Develop Reference

Python dictionary keys seem to vanish before I delete them

I'm a newbie programmer working on an idea for a small game. I wanted my play space to be a grid for various reasons. Without a lot of good reason, I decided to create a class of GridSquare objects, each object having properties like size, an index to describe what (x,y) coordinates they represented, and some flags to determine if the grid squares were on land or empty space, for example. My grid is a dictionary of these objects, where each GridSquare is a key. The values in the dictionary are going to be various objects in the place space, so that I can easily look up which objects are on each grid square.
Just describing this I feel like a complete lunatic. Please bear in mind that I've only been at this a week.
My problem appears when I try to change the GridSquare objects. For example, I want to use a list to generate the land on each level. So I iterate over the list, and for each value I look through my grid squares using a for loop until I find one with the right index, and flip the GridSquare.land property. But I found that this caused a runtime error, since I was changing keys in a dictionary I was looping through. OK.
Now what I'm trying to do is to create a list of the keys I want to change. For each item in my level-generating list, I go through all the GridSquares in my grid dictionary until I find the one with the index I'm looking for, then I append that GridSquare to a list of old GridSquares that need updating. I then make another copy of the GridSquare, with some properties changed, in a list of altered GridSquares. Finally, I delete any keys from my grid dictionary which match my list of "old" GridSquares, and then add all of the altered ones into my grid dictionary.
The problem is that when I delete keys from my grid dictionary which match my list of "old" keys, I run into keyerrors. I can't understand what is happening to my keys before I can delete them. Using try/except, I can see that it's only a small number of the keys, which seems to vary kind of arbitrarily when I change parts of my code.
I would appreciate any insight into this behaviour.
Here is code for anyone still reading:
aspect_ratio = (4, 3)
screen_size = (1280, 720)
#defining a class of objects called GridSquares
class GridSquare:
def __init__(self, x, y):
self.index = (x, y)
self.land = 0
#creates a dictionary of grid squares which I hope will function as a grid......
grid = {}
for x_index in range(1, (aspect_ratio[0] + 1)):
for y_index in range (1, (aspect_ratio[1] + 1)):
new_square = GridSquare(x_index, y_index)
grid[new_square] = None
#these are lists to hold changes I need to make to the dictionary of grid squares
grid_changes = []
old_gridsquares = []
#this unweildly list is meant to be used to generate a level. Numbers represent land, spaces are empty space.
for number_no, number in enumerate(["1", "1", "1", "1",
" ", " ", " ", " ",
"1", "1", "1", "1"]):
#makes grid squares land if they are designated as such in the list
for gridsquare in grid.keys():
#this if statement is meant to convert each letter's position in the list into an index like the grid squares have.
if gridsquare.index == ((number_no + 1) % (aspect_ratio[0]), ((number_no + 1) // (aspect_ratio[0] + 1)) + 1):
#create a list of squares that need to be updated, and a list of squares to be deleted
old_gridsquares.append(gridsquare)
flagged_gridsquare = GridSquare((number_no + 1) % (aspect_ratio[0]), ((number_no + 1) // (aspect_ratio[0] + 1)) + 1)
flagged_gridsquare.land = 1
#this part is meant to set the flag for the gridsquare that indicates if it is on the far side or the near side,
#if it is land
if number == "1":
flagged_gridsquare.near = 1
grid_changes.append(flagged_gridsquare)
#deletes from grid any items with a key that matches the old squares, and adds updated versions.
for old_gridsquare in old_gridsquares:
try:
del grid[old_gridsquare]
except:
print(old_gridsquare.index)
print(old_gridsquare.land)
for grid_change in grid_changes:
grid[grid_change] = None

Monte Carlo simulation of Birthday paradox in python 3

The birthday paradox is that everyone has equal probability of having a birthday on any given of 365 days. We start adding people in a room. What is the probability that 2 people have birthdays on same day as a function of number of people in the room? The code I wrote is as follows:
import numpy as np
import matplotlib.pyplot as plt
x=[0]
y=[0]
for j in range(1000):
if j!=0:
freq = []
L1 = list(np.random.randint(low = 1, high=366, size = j))
result = list((i, L1.count(i)) for i in L1)
for a_tuple in result:
freq.append(a_tuple[1])
print(freq)
rep = j - freq.count(1)
prob = rep/j
y = y + [prob]
x = x + [j]
print(prob)
plt.plot(x,y)
Here, in L1 = list(np.random.randint(low = 1, high=366, size = j)) I select the day on which someone would have a birthday and in result = list((i, L1.count(i)) for i in L1) I calculate the frequency of birthdays on each day. The entire thing is looped over to account for increasing number of people.
In the following for loop, I isolate the unique events and find repetitions and store the value in rep.
Next I calculated the probability as fraction of people sharing birthdays and plotted them as a function of number.
However, the question requires me to find the probability of just one shared birthday. How do I calculate that? I think I have to loop this entire thing for number of trials but that just gives an accurate solution with less variations of the same program. Currently my program gives fraction of people having shared birthdays I think.
Birthday problem Wikipedia for better reference

NOTE
I assume that when n persons have been in the room, they are all thrown out of the room and then n+1 persons enter the room.
========================================
I would think of it this way;
First, set probs = [0]*365. Now, say 2 persons get in the room - we then write their birthdays onto a piece of paper and check, if those two dates are equal. If they are, we increase probs[2] by 1 (yes, theres some indexes that we don't need, and Python is 0-indexed etc. but to keep it simple).
Now do the same for 3 persons, for 4 persons, for 5 persons ... all the way up to 365.
Your array might look something like probs==[0,0,0,0,0,1,0,1,1,0,1,1,1,1,0,1....].
You can now start over from 2 persons (still keeping the same array as before i.e don't create a new one with 0's!), then 3 persons etc. and start over 1000 times. Your array might look like
probs==[0,0,2,0,4,1,5,2,9,12,10,17....,967,998]
If you divide that array by 1000 (elementwise) you now have your simulated probability as a function of n persons.
import numpy as np
import matplotlib.pyplot as plt
N_TOTAL_PERS= 366
N_SIM = 10000 #number of simulations
counts = np.zeros(N_TOTAL_PERS)
for _ in range(N_SIM):
for n in range(2,N_TOTAL_PERS):
b_days = np.random.randint(1,366,size=n) #Get each persons birth-day
counts [n] += len(b_days) != len(set(b_days)) #Increment if some birthdays are equal
total_probs = counts/N_SIM #convert to probabilities
total_probs[70] #Get the probability when 70 persons are together (0.9988)
plt.plot(range(N_TOTAL_PERS),total_probs)
which generates a plot that looks like

You should run multiple experiments for different number of people in the room. Note that for N_people > 365, the probability should compute equal to 1.
Refactoring your code, and changing the logic a bit, I came up with the following:
import numpy as np
import matplotlib.pyplot as plt
def random_birthdays(n_people):
return list(np.random.randint(low=1, high=366, size=n_people))
def check_random_room(n_people):
"""
Generates a random sample of `n_people` and checks if at least two of them
have the same birthday
"""
birthdays = random_birthdays(n_people)
return len(birthdays) != len(set(birthdays))
def estimate_probability(n_people, n_experiments):
results = [check_random_room(n_people) for _ in range(n_experiments)]
return sum(results)/n_experiments
N_EXPERIMENTS = 1000
x = list(range(1, 400))
y = [estimate_probability(x_i, N_EXPERIMENTS) for x_i in x]
plt.plot(x, y)
plt.show()

Creating a list of dictionaries where each dictionary contains another dictionary as values

Im writing an algorithm in Python which plays this game.
The current state of the board of tiles in the game is a dictionary in the form of:
{
<tile_id>: {
'counters': <number of counters on tile or None>,
'player': <player id of the player who holds the tile or None>,
'neighbours': <list of ids of neighbouring tile>
},
...
}
I have another dictionary which stores all of my tiles which are 'full' (i.e. a tile which has one less counter than its number of neighbours and where the player is me) This dictionary, full_tiles, is in the same form as the board dictionary above.
I am now trying to create a list, chains, where each element in the list is a dictionary of my full tiles that are neighbouring at least one other full tile (i.e a chain of full tiles). So this will be a list of all my seperate chains on the board.
Here is my code so far:
for tile_id, tile in full_tiles.items(): #iterates through all full tiles
current_tile = {tile_id : tile} #temporarily stores current tile
if not chains: #if chains list is empty
chains.append(current_tile) #begin list
else: #if list is not empty
for index, chain in enumerate(chains): #iterate though list of chains
if not (current_tile in chain): #if current tile is not in current chain
for tile_id2, tile2 in chain.items(): #iterate through tiles in current chain
for neighbour in tile2["neighbours"]: #iterate through each neighbour of current tile
#neighbour is in the form of tile_id
if neighbour in chain: #if current tile's neighbour is in chain
chain[tile_id] = tile #add tile to chain
It is very difficult for me to test and debug my code and check if it is working correctly as the code can only be run in an application that simulates the game. As you can see, there is quite a lot going on in this block of code with all of the nested loops which are difficult to follow. I cannot seem to think straight at the minute and so I cannot determine if this mess, in all honesty, will function as I hope.
While I am writing this, I have just realised that - on line 7 of this code - I am only checking if the current tile is not in the current chain and so there will be intersecting chains which, of course, will be a mess. Instead of this, I need to first check if the current tile is in not in any of the chains, not just one.
Apart from this error, will my code achieve what I am attempting? Or can you recommend a simpler, neater way to do it? (There has to be!)
Also, let me know if I have not given enough information on how the code is run or if I need to explain anything further, such as the contents of the board dictionary.
Thank you for any help in advance.
EDIT: Unfortunately, I was under a time constraint to complete this project, and as it was my first time ever working with Python, I currently lack the knowledge in the language to optimise my solution using the sources given below. Here is my final extremely ugly and messy solution to this problem which, in the end, worked fine and wasn't terribly inefficient given the small data set that the code works on.
for x in range(0, len(my_hexplode_chains) - 1):
match_found = False
for y in range(x + 1, len(my_hexplode_chains)):
for tile_id_x, tile_x in my_hexplode_chains[x].items(): #compare each chain in list
for tile_id_y, tile_y in my_hexplode_chains[y].items(): #to every other chain
for neighbour in tile_x["neighbours"]: #if tiles in different lists
if neighbour == tile_id_y: #are neighbours
match_found = True
my_hexplode_chains[x].update(my_hexplode_chains[y]) #append one chain to the other
del my_hexplode_chains[y] #delete appended chain
if match_found: #continue loop at next chain
break #very ugly way to do this
if match_found:
break
if match_found:
break
if match_found:
break

How about this optimization?
def find_match (my_hexplode_chains):
x = 0
len_chain = len(my_hexplode_chains)
while x <= len_chain:
y = x + 1
for tile_id_x, tile_x in my_hexplode_chains[x].items():
for tile_id_y, tile_y in my_hexplode_chains[y].items():
if tile_id_y in tile_x["neighbours"]:
my_hexplode_chains[x].update(my_hexplode_chains[y])
del my_hexplode_chains[y]
return True
x += 1
return False
You could pass this function after each move in your game and trace the output.

Python Data Structure Selection

Let's say I have a list of soccer players. For now, I only have four players. [Messi, Iniesta, Xavi, Neymar]. More players will be added later on. I want to keep track of the number of times these soccer players pass to each other during the course of a game. To keep track of the passes, I believe I'll need a data structure similar to this
Messi = {Iniesta: 4, Xavi: 5 , Neymar: 8}
Iniesta = {Messi: 4, Xavi: 10 , Neymar: 5}
Xavi = {Messi: 5, Iniesta: 10 , Neymar: 6}
Neymar = {Messi: 8, Iniesta: 5 , Xavi: 6}
Am I right to use a dictionary? If not, what data structure would be better suited? If yes, how do I approach this using a dictionary though? How do I address the issue of new players being included from time to time, and creating a dictionary for them as well.
As an example, If I get the first element in the list, List(i) in the first iteration is Messi, how do i use the value stored in it to create a dictionary with the name Messi. That is how do i get the line below.
Messi = [Iniesta: 4, Xavi: 5 , Neymar: 8]
It was suggested I try something like this
my_dynamic_vars = dict()
string = 'someString'
my_dynamic_vars.update({string: dict()})
Python and programming newbie here. Learning with experience as I go along. Thanks in advance for any help.

This is a fun question, and perhaps a good situation where something like a graph might be useful. You could implement a graph in python by simply using a dictionary whose keys are the names of the players and whose values are lists players that have been passed the ball.
passes = {
'Messi' : ['Iniesta', 'Xavi','Neymar', 'Xavi', 'Xavi'],
'Iniesta' : ['Messi','Xavi', 'Neymar','Messi', 'Xavi'],
'Xavi' : ['Messi','Neymar','Messi','Neymar'],
'Neymar' : ['Iniesta', 'Xavi','Iniesta', 'Xavi'],
}
To get the number of passes by any one player:
len(passes['Messi'])
To add a new pass to a particular player:
passes['Messi'].append('Xavi')
To count the number of times Messi passed to Xavi
passes['Messi'].count('Xavi')
To add a new player, just add him the first time he makes a pass
passes['Pele'] = ['Messi']
Now, he's also ready to have more passes 'appended' to him
passes['Pele'].append['Xavi']
What's great about this graph-like data structure is that not only do you have the number of passes preserved, but you also have information about each pass preserved (from Messi to Iniesta)
And here is a super bare-bones implementation of some functions which capture this behavior (I think a beginner should be able to grasp this stuff, let me know if anything below is a bit too confusing)
passes = {}
def new_pass(player1, player2):
# if p1 has no passes, create a new entry in the dict, else append to existing
if player1 not in passes:
passes[player1] = [player2]
else:
passes[player1].append(player2)
def total_passes(player1):
# if p1 has any passes, return the total number; otherewise return 0
total = len(passes[player1]) if player1 in passes else 0
return total
def total_passes_from_p1_to_p2(player1, player2):
# if p1 has any passes, count number of passes to player 2; otherwise return 0
total = passes[player1].count(player2) if player1 in passes else 0
return total
Ideally, you would be saving passes in some database that you could continuously update, but even without a database, you can add the following code and run it to get the idea:
# add some new passes!
new_pass('Messi', 'Xavi')
new_pass('Xavi', 'Iniesta')
new_pass('Iniesta', 'Messi')
new_pass('Messi', 'Iniesta')
new_pass('Iniesta', 'Messi')
# let's see where we currently stand
print total_passes('Messi')
print total_passes('Iniesta')
print total_passes_from_p1_to_p2('Messi', 'Xavi')
Hopefully you find this helpful; here's some more on python implementation of graphs from the python docs (this was a fun answer to write up, thanks!)

I suggest you construct a two dimensional square array. The array should have dimensions N x N. Each index represents a player. So the value at passes[i][j] is the number of times the player i passed to player j. The value passes[i][i] is always zero because a player can't pass to themselves
Here is an example.
players = ['Charles','Meow','Rebecca']
players = dict( zip(players,range(len(players)) ) )
rplayers = dict(zip(range(len(players)),players.keys()))
passes = []
for i in range(len(players)):
passes.append([ 0 for i in range(len(players))])
def pass_to(f,t):
passes[players[f]][players[t]] += 1
pass_to('Charles','Rebecca')
pass_to('Rebecca','Meow')
pass_to('Charles','Rebecca')
def showPasses():
for i in range(len(players)):
for j in range(len(players)):
print("%s passed to %s %d times" % ( rplayers[i],rplayers[j],passes[i][j],))
showPasses()

Python Set Birthday

So I am trying to make a program that creates the probability of a bunch of people in a room to have the same birthday... I can't figure out how to create the function. Here is what I have so far
def birthday():
mySet = set()
x = 1
for item in mySet:
if item in mySet:
return x
else:
mySet().append() # don't know what to do here.
Edit:
Alright so what I am trying to accomplish is to make a function using a set that stores birthdays using numbers 1 through 365...For example, if you randomly pick a room with 30 people in it, they may not have the same birthday. Although, if you have twins in the same room, you only need 2 people
in the room to have the same birthday. So eventually I want a parameter that tests this function several times and averages it all up. Unfortunately I can't figure out how to make this. I want x to be a counter of how many people are in the room and when there is a match the loop stops and it stops. I also don't know what to append to.

Is there a reason why you're trying to simulate this rather than using the closed form solution to this problem? There's a pretty decent approximation that's fast and easy to code:
import math
def closed_form_approx_birthday_collision_probability(num_people):
return 1 - math.exp(-num_people * (num_people - 1) / (2 * 365.0))
You could also implement an very good "exact" solution (in quotes because some fidelity is lost when converting to float):
import operator
import functools
import fractions
def slow_fac(n):
return functools.reduce(operator.mul, range(2, n+1), 1)
def closed_form_exact_birthday_collision_probability(num_people):
p_no_collision = fractions.Fraction(slow_fac(365), 365 ** num_people * slow_fac(365 - num_people))
return float(1 - p_no_collision)
To do a simulation, you'd do something like this. I'm using a list rather than a set because the number of possibilities is small and this avoids some extra work that using a set would do:
import random
def birthday_collision_simulate_once(num_people):
s = [False] * 365
for _ in range(num_people):
birthday = random.randint(0, 364)
if s[birthday]:
return True
else:
s[birthday] = True
return False
def birthday_collision_simulation(num_people, runs):
collisions = 0
for _ in range(runs):
if birthday_collision_simulate_once(num_people):
collisions += 1
return collisions / float(runs)
The numbers I get from the simulation and the closed form solution look similar to the table at http://en.wikipedia.org/wiki/Birthday_problem
>>> closed_form_approx_birthday_collision_probability(20)
0.40580512747932584
>>> closed_form_exact_birthday_collision_probability(20)
0.41143838358058
>>> birthday_collision_simulation(20, 100000)
0.41108
Of course the simulation with that many runs is closer to the actual 41.1%, it's much slower to calculate. I'd choose one of the closed form solutions, depending on how accurate it needs to be.