graded randomness

Avi Gross avigross at verizon.net
Fri Dec 28 18:01:57 EST 2018


Abdur-Rahman

I am sure various modules available have ready-made solutions and I see
others have replied to your question. The usual disclaimers apply. This is
an academic discussion and not a statement of the right or only way to do an
abstract task.

So just a thought. You seem interested in a GENERAL situation where you have
N situations with each having a specific probability and the probabilities
sum to 1.0. If that is right, you are creating a partition where you can
make a data structure listing N ordered items along with their individual
probability. Given such a list, you can create another item that is a
cumulative sum.

In your example, your individual probabilities for ['green', 'red', 'blue']
are [0.4, 0.4, 0.2] but the use of lists is just for illustration. You might
use a Numpy array as they have a cumulative sum function:

import numpy as np
np.cumsum([0.4, 0.4, 0.2])
Returns:
array([0.4, 0.8, 1. ])

or viewed vertically:

np.cumsum([0.4, 0.4, 0.2]).reshape(3,1)
array([[0.4],
       [0.8],
       [1. ]])

Again, we are talking about a GENERAL solution but using this example to
illustrate. To get a weighted probability now, you use the random module (or
anything else) to generate a random number between 0 and 1. You search in
the cumulative sum data structure to find the right range. A random value
less than 0.4 should direct you to using green. If above that but less than
0.8, use red. Else, use blue.

To do this properly, you can decide what data structures makes this easy to
do. Maintaining three independent lists or arrays may not be optimal. 

The main idea is to find a way to segment your choices. So consider a
different common example of rolling a pair of (six sided) standard dice. You
know there are 6**2 possible outcomes. There is only one way to get a sum of
2 by rolling a one and another one. So the probability is 1/36 or .0277...
and you can calculate the probabilities of all the other sums with a 7 being
the most common roll at .1667. In this example your choices are rolls of 2
through 12 or 11 choices. The same logic applies. Generate 11 data measures
and a cumulative sum. Just as illustration, I show code using a Pandas
DataFrame object.

import numpy as np
import pandas as pd
diceSumText = np.array(["two", "three", "four", "five", "six", "seven",
"eight", "nine", "ten", "eleven", "twelve"])
diceSumVal = np.array(range(2,13))
diceProbability = np.array([1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]) / 36
diceProbCumSum = np.cumsum(diceProbability)

Now combine those:

mydata = pd.DataFrame({"Text": diceSumText, "Value": diceSumVal, "Prob":
diceProbability, "Cum": diceProbCumSum})
print(mydata)

      Text  Value      Prob       Cum
0      two      2  0.027778  0.027778
1    three      3  0.055556  0.083333
2     four      4  0.083333  0.166667
3     five      5  0.111111  0.277778
4      six      6  0.138889  0.416667
5    seven      7  0.166667  0.583333
6    eight      8  0.138889  0.722222
7     nine      9  0.111111  0.833333
8      ten     10  0.083333  0.916667
9   eleven     11  0.055556  0.972222
10  twelve     12  0.027778  1.000000

Again, you can do something any number of ways. This is just one. And in
this format the indentation is not great. 

But it lets you write an algorithm that finds the highest 'index" that still
is below the random number chosen and then select either the text or value
that fits in that partition. Not to repeat, there are many other ways so
feel free to innovate. 


-----Original Message-----
From: Python-list <python-list-bounces+avigross=verizon.net at python.org> On
Behalf Of Abdur-Rahmaan Janhangeer
Sent: Friday, December 28, 2018 2:45 PM
To: Python <python-list at python.org>
Subject: Re: graded randomness

well i wanted that to improve the following code:

https://www.pythonmembers.club/2018/12/28/reviving-bertrand-russell-through-
python/

that one i used the random list technique

Abdur-Rahmaan Janhangeer
http://www.pythonmembers.club | https://github.com/Abdur-rahmaanJ Mauritius

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