3
\$\begingroup\$

I wrote a script that gives a "generalized" function of an arbitrary dice. Is this any good? How can it improve so to make it cleaner?

import random
def roll(die):
    number = random.randint(0,len(die)-1)
    b = die[number]
    return b

Die1 = [1,2,3,4]
Die2 = [1,2,3,4,5,6] #num lists

def inptchacc():
    ending_conditions = ['stop','Stop','quit','Quit']
    end = False
    inpu = input('what number would you like to add to the new face of the Die? (to end the die please type "stop or Stop or quit or Quit to finish the DIE" )')
    while end == False:
        if not(inpu in ending_conditions):
            try:
                retr = int(inpu)
                return retr
            except:
                string = 'invalid input please try again'
                inpu = input('invalid input please try again ')
        else:
            stop = 'stop'
            return stop

def deeper(IDie):
    list = []
    Adding = True
    while Adding:
        print('The Die ' + IDie + ' is currently ' + str(list) )
        toadd = inptchacc()
        if toadd != 'stop':
            list.append(toadd)
        else:
            Adding = False
    return list


def chance_overlap(Die1,Die2):
    highnumber = 1000
    counter = 0
    for n in range(highnumber):
        x = roll(Die1)
        y = roll(Die2)
        if x == y:
            counter += 1
    chance = counter/highnumber
    return chance

chance = chance_overlap(Die1,Die2)

print(chance)
Doing = True
while Doing:
    try:
        IDie1 = deeper('IDie1')
        IDie2 = deeper('IDie1')
        chance2 = chance_overlap(IDie1,IDie2)
        Doing = False
    except:
        print ('incompatible options selected returning....... ')

print(chance2) 
\$\endgroup\$
  • \$\begingroup\$ What is the overall function of this, just rolling some arbitrary dice? It looks awfully complicated for such a simple task if that's it. \$\endgroup\$ – Mast Jan 7 at 12:15
  • \$\begingroup\$ " The assignment deals with a game in which two players both have a set of dice. In the game, each player casts his/her dice and sums the outcomes of the dice. The player with the highest sum wins. If the sums are equal, it's a draw. Usually the dice within one set are equal, but different from the kind of dice in the other set." This is the assignment \$\endgroup\$ – user183738 Jan 7 at 12:22
  • 1
    \$\begingroup\$ As we all want to make our code more efficient or improve it in one way or another, try to write a title that summarizes what your code does, not what you want to get out of a review. Please see How to get the best value out of Code Review - Asking Questions for guidance on writing good question titles. \$\endgroup\$ – BCdotWEB Jan 7 at 12:24
  • 3
    \$\begingroup\$ @Beginner-Coder123: You should add that description to the question body. \$\endgroup\$ – Graipher Jan 7 at 13:28
  • 2
    \$\begingroup\$ Please do not vandalize your posts. By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, any vandalism will be reverted. \$\endgroup\$ – Glorfindel Jan 8 at 19:07
9
\$\begingroup\$

Since you need to roll many times in the chance_overlap function, you might want to optimize making n rolls, using random.choices (Python 3.6+):

from itertools import groupby

def roll(die, n=1):
    if n == 1:
        return random.choice(die)
    return random.choices(die, k=n)

def all_equal(iterable):
    "Returns True if all the elements are equal to each other"
    g = groupby(iterable)
    return next(g, True) and not next(g, False)

def overlap_chance(*dice, n=1000):
    rolls = [roll(die, n) for die in dice]
    equal_rolls = sum(all_equal(roll) for roll in zip(*rolls))
    return equal_rolls / n

Here I chose to include it in your roll function, which is nice because you only have on function, but you do have different return types depending on the value of k, which is not so nice. If you want to you can make it into two separate functions instead.

I made chance_overlap take a variable number of dice so it even works for more than two (and also for one, which is a bit boring).

In addition, I followed Python's official style-guide, PEP8 for variable names (lower_case).

The all_equal function is directly taken from the itertools recipes.


Using a Monte-Carlo method to determine the chance for the dice to get the same values is fine, but you could just use plain old math.

Each distinct value \$j\$ on each die \$i\$ has probability \$p^i_j = n^i_j / k_i\$, where \$k_i\$ is the number of faces of die \$i\$ and \$n^i_j\$ the number of times the value \$j\$ appears on that die. Then the chance to have an overlap is simply given by

$$ P(overlap) = \sum\limits_j \prod\limits_i p^i_j = \sum\limits_j \prod\limits_i n^i_j / k_i, $$

where \$i\$ goes over all dice and \$j\$ over all values present on any dice (with \$n^i_j = 0\$ if value \$j\$ does not appear on die \$i\$).

In other words dice rolls are independent events and e.g. the chance to get two heads or two tails with a fair coin (dice = [["H", "T"], ["H", "T"]]) are \$P(HH \vee TT) = 0.5\cdot0.5 + 0.5\cdot0.5 = 0.5\$.

from collections import Counter
from functools import reduce
from itertools import chain
from operator import mul

def overlap_chance_real(*dice):
    all_values = set(chain.from_iterable(dice))
    counters = [Counter(die) for die in dice]
    lengths = [len(die) for die in dice]
    return sum(reduce(mul, [counter[val] / length
                            for counter, length in zip(counters, lengths)])
               for val in all_values)

The nice thing about this is that we don't need to worry if not all dice have the same values, since Counter objects return a count of zero for non-existing keys.

For dice = [[1,2,3], [1,1,1]] it returns the correct (and precise) value of 0.3333333333333333. and 0.5 for dice = [["H", "T"], ["H", "T"]].

The execution time of the first function (overlap_chance) increases linearly with the number of dice (all six sided) and it is in general slower than the second function (overlap_chance_real):

enter image description here

With two dice with increasing number of faces the first function is slower but basically constant in time while the second function is faster but the time increases with the number of faces. This plot also contains your function (since it can only deal with two dice it was not included in the previous plot), which is slower than both:

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ What did you use to make those graphs? \$\endgroup\$ – esote Jan 7 at 18:24
  • 1
    \$\begingroup\$ @esote timeit and matplotlib, as detailed in this question (and it's answer). \$\endgroup\$ – Graipher Jan 7 at 18:38
  • \$\begingroup\$ I tried to implement this, but I broke the code. \$\endgroup\$ – user183738 Jan 8 at 16:56
  • \$\begingroup\$ I just don't seem to know what functions you're specifying with the "all_equal", and "overlap_chance_real" If I intend to replace them with some of my written functions, it doesn't work :/. What changes have to be made exactly when reviewing the answer from you? I'm a newbie to be honest \$\endgroup\$ – user183738 Jan 8 at 17:01
  • 1
    \$\begingroup\$ Haha, yes I made the mistake with specifying who it was. but alright, I'll give it a try. Much appreciated! \$\endgroup\$ – user183738 Jan 8 at 17:05
8
\$\begingroup\$

The right functions

def roll(die):
    number = random.randint(0,len(die)-1)
    b = die[number]
    return b

Can be:

def roll(die):
    return random.choice(die)

inptchacc()

You could instead check only the first character of the input (case insensitive), rather than expecting the whole keyword.

Your return variable stop isn't needed. Just return directly.

stop = 'stop'
return stop

Should be:

return 'stop'

Likewise elsewhere in your code.

Use __main__

Rather than putting code between functions and also at the end of the file, instead use:

if __name__ == '__main__':

See here for more details.

\$\endgroup\$
3
\$\begingroup\$

Make the computer count

These:

Die1 = [1,2,3,4]
Die2 = [1,2,3,4,5,6]

should be

die1 = list(range(1, 5))
die2 = list(range(1, 7))

Only compare once

This list:

ending_conditions = ['stop','Stop','quit','Quit']

has two problems. First, don't store multiple cases for one word - just store lower (or upper) case - I prefer lower, because no shouting.

Also, it shouldn't be a list. It's getting tested for membership, so make it a set. In other words,

ending_conditions = {'stop', 'quiet'}

then for use, write "not in", instead of "not x in"; i.e.

if inpu.lower() not in ending_conditions:

PEP8

Run your code through a linter. It will catch several things, including

  • Don't capitalize your variables (Die1, Die2, Adding, Doing) if they're local. If they're global constants, they need to be all-uppercase. In the case of Die*, they probably shouldn't be global at all.
  • Add a couple of newlines after your imports

Grammar

Write:

print('Incompatible options selected. Returning...')
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy