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?

• superb rain: What on earth is "torai" supposed to mean?
• yfr: It's just an arbitrary name :), from トライ (try). I'll change the function name to be more understandable and edit other things based on Aryan's input.
• Mast: Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers .

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}")

• lights0123: I disagree with all of your removal of comments—instead, I would take the comment and make that the variable name. For example, curlongstr = 0 # Current longest streak could be replaced with current_longest_streak.
• user228914: @lights0123 A few comments here and there are fine and sometimes needed but in this case, it serves very less purpose, more in making the code less readable.