Weighted Probabilities
Introduction
In our previous chapter we ha a look at the following examples:
from random import choice professions = ["scientist", "philosopher", "engineer", "priest"] print(choice("abcdefghij")) print(choice(professions)) print(choice(("beginner", "intermediate", "advanced")))
i philosopher beginner
'choice' returns an object from a sequence. The chances for the elements of the sequence to be chosen are evenly distributed. So the chances for getting 'scientist' as a return value of the call choice(professions) is 1/4. This is out of touch with reality. There are surely more scientists and engineers in the world than there are priests and philosophers. Just like with the loaded die, we have again the need of a weighted choice.
We will devise a function "weighted_choice", which returns a random element from a sequence like random.choice, but the elements of the sequence are weighted.
Random Intervals
Before we can start with the design of the weighted version of choice, we will define a function find_interval(x, partition). We will need this function in weighted_choice. The find_interval function takes two arguments:
 a numerical value x
 a list or a tuple of numerical values p_{0}, p_{1}, p_{2}, ... p_{n}
The function will return i, if p_{i} < x < p_{i+1}. 1 will be returned, if x is less than p_{0} or if x is greater or equal than p_{n}
def find_interval(x, partition): """ find_interval > i partition is a sequence of numerical values x is a numerical value The return value "i" will be the index for which applies partition[i] < x < partition[i+1], if such an index exists. 1 otherwise """ for i in range(0, len(partition)): if x < partition[i]: return i1 return 1 I = [0, 3, 5, 7.8, 9, 12, 13.8, 16] for x in [1.3, 0, 0.1, 3.2, 5, 6.2, 7.9, 10.8, 13.9, 15, 16, 16.5]: print(find_interval(x, I), end=", ")
1, 0, 0, 1, 2, 2, 3, 4, 6, 6, 1, 1,
Weighted Random Choices
We can now define the weighted choice function. Let's assume that you have three weights, e.g. 1/5, 1/2, 3/10. We can build the cumulative sum of the weights with np.cumsum(weights).
import numpy as np weights = [0.2, 0.5, 0.3] cum_weights = [0] + list(np.cumsum(weights)) print(cum_weights)
[0, 0.20000000000000001, 0.69999999999999996, 1.0]
If we create a random number x between 0 and 1 by using random.random(), the probability for x to lie within the interval [0, cum_weights[0]) is equal to 1/5. The probability for x to lie within the interval [cum_weights[0], cum_weights[1]) is equal to 1/2 and finally, the probability for x to lie within the interval [cum_weights[1], cum_weights[2]) is 3/10.
Now you are able to understand the basic idea of how weighted_choice operates:
import numpy as np import random def weighted_choice(sequence, weights, secure=True): """ weighted_choice selects a random element of the sequence according to the list of weights """ if secure: crypto = random.SystemRandom() x = crypto.random() else: x = np.random.random() cum_weights = [0] + list(np.cumsum(weights)) index = find_interval(x, cum_weights) return sequence[index]
Example:
We can use the function weighted_choice for the following task:
Suppose, we have a "loaded" die with P(6)=3/12 and P(1)=1/12. The probability for the outcomes of all the other possibilities is equally likely,
i.e. P(2) = P(3) = P(4) = P(5) = p.
We can calculate p with
1  P(1)  P(6) = 4 x p
that means
p = 1 / 6
How can we simulate this die with our weighted_choice function?
We call weighted_choice with 'faces_of_die' and the 'weights' list. Each call correspondents to a throw of the loaded die.
We can show that if we throw the die a large number of times, for example 10,000 times, we get roughly the probability values of the weights:
from collections import Counter faces_of_die = [1, 2, 3, 4, 5, 6] weights = [1/12, 1/6, 1/6, 1/6, 1/6, 3/12] outcomes = [] n = 10000 for _ in range(n): outcomes.append(weighted_choice(faces_of_die, weights)) c = Counter(outcomes) for key in c: c[key] = c[key] / n print(sorted(c.values()))
[0.0848, 0.1663, 0.168, 0.1682, 0.1688, 0.2439]
So far, the values of the partition list defined the subpartitions, in which we expected the value x to be. If the value x was less than p_{0} or greater or equal than p_{n}, we returned 1.
We could also define our first subinterval to be the interval from ∞ to p_{0}, and we could return 0 in this case. The last subinterval could be correspondingly p_{n} to ∞ (infinity).
We will extend our function find_interval with this different way of interval mapping. To distinguish between these two cases, we will introduce an additional parameter "endpoints". "True" corresponds to our first approach and "False" to the previously described "infinity" case.
In other words: If we set 'endpoints' to 'False', we will get the following behaviour:
 i will be returned if x is smaller than p_{i}
 len(partition) will be returned, if x is larger or equal to p_{len(partition)1}
We demonstrate this in the following diagram:
The new function looks like this:
def find_interval(x, partition, endpoints=True): """ find_interval > i If endpoints is True, "i" will be the index for which applies partition[i] < x < partition[i+1], if such an index exists. 1 otherwise If endpoints is False, "i" will be the smallest index for which applies x < partition[i]. If no such index exists "i" will be set to len(partition) """ for i in range(0, len(partition)): if x < partition[i]: return i1 if endpoints else i return 1 if endpoints else len(partition) I = [0, 3, 5, 7.8, 9, 12, 13.8, 16] print("Endpoints are included:") for x in [1.3, 0, 0.1, 3.2, 5, 6.2, 7.9, 10.8, 13.9, 15, 16, 16.5]: print(find_interval(x, I), end=", ") print("\nEndpoints are not included:") for x in [1.3, 0, 0.1, 3.2, 5, 6.2, 7.9, 10.8, 13.9, 15, 16, 16.5]: print(find_interval(x, I, endpoints=False), end=", ")
Endpoints are included: 1, 0, 0, 1, 2, 2, 3, 4, 6, 6, 1, 1, Endpoints are not included: 0, 1, 1, 2, 3, 3, 4, 5, 7, 7, 8, 8,
Weighted Random Choice with Numpy
To produce a weighted choice of an array like object, we can also use the choice function of the numpy.random package. This way, there is no necessity to create a custommade function like we did in our previous chapter:
from numpy.random import choice professions = ["scientist", "philosopher", "engineer", "priest", "programmer"] probabilities = [0.2, 0.05, 0.3, 0.15, 0.3] choice(professions, p=probabilities)This gets us the following result:
'engineer'
Let's experiment with choice:
from collections import Counter c = Counter() for _ in range(1000): profession = choice(professions, p=probabilities) c[profession] += 1 print(c) s = sum(c.values()) for el in c: c[el] /= s print(c)
Counter({'programmer': 306, 'engineer': 305, 'scientist': 202, 'priest': 131, 'philosopher': 56}) Counter({'programmer': 0.306, 'engineer': 0.305, 'scientist': 0.202, 'priest': 0.131, 'philosopher': 0.056})
There is still one benifit for our custommade approach for the weighted choice: We use the cryptographicsecure method of the SystemRandom class, whereas Numpy choice is deterministic and therefor not secure.
Weighted Sample
In the previous chapter on random numbers and probability, we introduced a method to randomly extract a population or sample from a group of objects liks lists or tuples. Every object had the same likelikhood to be drawn, i.e. to be part of the sample.
In real life situation there will be of course situation in which every or some objects will have different probabilities. We define a function for a weighted sample in the following, which makes use of our previously defined weighted_choice function:
def weighted_sample(population, weights, k): """ This function draws a random sample of length k from the sequence 'population' according to the list of weights """ sample = set() population = list(population) weights = list(weights) while len(sample) < k: choice = weighted_choice(population, weights) sample.add(choice) index = population.index(choice) weights.pop(index) population.remove(choice) weights = [ x / sum(weights) for x in weights] return list(sample)
def weighted_sample_alternative(population, weights, k): """ Alternative way to previous implementation. This function draws a random sample of length k from the sequence 'population' according to the list of weights """ sample = set() population = list(population) weights = list(weights) while len(sample) < k: choice = weighted_choice(population, weights) if choice not in sample: sample.add(choice) return list(sample)
Example:
Let's assume we have eight candies, coloured "red", "green", "blue", "yellow", "black", "white", "pink", and "orange". Our friend Peter will have the "weighted" preference 1/24, 1/6, 1/6, 1/12, 1/12, 1/24, 1/8, 7/24 for thes colours. He is allowed to take 3 candies:
balls = ["red", "green", "blue", "yellow", "black", "white", "pink", "orange"] weights = [ 1/24, 1/6, 1/6, 1/12, 1/12, 1/24, 1/8, 7/24] for i in range(10): print(weighted_sample(balls, weights, 3))
['pink', 'black', 'white'] ['pink', 'green', 'orange'] ['pink', 'green', 'blue'] ['yellow', 'red', 'black'] ['yellow', 'green', 'black'] ['orange', 'blue', 'white'] ['pink', 'black', 'orange'] ['green', 'blue', 'orange'] ['yellow', 'black', 'orange'] ['pink', 'orange', 'white']
Let's approximate the likelihood for an orange candy to be included in the sample:
n = 100000 orange_counter = 0 orange_counter_alternative = 0 for i in range(n): if "orange" in weighted_sample(balls, weights, 3): orange_counter += 1 if "orange" in weighted_sample_alternative(balls, weights, 3): orange_counter_alternative += 1 print(orange_counter / n) print(orange_counter_alternative / n)
0.71031 0.71084
Cartesian Choice
The function cartesian_choice is named after the Cartesian product from set theory
Cartesian product
The Cartesian product is an operation which returns a set from multiple sets. The result set from the Cartesian product is called a "product set" or simply the "product".
For two sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B:
A x B = { (a, b)  a ∈ A and b ∈ B }
If we have n sets A_{1}, A_{2}, ... A_{n}, we can build the Cartesian product correspondingly:
A_{1} x A_{2} x ... x A_{n} = { (a_{1}, a_{2}, ... a_{n})  a_{1} ∈ A_{1}, a_{2} ∈ A_{2}, ... a_{n} ∈ A_{n}]
The Cartesian product of n sets is sometimes called an nfold Cartesian product.
Cartesian Choice: cartesian_choice
We will write now a function cartesian_choice, which takes an arbitrary number of iterables as arguments and returns a list, which consists of random choices from each iterator in the respective order.
Mathematically, we can see the result of the function cartesian_choice as an element of the Cartesian product of the iterables which have been passed as arguments.
import random def cartesian_choice(*iterables): """ A list with random choices from each iterable of iterables is being created in respective order. The result list can be seen as an element of the Cartesian product of the iterables """ res = [] for population in iterables: res.append(random.choice(population)) return res cartesian_choice(["The", "A"], ["red", "green", "blue", "yellow", "grey"], ["car", "house", "fish", "light"], ["smells", "dreams", "blinks"])The previous Python code returned the following:
['A', 'red', 'light', 'smells']
We define now a weighted version of the previously defined function:
import random def weighted_cartesian_choice(*iterables): """ A list with weighted random choices from each iterable of iterables is being created in respective order """ res = [] for population, weight in iterables: lst = weighted_choice(population, weight) res.append(lst) return res determiners = (["The", "A", "Each", "Every", "No"], [0.3, 0.3, 0.1, 0.1, 0.2]) colours = (["red", "green", "blue", "yellow", "grey"], [0.1, 0.3, 0.3, 0.2, 0.2]) nouns = (["water", "elephant", "fish", "light", "programming language"], [0.3, 0.2, 0.1, 0.1, 0.3]) nouns2 = (["of happiness", "of chocolate", "of wisdom", "of challenges", "of air"], [0.5, 0.2, 0.1, 0.1, 0.1]) verb_phrases = (["smells", "dreams", "thinks", "is made of"], [0.4, 0.3, 0.3]) print("It may or may not be true:") for i in range(10): res = weighted_cartesian_choice(determiners, colours, nouns, verb_phrases, nouns2) print(" ".join(res) + ".")
It may or may not be true: A green elephant dreams of wisdom. No grey elephant smells of happiness. The green water thinks of happiness. A blue programming language smells of chocolate. Every yellow fish dreams of happiness. A yellow water dreams of happiness. A yellow programming language thinks of chocolate. Each yellow water dreams of happiness. Each yellow water dreams of happiness. The green programming language dreams of happiness.
We check in following version, if the "probabilities" are all right:
import random def weighted_cartesian_choice(*iterables): """ A list with weighted random choices from each iterable of iterables is being created in respective order """ res = [] for population, weight in iterables: lst = weighted_choice(population, weight) res.append(lst) return res determiners = (["The", "A", "Each", "Every", "No"], [0.3, 0.3, 0.1, 0.1, 0.2]) colours = (["red", "green", "blue", "yellow", "grey"], [0.1, 0.3, 0.3, 0.2, 0.2]) nouns = (["water", "elephant", "fish", "light", "programming language"], [0.3, 0.2, 0.1, 0.1, 0.3]) nouns2 = (["of happiness", "of chocolate", "of wisdom", "of challenges", "of air"], [0.5, 0.2, 0.1, 0.1, 0.1]) verb_phrases = (["smells", "dreams", "thinks", "is made of"], [0.4, 0.3, 0.2, 0.1]) print("It may or may not be true:") sentences = [] for i in range(10000): res = weighted_cartesian_choice(determiners, colours, nouns, verb_phrases, nouns2) sentences.append(" ".join(res) + ".") words = ["smells", "dreams", "thinks", "is made of"] from collections import Counter c = Counter() for sentence in sentences: for word in words: if word in sentence: c[word] += 1 wsum = sum(c.values()) for key in c: print(key, c[key] / wsum)
It may or may not be true: thinks 0.2015 is made of 0.1039 smells 0.4016 dreams 0.293
Random Seed
A random seed,  also called "seed state", or just "seed"  is a number used to initialize a pseudorandom number generator. When we called random.random() we expected and got a random number between 0 and 1. random.random() calculates a new random number by using the previously produced random number. What about the first time we use random in our program? Yes, there is no previously created random number. If a random number generator is called for the first time, it will have to create a first "random" number.
If we seed a pseudorandom number generator, we provide a first "previous" value. A seed value corresponds to a sequence of generated values for a given random number generator. If you use the same seed value again, you get and you can rely on getting the same sequence of numbers again.
The seed number itself doesn't need to be randomly chosen so that the algorithm creates values which follow a probability distribution in a pseudorandom manner. Yet, the seed matters in terms of security. If you know the seed, you could for example generate the secret encryption key which is based on this seed.
Random seeds are in many programming languages generated from the state of the computer system, which is in lots of cases the system time.
This is true for Python as well. Help on random.seed says that if you call the function with None or no argument it will seed "from current time or from an operating system specific randomness source if available."
import random help(random.seed)
Help on method seed in module random: seed(a=None, version=2) method of random.Random instance Initialize internal state from hashable object. None or no argument seeds from current time or from an operating system specific randomness source if available. For version 2 (the default), all of the bits are used if *a* is a str, bytes, or bytearray. For version 1, the hash() of *a* is used instead. If *a* is an int, all bits are used.
The seed functions allows you to get a determined sequence of random numbers. You can repeat this sequence, whenever you need it again, e.g. for debugging purposes.
import random random.seed(42) for _ in range(10): print(random.randint(1, 10), end=", ") print("\nLet's create the same random numbers again:") random.seed(42) for _ in range(10): print(random.randint(1, 10), end=", ")
2, 1, 5, 4, 4, 3, 2, 9, 2, 10, Let's create the same random numbers again: 2, 1, 5, 4, 4, 3, 2, 9, 2, 10,
Random Numbers in Python with Gaussian and Normalvariate Distribution
We want to create now 1000 random numbers between 130 and 230 that have a gaussian distribution with the mean value mu set to 550 and the standard deviation sigma is set to 30.
from random import gauss n = 1000 values = [] frequencies = {} while len(values) < n: value = gauss(180, 30) if 130 < value < 230: frequencies[int(value)] = frequencies.get(int(value), 0) + 1 values.append(value) print(values[:10])
[173.49123947564414, 183.47654360102564, 186.96893210720162, 214.90676059797428, 199.69909520396007, 183.31521532331496, 157.85035192965537, 149.56012897536849, 187.39026585633607, 219.33242481612143]
The following program plots the random values, which we have created before. We haven't covered matplotlib so far, so it's not necessary to understand the code:
%matplotlib inline import matplotlib.pyplot as plt freq = list(frequencies.items()) freq.sort() plt.plot(*list(zip(*freq)))The code above returned the following:
[<matplotlib.lines.Line2D at 0x7f282554a828>]
We do the same now with normvariate instead of gauss:
from random import normalvariate n = 1000 values = [] frequencies = {} while len(values) < n: value = normalvariate(180, 30) if 130 < value < 230: frequencies[int(value)] = frequencies.get(int(value), 0) + 1 values.append(value) freq = list(frequencies.items()) freq.sort() plt.plot(*list(zip(*freq)))The previous Python code returned the following:
[<matplotlib.lines.Line2D at 0x7f2824fef1d0>]
Exercise With Zeros and Ones
It might be a good idea to write the following function as an exercise yourself. The function should be called with a parameter p, which is a probabilty value between 0 and 1. The function returns a 1 with a probability of p, i.e. ones in p percent and zeros in (1  p) percent of the calls:
import random def random_ones_and_zeros(p): """ p: probability 0 <= p <= 1 returns a 1 with the probability p """ x = random.random() if x < p: return 1 else: return 0
Let's test our little function:
n = 1000000 sum(random_ones_and_zeros(0.8) for i in range(n)) / nThe above code returned the following:
0.800609
It might be a great idea to implement a task like this with a generator. If you are not familar with the way of working of a Python generator, we recommend to consult our chapter on generators and iterators of our Python tutorial.
import random def random_ones_and_zeros(p): while True: x = random.random() yield 1 if x < p else 0 def firstn(generator, n): for i in range(n): yield next(generator)
n = 1000000 sum(x for x in firstn(random_ones_and_zeros(0.8), n)) / nThis gets us the following output:
0.799762
Our generator random_ones_and_zeros can be seen as a sender, which emits ones and zeros with a probability of p and (1p) respectively.
We will write now another generator, which is receiving this bitstream. The task of this new generator is to read the incoming bitstream and yield another bitstream with ones and zeros with a probability of 0.5 without knowing or using the probability p. It should work for an arbitrary probability value p.^{1}
def ebitter(bitstream): while True: bit1 = next(bitstream) bit2 = next(bitstream) if bit1 + bit2 == 1: bit3 = next(bitstream) if bit2 + bit3 == 1: yield 1 else: yield 0
def ebitter2(bitstream): bit1 = next(bitstream) bit2 = next(bitstream) bit3 = next(bitstream) while True: if bit1 + bit2 == 1: if bit2 + bit3 == 1: yield 1 else: yield 0 bit1, bit2, bit3 = bit2, bit3, next(bitstream)
n = 1000000 sum(x for x in firstn(ebitter(random_ones_and_zeros(0.8)), n)) / nThe previous Python code returned the following:
0.49975
n = 1000000 sum(x for x in firstn(ebitter2(random_ones_and_zeros(0.8)), n)) / nThe above code returned the following output:
0.500011
Underlying theory:
Our first generator emits a bitstream B_{0}, B_{1}, B_{2},...
We check now an arbitrary pair of consecutive Bits B_{i}, B_{i+1}, ...
Such a pair can have the values 01, 10, 00 or 11. The probability P(01) = (p1) x p and probability P(10) = p x (p1), so that the combined probabilty that the two consecutive bits are either 01 or 10 (or the sum of the two bits is 1) is 2 x (p1) x p
Now we look at another bit B_{i+2}. What is the probability that both
B_{i} + B_{i+1} = 1
and
B_{i+1} + B_{i+2} = 1?
The possible outcomes satisfying these conditions and their corresponding probabilities can be found in the following table:
Probability  B_{i}  B_{i+1}  B_{i+2} 

p^{2} x (1p)  0  1  0 
p x (1  p)^{2}  1  0  1 
We will denote the outcome sum(B_{i}, B_{i+1})=1 asX_{1} and correspondingly the outcome sum(B_{i+1}, B_{i+2})=1 as X_{2}
So, the joint probability P(X_{1}, X_{2}) = p^{2} x (1p) + p x (1  p)^{2} which can be rearranged to p x (1p)
The conditional probability of X_{2} given X_{1}:
P(X_{2}  X_{1}) = P(X_{1}, X_{2}) / P(X_{2})
P(X_{2}  X_{1}) = p x (1p) / 2 x p x (1p) = 1 / 2
Synthetical Sales Figures
In this subchapter we want to create a data file with sales figures. Imagine that we have a chain of shops in various European and Canadian cities: Frankfurt, Munich, Berlin, Zurich, Hamburg, London, Toronto, Strasbourg, Luxembourg, Amsterdam, Rotterdam, The Hague
We start with an array 'sales' of sales figures for the year 1997:
import numpy as np sales = np.array([1245.89, 2220.00, 1635.77, 1936.25, 1002.03, 2099.13, 723.99, 990.37, 541.44, 1765.00, 1802.84, 1999.00])
The aim is to create a comma separated list like the ones you get from Excel. The file should contain the sales figures, we don't know, for all the shops, we don't have, spanning the year from 1997 to 2016.
We will add random values to our sales figures year after year. For this purpose we construct an array with growthrates. The growthrates can vary between a minimal percent value (min_percent) and maximum percent value (max_percent):
min_percent = 0.98 # corresponds to 1.5 % max_percent = 1.06 # 6 % growthrates = (max_percent  min_percent) * np.random.random_sample(12) + min_percent print(growthrates)
[ 1.03476561 1.00885095 1.00614899 1.05164581 1.0307091 0.98822763 0.99366872 1.05810125 1.04798573 1.02784796 1.05035899 1.02262023]
To get the new sales figures after a year, we multiply the sales array "sales" with the array "growthrates":
sales * growthratesThe previous Python code returned the following output:
array([ 1289.20412146, 2239.649113 , 1645.82833205, 2036.24919608, 1032.80143634, 2074.41825668, 719.40621318, 1047.91173209, 567.42139317, 1814.15165757, 1893.62920988, 2044.21783979])
To get a more sustainable sales development, we change the growthrates only every four years.
This is our complete program, which saves the data in a file called sales_figures.csv:
import numpy as np fh = open("sales_figures.csv", "w") fh.write("Year, Frankfurt, Munich, Berlin, Zurich, Hamburg, London, Toronto, Strasbourg, Luxembourg, Amsterdam, Rotterdam, The Hague\n") sales = np.array([1245.89, 2220.00, 1635.77, 1936.25, 1002.03, 2099.13, 723.99, 990.37, 541.44, 1765.00, 1802.84, 1999.00]) for year in range(1997, 2016): line = str(year) + ", " + ", ".join(map(str, sales)) fh.write(line + "\n") if year % 4 == 0: min_percent = 0.98 # corresponds to 1.5 % max_percent = 1.06 # 6 % growthrates = (max_percent  min_percent) * np.random.random_sample(12) + min_percent #growthrates = 1 + (np.random.rand(12) * max_percent  negative_max) / 100 sales = np.around(sales * growthrates, 2) fh.close()
The result is in the file sales_figures.csv.
We will use this file in our chapter on reading and writing in Numpy.
Exercises
 Let's do some more die rolling. Prove empirically  by writing a simulation program  that the probability for the combined events "an even number is rolled" (E) and "A number greater than 2 is rolled" is 1/3.
 The file ["universities_uk.txt"](universities_uk.txt) contains a list of universities in the United Kingdom by enrollment from 20132014 (data from ([Wikepedia](https://en.wikipedia.org/wiki/List_of_universities_in_the_United_Kingdom_by_enrollment#cite_note1)). Write a function which returns a tuple (universities, enrollments, total_number_of_students) with  universities: list of University names  enrollments: corresponding list with enrollments  total_number_of_students: over all universities Now you can enroll a 100,000 fictional students with a likelihood corresponding to the real enrollments.

Let me take you back in time and space in our next exercise. We will travel back into ancient Pythonia (Πηθωνια). It was the time when king Pysseus ruled as the benevolent dictator for live. It was the time when Pysseus sent out his messengers throughout the world to announce that the time has come for his princes Anacondos (Ανακονδος), Cobrion (Κομπριον), Boatos (Μποατος) and Addokles (Ανδοκλης) to merry. So, they organized the toughest programming contests amongst the fair and brave amazons, better known as Pythonistas of Pythonia. Finally, only eleven amazons were left to choose from:
1) The ethereal Airla (Αιρλα) 2) Barbara (Βαρβάρα), the one from a foreign country. 3) Eos (Ηως), looking divine in dawn 4) The sweet Glykeria (Γλυκερία) 5) The gracefull Hanna (Aννα) 6) Helen (Ελενη), the light in the dark 7) The good angel Agathangelos (Αγαθάγγελος) 8) the violet tinted cloud Iokaste (Ιοκάστη) 9) Medousa (Μέδουσα), the guardian 10) the selfcontrolled Sofronia (Σωφρονία) 11) Andromeda (Ανδρομεδα), the one who thinks man or a warrior.
On the day they arrived the chances to be drawn in the lottery are the same for every amazon, but Pysseus wants the lottery to be postponed to some day in the future. The probability changes every day: It will be lowered by 1/13 for the first seven amazones and it will be increased by 1/12 for the last four amazones.
How long will the king have to wait until he can be more than 90 percent sure that his princes Anacondos, Cobrion, Boatos and Addokles will be married to Iokaste, Medousa, Sofronia and Andromeda?
</li>
</ol>
Solutions to our exercises

from random import randint outcomes = [ randint(1, 6) for _ in range(10000)] even_pips = [ x for x in outcomes if x % 2 == 0] greater_two = [ x for x in outcomes if x > 2] combined = [ x for x in outcomes if x % 2 == 0 and x > 2] print(len(even_pips) / len(outcomes)) print(len(greater_two) / len(outcomes)) print(len(combined) / len(outcomes))
0.5061 0.6719 0.3402

At first we will write the function "process_datafile" to process our data file:
def process_datafile(filename): """ process_datafile > (universities, enrollments, total_number_of_students) universities: list of University names enrollments: corresponding list with enrollments total_number_of_students: over all universities """ universities = [] enrollments = [] with open(filename) as fh: total_number_of_students = 0 fh.readline() # get rid of descriptive first line for line in fh: line = line.strip() *praefix, undergraduates, postgraduates, total = line.rsplit() university = praefix[1:] total = int(total.replace(",", "")) enrollments.append(total) universities.append(" ".join(university)) total_number_of_students += total return (universities, enrollments, total_number_of_students)
Let's start our function and check the results:
universities, enrollments, total_students = process_datafile("universities_uk.txt") for i in range(14): print(universities[i], end=": ") print(enrollments[i]) print("Total number of students onrolled in the UK: ", total_students)
Open University in England: 123490 University of Manchester: 37925 University of Nottingham: 33270 Sheffield Hallam University: 33100 University of Birmingham: 32335 Manchester Metropolitan University: 32160 University of Leeds: 30975 Cardiff University: 30180 University of South Wales: 29195 University College London: 28430 King's College London: 27645 University of Edinburgh: 27625 Northumbria University: 27565 University of Glasgow: 27390 Total number of students onrolled in the UK: 2299380
We want to enroll now a virtual student randomly to one of the universities. To get a weighted list suitable for our weighted_choice function, we have to normalize the values in the list enrollments:
normalized_enrollments = [ students / total_students for students in enrollments] # enrolling a virtual student: print(weighted_choice(universities, normalized_enrollments))
University of Dundee
We have been asked by the exercise to "enroll" 100,000 fictional students. This can be easily accomplished with a loop:
from collections import Counter outcomes = [] n = 100000 for i in range(n): outcomes.append(weighted_choice(universities, normalized_enrollments)) c = Counter(outcomes) print(c.most_common(20))
[('Open University in England', 5529), ('University of Manchester', 1574), ('University of Nottingham', 1427), ('University of Birmingham', 1424), ('Sheffield Hallam University', 1410), ('Manchester Metropolitan University', 1408), ('Cardiff University', 1334), ('University of Leeds', 1312), ('University of South Wales', 1264), ('University of Plymouth', 1218), ('University College London', 1209), ('Coventry University', 1209), ('University of the West of England', 1197), ('University of Edinburgh', 1196), ("King's College London", 1183), ('University of Glasgow', 1181), ('University of Central Lancashire', 1176), ('Nottingham Trent University', 1174), ('University of Sheffield', 1160), ('Northumbria University', 1154)]
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The bunch of amazons is implemented as a list, while we choose a set for Pysseusses favorites. The weights at the beginning are 1/11 for all, i.e. 1/len(amazons).
Every loop cycle corresponds to a new day. Every time we start a new loop cycle, we will draw "n" samples of Pythonistas to calculate the ratio of the number of times the sample is equal to the king's favorites divided by the number of times the sample doesn't match the king's idea of daughterinlaws. This corresponds to the probability "prob". We stop the first time, the probability is equal or larger than 0.9.
We can use both the function "weighted_same" and "weighted_sample_alternative" to do the drawing.
import time amazons = ["Airla", "Barbara", "Eos", "Glykeria", "Hanna", "Helen", "Agathangelos", "Iokaste", "Medousa", "Sofronia", "Andromeda"] weights = [ 1/len(amazons) for _ in range(len(amazons)) ] Pytheusses_favorites = {"Iokaste", "Medousa", "Sofronia", "Andromeda"} n = 1000 counter = 0 prob = 1 / 330 days = 0 factor1 = 1 / 13 factor2 = 1 / 12 start = time.clock() while prob < 0.9: for i in range(n): the_chosen_ones = weighted_sample_alternative(amazons, weights, 4) if set(the_chosen_ones) == Pytheusses_favorites: counter += 1 prob = counter / n counter = 0 weights[:7] = [ p  p*factor1 for p in weights[:7] ] weights[7:] = [ p + p*factor2 for p in weights[7:] ] weights = [ x / sum(weights) for x in weights] days += 1 print(time.clock()  start) print("Number of days, he has to wait: ", days)
2.870792999999999 Number of days, he has to wait: 33
Teh value for the number of days differs, if n is not large enough.
The following is a solutions without roundoff errors. We will use Fraction from the module fractions.
import time from fractions import Fraction amazons = ["Airla", "Barbara", "Eos", "Glykeria", "Hanna", "Helen", "Agathangelos", "Iokaste", "Medousa", "Sofronia", "Andromeda"] weights = [ Fraction(1, 11) for _ in range(len(amazons)) ] Pytheusses_favorites = {"Iokaste", "Medousa", "Sofronia", "Andromeda"} n = 1000 counter = 0 prob = Fraction(1, 330) days = 0 factor1 = Fraction(1, 13) factor2 = Fraction(1, 12) start = time.clock() while prob < 0.9: #print(prob) for i in range(n): the_chosen_ones = weighted_sample_alternative(amazons, weights, 4) if set(the_chosen_ones) == Pytheusses_favorites: counter += 1 prob = Fraction(counter, n) counter = 0 weights[:7] = [ p  p*factor1 for p in weights[:7] ] weights[7:] = [ p + p*factor2 for p in weights[7:] ] weights = [ x / sum(weights) for x in weights] days += 1 print(time.clock()  start) print("Number of days, he has to wait: ", days)
35.920345 Number of days, he has to wait: 33
We can see that the solution with fractions is beautiful but very slow. Whereas the greater precision doesn't play a role in our case.
So far, we haven't used the power of Numpy. We will do this in the next implementation of our problem:
import time import numpy as np amazons = ["Airla", "Barbara", "Eos", "Glykeria", "Hanna", "Helen", "Agathangelos", "Iokaste", "Medousa", "Sofronia", "Andromeda"] weights = np.full(11, 1/len(amazons)) Pytheusses_favorites = {"Iokaste", "Medousa", "Sofronia", "Andromeda"} n = 1000 counter = 0 prob = 1 / 330 days = 0 factor1 = 1 / 13 factor2 = 1 / 12 start = time.clock() while prob < 0.9: for i in range(n): the_chosen_ones = weighted_sample_alternative(amazons, weights, 4) if set(the_chosen_ones) == Pytheusses_favorites: counter += 1 prob = counter / n counter = 0 weights[:7] = weights[:7]  weights[:7] * factor1 weights[7:] = weights[7:] + weights[7:] * factor2 weights = weights / np.sum(weights) #print(weights) days += 1 print(time.clock()  start) print("Number of days, he has to wait: ", days)
4.930090000000007 Number of days, he has to wait: 33
Footnotes:
In [ ]:
^{1} I am thankful to Dr. Hanno Baehr who introduced me to the problem of "Random extraction" when participating in a Python training course in Nuremberg in January 2014. Hanno outlined some bits of the theoretical framework. During a night session in a pub called "Zeit & Raum" (english: "Time & Space") I implemented a corresponding Python program to back the theoretical solution empirically.
