Coin Flip Streak from Automate the Boring Stuff with Python

Question

I apologize beforehand if the question has been repeated so many times. This is a practice task from Automate the Boring Stuff with Python. In brief, the task entails writing a code that carries out an experiment of checking if there is a streak of 6 'heads' or 'tails' in 100 coin tosses, then replicates it 10,000 times and gives a percentage of the success rate.

When I wrote the code, I tried to be different by making the code applicable to any streaks in a number of predetermined experiments (in my case, the sample was 1 million coin tosses). I also tried to find the longest streak possible in that said experiment.

I also want to apologize beforehand that the comments were awfully verbose.

import random, copy, time

def torai(seq,pop): # seq is for #=streak, pop is for total sample/population/experiment
    # Creating a random chance of heads and tails
    tosses = []
    for i in range(pop):
        tosses.append(random.randint(1,2)) # 1 and 2 for head and tail, and vice versa

    # Defining initial values for the main loop
    streak = 0 # Iterated streak
    curlongstr = 0 # Current longest streak
    longeststr = 0 # Longest streak evaluated
    peak = [] # Record local streaks from 'tosses' list

    # The main loop
    for i in range(len(tosses)): # Looping based on list indexes
        if i == 0:  # Conditional for preventing tosses[0] == tosses[-1]
            continue

        elif tosses[i] == tosses[i-1]: # Conditional for checking if an i element has the same value as the previous element value, i-1
            streak += 1 # Adding tally mark if the line above is fulfilled
            if i == len(tosses)-1: # A nested conditional for adding the last tally mark from 'tosses' into the overall list of steaks 'peak', see lines 27-33
                peak.append(streak)

        elif tosses[i] != tosses[i-1]: # Conditional for checking if an i element value is different than the previous element value, i-1
            curlongstr = copy.copy(streak) # Creating a variable by returning a copy of streak before it resets to 0, see line 31
            if curlongstr > longeststr: # A nested conditional for comparing the current longest streak and the longest streak that has happened when looping the 'tosses' list
                longeststr = curlongstr
            streak = 0 # This is where streaks ended and then resets to 0, so before that, the value of the streak is copied first, see line 28
            if curlongstr > streak: # After streak is reset to 0, the value of current long streak is compared to 0, so that we create a list of streaks from 'tosses' list
                peak.append(curlongstr)

    truepeak = []
    for i in peak: # Example: a 2-streak is equal to either [1,1,1] or [2,2,2], a 4-streak is either [1,1,1,1,1] or [2,2,2,2,2]
        truepeak.append(i+1)

    apr = []
    # Loop for finding how many #-streaks happened
    for i in truepeak:
        if i == seq:
            apr.append(i)

    print('%s-streak count: ' %seq, len(apr)) # Total of #-streaks happened in 'tosses' list
    print('%s-streak prob (percent): ' %seq, (len(apr)/pop)*100) # Calculating probability if how many #-streak happened in given n times tosses
    print('longest streak: ',longeststr + 1) # Similar reason as line 36
    print('process time: ',time.process_time(), 'second\n')

    return (len(apr)/pop)*100

x = torai(2,1000000)
y = torai(6,1000000)
z = torai(10,1000000)
print(x, y, z)

I tried to increase the sample to 10 million coin tosses. However, the program will run 9-10 slower each time the function was called.

My request are can anyone check whether if the result (probability of n-streak) is right or not and are there any ways to make the code and process time shorter?

Answer 1

Bugs

torai(1, 10000)

This should print something around 50 %, since it's the individual count. But instead, it prints

1-streak count:  0
1-streak prob (percent):  0.0
longest streak:  19
process time:  0.046875 second

---

Avoid too many comments

There are too many comments in your code, which makes the code look unnecessarily convoluted. What I recommend is the usage of docstrings. IMO It isn't very important here, but its better than a million comments

def torai(seq,pop): 
    tosses = []
    for i in range(pop):
        tosses.append(random.randint(1,2))
    streak = 0
    curlongstr = 0
    longeststr = 0
    peak = []
    for i in range(len(tosses)): 
        if i == 0:  
            continue
        elif tosses[i] == tosses[i-1]: 
            streak += 1 
            if i == len(tosses)-1: 
                peak.append(streak)

        elif tosses[i] != tosses[i-1]: 
            curlongstr = copy.copy(streak) 
            if curlongstr > longeststr: 
                longeststr = curlongstr
            streak = 0 
            if curlongstr > streak: 
                peak.append(curlongstr)

    truepeak = []
    for i in peak: 
        truepeak.append(i+1)

    apr = []
    

    for i in truepeak:
        if i == seq:
            apr.append(i)

    print('%s-streak count: ' %seq, len(apr)) 
    print('%s-streak prob (percent): ' %seq, (len(apr)/pop)*100) 
    print('longest streak: ',longeststr + 1) 
    print('process time: ',time.process_time(), 'second\n')

    return (len(apr)/pop)*100

---

Simplify #1

    for i in range(len(tosses)): 
        if i == 0:  
            continue

It's clear to me that you want to skip the first element. In that case, you can specify the starting point for range()

    for i in range(1, len(tosses)): 

---

Simplify #2

    for i in range(pop):
        tosses.append(random.randint(1,2))

Since this is going to be an immutable sequence, use a tuple, with a generator

tosses = tuple(random.randint(1, 2) for _ in range(pop)

---

Simplify #3

            if curlongstr > longeststr:
                longeststr = curlongstr

Your condition is simple. The new value is always the larger of the two
Just use the max() function

            longeststr = max(longeststr, curlongstr)

---

Simplify #4

truepeak = []
    for i in peak:
        truepeak.append(i+1)

You're creating an entirely new list, and fill it up with the exact same elements as peak except with a constant 1 added to them. Very inefficient. Either add the values with the +1 from the beginning or use the +1 where necessary.

    for i in peak:
        if i + 1 == seq:
            apr.append(i + 1)

But again, all you do with apr is get its length, so there's absolutely no point in maintaining so many lists when all you have to do is keep a counter. That also removes the need to maintain peak

---

Calculate tosses as you go

After removing all of the previous loops, there will still be 2 left. One for calculating the tosses and the other goes through them to calculate them. What I propose is, go through it only once, and keep track of two things. The current flip and the previous flip

def torai(seq, iterations ):
    total_streaks = 0

    previous_flip = random.randint(1, 2)
    for _ in range(1, iterations):
        current_flip = random.randint(1, 2)

        if current_flip == previous_flip:
            total_streaks += 1

        # other calculations

        current_flip = previous_flip

    print(f"Total streaks: {total_streaks}")