timeit.timeit('a = range(50);b=range(25, 75);set(a)&set(b)')
# takes about 5 seconds
timeit.timeit('a = range(50);b=range(25, 75);[x for x in a if x in b]')
# takes about 15 seconds
Anik's Tech Tutorials
Saturday, June 10, 2017
Snippet: List Comprehension vs Set AND operation
Sunday, April 30, 2017
Interesting Plot Snippet: Bitwise Or and Addition Operation
If the sum and the result of bitwise or operation of two numbers are the same, plotting these numbers in a scatter diagram produces a beautiful shape.
import matplotlib.pyplot as plt l1 = [] l2 = [] for i in xrange(1000): for j in xrange(1000): if (i + j) == (i | j): l1.append(i) l2.append(j) plt.scatter(l1,l2) plt.show()This is the figure that comes up:
Monday, January 9, 2017
Value Iteration in Gridworld, Reinforcement Learning
import operator
import copy
import math
class State:
def __init__(self, rowid, colid):
self.row = rowid
self.col = colid
def __str__(self):
return str(self.row) + " " + str(self.col)
def get_value_from_state(state):
return state_values[state.row][state.col]
def get_reward_from_state(state):
return reward_values[state.row][state.col]
def move_to_new_state(state, action):
if action == 'left': # state will only change columnwise
if state.col == 0 or (state.col == 2 and state.row == 1): # then we can't move left
return copy.deepcopy(state)
else:
return State(state.row, (state.col - 1))
if action == 'right': # state will only change columnwise
if state.col == 3 or (
state.col == 0 and state.row == 1): # can't move right when on the right boundary or at the first column
return copy.deepcopy(state)
else:
return State(state.row, (state.col + 1))
if action == 'up':
if state.row == 0 or (state.row == 2 and state.col == 1):
return copy.deepcopy(state)
else:
return State(state.row - 1, state.col)
if action == 'down':
if state.row == 2 or (state.row == 0 and state.col == 1):
return copy.deepcopy(state)
else:
return State(state.row + 1, state.col)
def moveWithProbabilites(state, action):
if action == 'left' or action == 'right':
thisstate = move_to_new_state(state, action)
upstate = move_to_new_state(state, 'up')
downstate = move_to_new_state(state, 'down')
return {thisstate: .8, upstate: .1, downstate: .1}
elif action == 'up' or action == 'down':
thisstate = move_to_new_state(state, action)
leftstate = move_to_new_state(state, 'left')
rightstate = move_to_new_state(state, 'right')
return {thisstate: .8, leftstate: .1, rightstate: .1}
state_values = [
[0, 0, 0, 1],
[0, -100, 0, -1], # -100 means we can't go this way
[0, 0, 0, 0]
]
reward_values = [ # in each state, the agent gets a reward of -.03
[-.03, -.03, -.03, -.03],
[-.03, -.03, -.03, -.03],
[-.03, -.03, -.03, -.03]
]
gamma = 1
states = [
State(0, 0),
State(0, 1),
State(0, 2),
State(1, 0),
State(1, 2),
State(2, 0),
State(2, 1),
State(2, 2),
State(2, 3),
]
actions = ['left', 'right', 'up', 'down']
prev = state_values[0][0]
if __name__ == "__main__":
while True:
for s in states:
mx = -9999999
for a in actions:
dic = moveWithProbabilites(s, a)
tempSum = get_reward_from_state(s) # to be added to the values later
tempSum += gamma * dic.items()[0][1] * get_value_from_state(dic.items()[0][0]) # .9 * .1 * v(s)
tempSum += gamma * dic.items()[1][1] * get_value_from_state(dic.items()[1][0])
tempSum += gamma * dic.items()[2][1] * get_value_from_state(dic.items()[2][0])
if tempSum > mx:
mx = tempSum
state_values[s.row][s.col] = mx
if abs(state_values[0][0] - prev) < .00003:
break
prev = state_values[0][0]
print state_values
Wednesday, December 28, 2016
Reinforcement Learning with Pygame Tutorial Part - 2
(Part 1: http://anixtech.blogspot.com/2016/12/reinforcement-learning-with-pygame-part.html)
I have created the second part of the tutorial as a video. It could be seen here:
https://www.youtube.com/watch?v=o5GiQkClbAY
The full code is also available in github:
https://github.com/hasanIqbalAnik/q-learning-python-example
I have created the second part of the tutorial as a video. It could be seen here:
https://www.youtube.com/watch?v=o5GiQkClbAY
The full code is also available in github:
https://github.com/hasanIqbalAnik/q-learning-python-example
Friday, December 23, 2016
Reinforcement Learning With Pygame Part - 1
Following is an example of a simple game which could be used to train agents.
import pygame as pg
from pygame.locals import *
import sys
import random
def new_rect_after_action(newrct, act):
if act == 'right':
if newrct.right + newrct.width > windowWidth:
return newrct
else:
return pg.Rect(newrct.left + newrct.width, newrct.top, newrct.width,
newrct.height) # Rect(left, top, width, height)
else: # action is left
if newrct.left - newrct.width < 0:
return newrct
else:
return pg.Rect(newrct.left - newrct.width, newrct.top, newrct.width,
newrct.height) # Rect(left, top, width, height)
def circle_falling(crclradius):
newx = random.randint(0 + crclradius, 200 - crclradius)
multiplier = random.randint(1, 4)
newx *= multiplier
return newx
windowWidth = 800
windowHeight = 400
FPS = 3 # frames per second setting
fpsClock = pg.time.Clock()
pg.init()
window = pg.display.set_mode((windowWidth, windowHeight)) # width, height
pg.display.set_caption('Catch the ball!')
# setup colors
RED = (255, 0, 0)
GREEN = (0, 255, 0)
WHITE = (255, 255, 255)
BLACK = (0, 0, 0)
# specify circle properties
crclCentreX = 50
crclCentreY = 50
crclRadius = 20
# specify rectangle properties
rctLeft = 400
rctTop = 350
rctWidth = 200
rctHeight = 50
rct = pg.Rect(rctLeft, rctTop, rctWidth, rctHeight) # Rect(left, top, width, height)
action = 'left'
score = 0
font = pg.font.Font(None, 30)
while True:
for event in pg.event.get():
if event.type == QUIT:
pg.quit()
sys.exit()
elif event.type == pg.KEYDOWN:
if event.key == pg.K_LEFT:
action = 'left'
rct = new_rect_after_action(rct, action)
elif event.key == pg.K_RIGHT:
action = 'right'
rct = new_rect_after_action(rct, action)
window.fill(BLACK)
# at this position, the rectangle should be here. else loses
if crclCentreY >= windowHeight - rctHeight - crclRadius:
if rct.left <= crclCentreX <= rct.right:
score += 1
else:
score -= 1
crclCentreX = circle_falling(crclRadius)
crclCentreY = 50
else:
crclCentreY += 20
pg.draw.circle(window, RED, (crclCentreX, crclCentreY),
crclRadius) # circle(Surface, color, pos(x, y), radius, width=0)
pg.draw.rect(window, GREEN, rct) # rect(Surface, color, Rect, width=0)
text = font.render('score: ' + str(score), True, (238, 58, 140))
window.blit(text, (windowWidth - 100, 10))
pg.display.update()
fpsClock.tick(FPS)
This will create a game window that looks like the following screenshot.
The green rectangle can be moved with left and right arrow of the keyboard. If the player can place it under the falling red circle, he gets a point, else loses one. That's all there is in this game.
In the second part of this tutorial, we will add intelligence and train an agent to play this game. Update: Second part is now available: http://anixtech.blogspot.com/2016/12/reinforcement-learning-with-pygame.html
Wednesday, July 27, 2016
Simplified Neural Network Backpropagation Example Python
This is an even simplified version of this fantastic tutorial: https://mattmazur.com/2015/03/17/a-step-by-step-backpropagation-example/
Let's consider a very simple neural network shown in the following image:
All we want is that given the input .1, the network would output the value .9.
We will be using $\frac{1}{1+e^{-x}}$ as our activation function. Initially both weight w1 and w2 are initialized to .5, though this can be any arbitrary value. However, different initialization shall converge to different final values.
First let's consider the net input in the hidden layer(neth). It's simply the input multiplied by w1. That is:(Here, $outh$ means the output of the hidden layer, $neto$ means the net input in the output layer, $outo$ is the output of the output layer. $E$ is the error. )
$neth = i1 * w1 = .1 * .5 = .05$
$outh = \frac{1}{1+e^{-neth}} = \frac{1}{1+e^{-.05}} = .51249$
$neto = outh * w2 = .51249 * .5 = .25624$
$outo = \frac{1}{1+e^{-neto}} = \frac{1}{1+e^{-.25624}} = .563711$
$E = \frac{1}{2}*(target - outo)^2 = \frac{1}{2}*(.9 - .563711)^2 = .056545$
This concludes our forward pass of the network. We've selected that particular error function for our convenience, as it's convex and differentiable. Now, we can see that our target value was $.9$ but we've got $.5637$ . We are going to update our weights $w1, w2$ so that the network produces that desired output. How would we do that is the reason for using the famous Backpropagation algorithm. The basic idea is, we will calculate how much responsibility should each of the weights take for producing that error. Then the weight would be modified accordingly and the next iteration would begin.
First let's see what happens in case of $w2$. The partial differentiation expression $\frac{\partial E}{\partial w2}$ says the rate of change in the error with respect to $w2$. As it can't be calculated directly, we would use the chain rule to find it's constituting parts.
$\frac{\partial E}{\partial w2} = \frac{\partial E}{\partial outo} * \frac{\partial outo}{\partial neto} * \frac{\partial neto}{\partial w2}$
Let's find out each of those parts separately:
$\frac{\partial E}{\partial outo} = \frac{\partial}{\partial outo}(\frac{1}{2}*(target - outo)^2) = \frac{1}{2} * 2 * ( target - outo) * (-1) $
$= (.9 - .563711) * (-1) = -.336289$
$\frac{\partial outo}{\partial neto} = \frac{\partial }{\partial neto} (\frac{1}{1+e^{-neto}}) = \frac{-1}{(1+e^{-neto})^2} * e^{-neto} * (-1) = \frac{e^{-neto}}{(1+e^{-neto})^2}$
$ = outo * (1-outo) = .563711 * ( 1- .563711) = .24594$
$\frac{\partial neto}{\partial w2} = \frac{\partial (w2*outh)}{\partial w2} = outh = .51249$
So finally we have found the responsibility of $w2$ for producing the error $E$:
$\frac{\partial E}{\partial w2} = -.33628 * .24594 * .51249 = -.042386$
Now let's turn our attention to $w1$, that is we want to calculate $\frac{\partial E}{\partial w1}$. Again, chain rule comes to our rescue.
$\frac{\partial E}{\partial w1} = \frac{\partial E}{\partial outo} * \frac{\partial outo}{\partial neto} *\frac{\partial neto}{\partial outh} * \frac{\partial outh}{\partial neth} * \frac{\partial neth}{\partial w1}$
We already have calculated $\frac{\partial E}{\partial outo}$ and $\frac{\partial outo}{\partial neto}$. Let's calculate the rest.
$\frac{\partial neto}{\partial outh} = \frac{\partial (w2*outh)}{\partial outh} = w2 = .5$
$\frac{\partial outh}{\partial neth} = outh * ( 1- outh) = .51249 * ( 1- .51249) = .24984$
$\frac{\partial neth}{\partial w1} = \frac{\partial (w1 * input)}{\partial w1} = input = .1 $
So,
$\frac{\partial E}{\partial w1} = -.336289 * .24594 * .5 * .24984 * .1 = -.001033$
Now it's time to update our initial guesses of $w1$ and $w2$. Let's say our learning rate $\eta = .9$
$w1 = w1 - (\eta * \frac{\partial E}{\partial w1}) = .5 + .9 * 001033 = 0.5009298654$
Similarly,
$w2 = w2 - ( \eta * \frac{\partial E}{\partial w2}) = .5 + .9 * 042386 = 0.538148$
We can continue this process until our network's output is very close to our target. The following is a python implementation of that process:
Sample output:
Let's consider a very simple neural network shown in the following image:
All we want is that given the input .1, the network would output the value .9.
We will be using $\frac{1}{1+e^{-x}}$ as our activation function. Initially both weight w1 and w2 are initialized to .5, though this can be any arbitrary value. However, different initialization shall converge to different final values.
First let's consider the net input in the hidden layer(neth). It's simply the input multiplied by w1. That is:(Here, $outh$ means the output of the hidden layer, $neto$ means the net input in the output layer, $outo$ is the output of the output layer. $E$ is the error. )
$neth = i1 * w1 = .1 * .5 = .05$
$outh = \frac{1}{1+e^{-neth}} = \frac{1}{1+e^{-.05}} = .51249$
$neto = outh * w2 = .51249 * .5 = .25624$
$outo = \frac{1}{1+e^{-neto}} = \frac{1}{1+e^{-.25624}} = .563711$
$E = \frac{1}{2}*(target - outo)^2 = \frac{1}{2}*(.9 - .563711)^2 = .056545$
This concludes our forward pass of the network. We've selected that particular error function for our convenience, as it's convex and differentiable. Now, we can see that our target value was $.9$ but we've got $.5637$ . We are going to update our weights $w1, w2$ so that the network produces that desired output. How would we do that is the reason for using the famous Backpropagation algorithm. The basic idea is, we will calculate how much responsibility should each of the weights take for producing that error. Then the weight would be modified accordingly and the next iteration would begin.
First let's see what happens in case of $w2$. The partial differentiation expression $\frac{\partial E}{\partial w2}$ says the rate of change in the error with respect to $w2$. As it can't be calculated directly, we would use the chain rule to find it's constituting parts.
$\frac{\partial E}{\partial w2} = \frac{\partial E}{\partial outo} * \frac{\partial outo}{\partial neto} * \frac{\partial neto}{\partial w2}$
Let's find out each of those parts separately:
$\frac{\partial E}{\partial outo} = \frac{\partial}{\partial outo}(\frac{1}{2}*(target - outo)^2) = \frac{1}{2} * 2 * ( target - outo) * (-1) $
$= (.9 - .563711) * (-1) = -.336289$
$\frac{\partial outo}{\partial neto} = \frac{\partial }{\partial neto} (\frac{1}{1+e^{-neto}}) = \frac{-1}{(1+e^{-neto})^2} * e^{-neto} * (-1) = \frac{e^{-neto}}{(1+e^{-neto})^2}$
$ = outo * (1-outo) = .563711 * ( 1- .563711) = .24594$
$\frac{\partial neto}{\partial w2} = \frac{\partial (w2*outh)}{\partial w2} = outh = .51249$
So finally we have found the responsibility of $w2$ for producing the error $E$:
$\frac{\partial E}{\partial w2} = -.33628 * .24594 * .51249 = -.042386$
Now let's turn our attention to $w1$, that is we want to calculate $\frac{\partial E}{\partial w1}$. Again, chain rule comes to our rescue.
$\frac{\partial E}{\partial w1} = \frac{\partial E}{\partial outo} * \frac{\partial outo}{\partial neto} *\frac{\partial neto}{\partial outh} * \frac{\partial outh}{\partial neth} * \frac{\partial neth}{\partial w1}$
We already have calculated $\frac{\partial E}{\partial outo}$ and $\frac{\partial outo}{\partial neto}$. Let's calculate the rest.
$\frac{\partial neto}{\partial outh} = \frac{\partial (w2*outh)}{\partial outh} = w2 = .5$
$\frac{\partial outh}{\partial neth} = outh * ( 1- outh) = .51249 * ( 1- .51249) = .24984$
$\frac{\partial neth}{\partial w1} = \frac{\partial (w1 * input)}{\partial w1} = input = .1 $
So,
$\frac{\partial E}{\partial w1} = -.336289 * .24594 * .5 * .24984 * .1 = -.001033$
Now it's time to update our initial guesses of $w1$ and $w2$. Let's say our learning rate $\eta = .9$
$w1 = w1 - (\eta * \frac{\partial E}{\partial w1}) = .5 + .9 * 001033 = 0.5009298654$
Similarly,
$w2 = w2 - ( \eta * \frac{\partial E}{\partial w2}) = .5 + .9 * 042386 = 0.538148$
We can continue this process until our network's output is very close to our target. The following is a python implementation of that process:
from __future__ import division
import math
ctr = 0
eta = .9
w1 = .5
w2 = .5
i1 = .1
target = .9
while True:
ctr += 1
neth = i1 * w1
outh = 1 / (1 + math.exp(-neth))
neto = w2 * outh
outo = 1 / (1 + math.exp(-neto))
e = .5 * math.pow((target - outo), 2)
print 'error now is: ', e
if e < .00005:
print '*' * 100
print 'target is: ', target
print 'output is now: ', outo
print 'w1 and w2 is: ', w1, ',', w2
print 'number of iterations: ', ctr
break
# update w2
de_by_douto = (-1) * (target - outo)
douto_by_dneto = outo * (1 - outo)
dneto_by_dw2 = outh
de_by_dw2 = de_by_douto * douto_by_dneto * dneto_by_dw2
# update w1
# de_by_dw1 = de_by_douto * douto_by_dneto * dneto_by_douth * douth_by_dneth * dneth_by_dw1
dneto_by_douth = w2
de_by_outh = de_by_douto * douto_by_dneto * dneto_by_douth
douth_by_dneth = outh * (1 - outh)
dneth_by_dw1 = i1
de_by_dw1 = de_by_outh * douth_by_dneth * dneth_by_dw1
# updated w1 and w2
w1 = w1 - (eta * de_by_dw1)
w2 = w2 - (eta * de_by_dw2)
Sample output:
Tuesday, July 26, 2016
Who Liked Your Last N Number of Facebook Posts Most
Previously it was possible to find out this information for any friend's profile too. But Facebook has recently revoked that access and apps now need to have this access directly from that friend. However, you can find out this info about yourself in the following way:
1. Go to this link: https://developers.facebook.com/tools/explorer/.
2. Click on 'Get Token' > 'Get User Access Token' > check user_posts > 'Get Access token'.
3. This would ask you for the permission to read your timeline. Click on 'Continue as your-name'.
4. You would get an access token like the following image. Copy the whole string.
5. Replace the 'your-token' part in the following code with that string.
7. Run that program, it might take a while depending on the response time of facebook server. The output would contain the list of likes in a sorted fashion. With the most liker at the top.
Sample output:
1. Go to this link: https://developers.facebook.com/tools/explorer/.
2. Click on 'Get Token' > 'Get User Access Token' > check user_posts > 'Get Access token'.
3. This would ask you for the permission to read your timeline. Click on 'Continue as your-name'.
4. You would get an access token like the following image. Copy the whole string.
5. Replace the 'your-token' part in the following code with that string.
from __future__ import division
import operator
import requests
from facepy import GraphAPI
large_dic = {}
token = 'your-token'
graph = GraphAPI(token)
n = 50 # number of posts in the history that we want to consider
ctr = 0
pcount = 0
res = graph.get('me/feed')
def update_progress(progress):
print '\r{0}% [{1}]'.format(int(progress), '#'*int(progress/10)),
def merge_two_dicts(dict1, dict2):
""" merges and sums up dict1 with the values from dict2. if key2 from dict2 is present in dict1, then dict1[key2] = dict1[key2] + dict2[key2]
else it appends key2 with the value as it is
example: if
d1 = {'a': 1, 'b': 1, 'c': 1} and
d2 = {'a': 4, 'd': 1}
then it returns {'a': 5, 'c': 1, 'b': 1, 'd': 1}
:param dict1, dict2
:return merged dict1 and dict2
"""
for key2 in dict2:
if key2 in dict1:
dict1[key2] = dict1[key2] + dict2[key2]
else:
dict1[key2] = dict2[key2]
return dict1
def get_likes_list(post_id):
"""given a post id, this method queries the graphapi to get details of the post, gets the likes list
and adds them into a dictionary by using the person's name as the key
:param post_id
:return dictionary containing likers
"""
global graph
global pcount
d = {} # to hold the liker's name with a count of 1
likes_obj = graph.get(post_id + '/likes')
while True:
try:
for liker in likes_obj['data']:
d[liker['name']] = 1
likes_obj = requests.get(likes_obj['paging']['next']).json()
except KeyError:
pcount += 1
update_progress((pcount/n)*100)
break
return d
for item in res['data']:
merge_two_dicts(large_dic, get_likes_list(item['id']))
while True:
if ctr == (n - 25) / 25: # as 25 is the default limit of /me/feed, to get the last n posts, set the value of ctr to (n-25)/25
break
try:
res = requests.get(res['paging']['next']).json()
for item in res['data']:
merge_two_dicts(large_dic, get_likes_list(item['id']))
ctr += 1
except KeyError:
print 'end of calculating posts'
sorted_x = sorted(large_dic.items(), key=operator.itemgetter(1))
print '\n'
print '*'*100
print 'total posts considered', pcount
print 'total likes generated', sum(large_dic.values())
print 'average likes per post', sum(large_dic.values()) / pcount
print '*'*100
for item in reversed(sorted_x):
print item[0], 'liked your posts', item[1], 'times'
6. Change the value of n in the program to your preference, this is the number of posts that we want to consider while counting the likes.7. Run that program, it might take a while depending on the response time of facebook server. The output would contain the list of likes in a sorted fashion. With the most liker at the top.
Sample output:
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