'')

#Constants:

INIT_STAKE = 50

ITEMS = ['CHERRY', 'LEMON', 'ORANGE', 'PLUM', 'BELL', 'BAR']

firstWheel = None

secondWheel = None

thirdWheel = None

stake = INIT_STAKE

def play():

global stake, firstWheel, secondWheel, thirdWheel

playQuestion = askPlayer()

while(stake != 0 and playQuestion True):

firstWheel = spinWheel()

secondWheel = spinWheel()

thirdWheel = spinWheel()

printScore()

playQuestion = askPlayer()

def askPlayer():

Asks the player if he wants to play again.

expecting from the user to answer with yes, y, no or n

No case sensitivity in the answer. yes, YeS, y, y, nO . . . all works

global stake

while(True):

answer = input('You have $' + str(stake) + '. Would you like to play? ')

answer = answer.lower()

if(answer 'yes' or answer 'y'):

return True

elif(answer 'no' or answer 'n'):

print('You ended the game with $' + str(stake) + ' in your hand.')

return False

else:

print('wrong input!')

def spinWheel():

returns a random item from the wheel

randomNumber = random.randint(0, 5)

return ITEMS[randomNumber]

def printScore():

prints the current score

global stake, firstWheel, secondWheel, thirdWheel

if((firstWheel 'CHERRY') and (secondWheel != 'CHERRY')):

win = 2

elif((firstWheel 'CHERRY') and (secondWheel 'CHERRY') and (thirdWheel != 'CHERRY')):

win = 5

elif((firstWheel 'CHERRY') and (secondWheel 'CHERRY') and (thirdWheel 'CHERRY')):

win = 7

elif((firstWheel 'ORANGE') and (secondWheel 'ORANGE') and ((thirdWheel 'ORANGE') or (thirdWheel 'BAR'))):

win = 10

elif((firstWheel 'PLUM') and (secondWheel 'PLUM') and ((thirdWheel 'PLUM') or (thirdWheel 'BAR'))):

win = 14

elif((firstWheel 'BELL') and (secondWheel 'BELL') and ((thirdWheel 'BELL') or (thirdWheel 'BAR'))):

win = 20

elif((firstWheel 'BAR') and (secondWheel 'BAR') and (thirdWheel 'BAR')):

win = 250

else:

win = -1

stake += win

if(win > 0):

print(firstWheel + 't' + secondWheel + 't' + thirdWheel + ' -- You win $' + str(win))

else:

print(firstWheel + 't' + secondWheel + 't' + thirdWheel + ' -- You lose')

play()

commented Dec 14, 2015

Ln t = natural logarithm of t. The idea behind this algorithm is that the value of the square root indicates the uncertainty of the value of action a. If an action is not chosen, t increases Nt (a) does not which will make the uncertainty of that action to increase, increasing its chances of being chosen (exploration).

Instead of;
if(answer 'yes' or answer 'y'):

Do;
if answer.lower() in ['yes',y']

commented Jun 2, 2017

I run it on python 2 ,it's need to modify the 43 line (input -> raw_input)

Reinforcement learning has yet to reach the hype levels of its Supervised and Unsupervised learning cousins. Albeit, it is an exceptionally powerful approach aimed to solve a variety of problems in a completely different way. To learn reinforcement learning, it is best to start from its building blocks and progress from there. In this article, we will go through a Multi-Armed Bandit using Python to solve a business problem.

What are Multi-Armed Bandits?

Imagine you have 3 slot machines and you are given a set of rounds. In each round, you have to choose 1 slot machine and pull its arm and receive the reward (or no at all) from that slot machine. Then, you do it again, and again… Eventually, you figure out which slot machine gives you the most rewards and keep pulling it for each round.

Machine Learning Algorithms Python Examples

This is the Multi-Armed Bandit problem also known as the k-Armed Bandit problem. K is the number of slot machines available and your reinforcement learning algorithm needs to figure out which slot machine to pull in order to maximize its rewards.

Multi-Armed bandits are a classical reinforcement learning example and it clearly exemplifies a well-known dilemma in reinforcement learning called the exploration-exploitation trade-off dilemma.

Keep in mind these important concepts. I will cover them in more detail in posts coming up. For now, let’s apply Multi-Armed bandits to a business problem.

Our Business Problem

We will apply the Multi-Armed Bandit method to a theoretical marketing business problem.

You are the Chief Marketing Officer promoting a new product and need to get the word out. To do this, you will be texting your target customers a message. Upon receiving the message, the customer will react one way or the other and generate a reward that you have access to.

You have 4 text messages to send but are not sure which one will work best.

Slot Machine Algorithm Python Compiler

Upper-Confidence-Bound Action Selection Formula

How do you choose the best action? Going back to my example in which you had to choose a slot machine at each round. Particularly at the beginning, you had to test them out correct? Either stick to one from the start (not the best option) or try a couple of them to find out the best one.

A common strategy is called the Upper-Confidence-Bound Action selection, in short, UCB.

If you are an