CODEWITHSUNDEEP
pythonpython-3.x
vamosrafa
vamosrafa

Reputation: 695

Exploring Python Exercise

I am learning Python from the book Exploring Python by Timothy Budd. One of the exercises from this chapter is this:

15. The function randint from the random module can be used to produce random numbers. A call on random.randint(1, 6), for example, will produce the values 1 to 6 with equal probability. Write a program that loops 1000 times. On each iteration it makes two calls on randint to simulate rolling a pair of dice. Compute the sum of the two dice, and record the number of times each value appears. Afterthe loop, print the array of sums. You can initialize the array using the idiom shown earlier in this chapter:

times = [0] * 12 # make an array of 12 elements, initially zero

I am able to print the sum in the array, but I have not understood the concept of recording the number of times each value appears. Also, what purpose would times = [0] serve? Here is my code for printing the sum:

 #############################################
#   Program to print the sum of dice rolls  #
#############################################

from random import randint
import sys


times = [0] * 12

summation = []

def diceroll():

    print "This program will print the"
    print "sum of numbers, which appears"
    print "after each time the dice is rolled."
    print "The program will be called 1000 times"



    for i in range(1,1000):

        num1 = randint(1,6)
        num2 = randint(1,6)

        sum = num1 + num2
        summation.append(sum)
        #times[i] = [i] * 12
    print summation
    #print times


diceroll()

Upvotes: 0

Views: 1264

Answers (2)

jfs
jfs

Reputation: 414875

To count number of occurrences:

times = [0]*13
for _ in range(1000):
    sum_ = randint(1, 6) + randint(1, 6)
    times[sum_] += 1

print(times[2:])

Possible values of sum_ are 2...12 including. Possible times indexes are 0...12 including.

times[i] corresponds to a number of occurrences of sum_ == i among 1000 tries.

Note: times[0] and times[1] are always zero because sum_ > 1

[x]*3 produces [x, x, x] It is a nicer version of:

L = []
L.append(x)
L.append(x)
L.append(x)

Additional comments about your code:

  • print something is an error on Python 3
  • for i range(1, 1000) is wrong. It produces i from 1 to 999 including (999 iterations instead of 1000 as required). range doesn't include upper limit as it customery for intervals in programming.

Upvotes: 0

AI Generated Response
AI Generated Response

Reputation: 8845

times[0] * 12 initiates a list of 12 elements with value 0. What you want to do next is

times[s] += 1 #adds 1 to the number of times s occurs

This is similar to using a dict to encode a value, but its simpler.

times = [0] initializes a list called times of length 1 and value 0. The idiom times = [0] * 12 means times is a list of 12 zeroes.

If you want to be able to use this without keyerror when num1==num2==6, you need to do times = [0]*13 because python is a 0 indexed system.

Sidenote: Don't use sum as a variable name because it is a builtin function (predefined) and you don't want to override it. Use times[sum([num1,num2])] += 1 or times[num1+num2] instead.

Upvotes: 1

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