Get count of occurrence of integers in a random number selection in a range N times

Question

When writing this code I found myself doing a lot of repetitive stuff and was wondering if there is an easier or simpler or shorter way to do a repetitive task such as this.

Here is the relevant code:

 from random import randint
    RandomNumber = Zeroes = Ones = Twos = Threes = Fours = Fives = Sixes = i = 0

    while i < 1000000:
        RandomNumber = (randint(0,6))
        if RandomNumber == 0:
            Zeroes = Zeroes + 1
        if RandomNumber == 1:
            Ones = Ones + 1
        if RandomNumber == 2:
            Twos = Twos + 1
        if RandomNumber == 3:
            Threes = Threes + 1
        if RandomNumber == 4:
            Fours = Fours + 1
        if RandomNumber == 5:
            Fives = Fives + 1
        if RandomNumber == 6:
            Sixes = Sixes + 1

        i = i + 1

Working code belongs on http://codereview.stackexchange.com/ — OneCricketeer, Aug 08 '16 at 17:26
Why use a ton of variables when you can create a `list` that could hold them? — ForceBru, Aug 08 '16 at 17:26
The answers so far mostly give an iterative solution, but in reality this can be accomplished in constant time. You don't need to iterate that much to get a distribution like the one you're looking for. — James, Aug 08 '16 at 17:46

karakfa · Answer 1 · 2016-08-08T17:34:46.490

1

here you go...

from random import randint
outcomes=[0]*7
for i in range(1000000):
    outcomes[randint(0,6)]+=1

edited Aug 08 '16 at 17:34

answered Aug 08 '16 at 17:31

karakfa

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outcomes will start out as [0,1,2,3,4,5,6], that's not what the asker wants. – James Aug 08 '16 at 17:31

score 1 · Accepted Answer · edited Aug 08 '16 at 18:33

1

Rather than taking the named variables for each random output, you can take dictionary with each possible value as key. This will shorten you code and made it extendable for any random range

from random import randint
randomMax = 6
randomList= {i:0 for i in range(0,randomMax+1)}
totalIterations = 10000
while totalIterations >0:
    randomList[randint(0,randomMax)]+=1
    totalIterations-=1

Sample Output:

{0: 1400, 1: 1400, 2: 1500, 3: 1400, 4: 1500, 5: 1000, 6: 1800}

edited Aug 08 '16 at 18:33

JRodDynamite

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answered Aug 08 '16 at 17:39

abhinsit

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Exactly what I was looking for! Thank you very much – Aug 08 '16 at 17:40
I have made it even more configurable, wherein you can just change the randomMax value and get results for 0->randomMax range – abhinsit Aug 08 '16 at 17:43
This works, but takes far, far longer than is actually necessary. See my answer for a constant time solution that doesn't require iterating 10000 times. – James Aug 08 '16 at 17:45

score 0 · Answer 3 · edited May 23 '17 at 10:28

First, when you find yourself using tons of variables that suit the same purpose (e.g. number of zeros, ones, twos, etc), you almost certainly need an array (a list in Python terminology).

import random

nums = [0] * 7 # number of digits from zero to six

for i in range(1000001):
    r = random.randint(0, 6) # you can get rid of this variable 
                             # and use random.randint(0, 6) as an array index
    nums[r] += 1 # each member of a list represents a number of some digit's occurrences

print(''.join("{} -> {}\n".format(a, b) for a, b in zip(range(7), nums)))

If you want something extremely short, fast and powerful:

import random, collections

print(''.join("{} -> {}\n".format(a, b) for a, b in collections.Counter(random.randint(0, 6) for _ in range(100)).items()))

Check this to see how one can count occurrences of items of a list with collections.Counter.

This code is fast and doesn't waste memory as it uses generators.

I'd make the first line `nums = [0] * 7`, it seems cleaner (I don't really understand why it's range(11) in the first place) — James, Aug 08 '16 at 17:34
@James, good idea! (This was to count the occurrences of each digit from zero to nine, for some reason :P) — ForceBru, Aug 08 '16 at 17:38

score 0 · Answer 4 · answered Aug 08 '16 at 17:33

0

in your special case you can do

from random import randint
results=[0]*7 # same as [0,0,0,0,0,0,0]

i=0
while i < 1000000:
    n = randint(0,6)
    results[n]+=1
    i+=1

for i in range(len(results)):
    print(i, results[i])

answered Aug 08 '16 at 17:33

janbrohl

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score 0 · Answer 5 · answered Aug 08 '16 at 17:33

This code might help, counter dictionary will hold numbers as keys, with number occurrences as values.

from random import randint

counter = {}

while i < 1000000:
    RandomNumber = (randint(0,6))
    if RandomNumber in counter:
        counter[RandomNumber] += 1
    else:
        counter[RandomNumber] = 1

    i += 1

score 0 · Answer 6 · edited May 23 '17 at 12:18

These answers all work, but they are going to take a very long time to loop when in reality you don't need to use a loop at all. Here's code that takes constant time rather than needs to loop over a million times:

from random import random

nums = [random() for _ in range(7)]
nums = [int((x / sum(nums)) * 1000000) for x in nums]

Admittedly this method will be a little off of 1000000 in total, but it's much, much faster, and you can also add a bit to random ones to make it actually 1000000.

For more info, see:

Getting N random numbers that the sum is M

Random numbers that add to 100: Matlab (For info on the statistical distribution that this generates)

Get count of occurrence of integers in a random number selection in a range N times

6 Answers6