循环数索引； 枚举（）

Question

I'm just going through the exercise in my python book and I'm struggling with the finish.

In the beginning, I was supposed to make a list with a series of 10 numbers and five letters. Randomly select 4 elements and print the message about the winning ticket.

 from random import choice, branding numbers_and_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', '1', '2', '3', '4', '5'] print("If you've got those 4 numbers or letters you've won,:!") for i in range(1, 5): print(choice(numbers_and_letters))

Then I suppose to make a loop to see how hard it might be to win the kind of lottery I just created. I need to make a list called my_ticket then write a loop that keeps pulling numbers until the ticket wins. Print a message reporting how many times the loop had to tun to give a winning ticket.

 from random import choice from itertools import count numbers_and_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', '1', '2', '3', '4', '5'] winning_numbers = [] my_ticket = ['a','1', '5', 'j'] for i in count(): #infinite loop for i in range(1, 5): # four elements from the numbers_and_letters list i = choice(numbers_and_letters) winning_numbers.append(i) print(winning_numbers) if sorted(winning_numbers):= sorted(my_ticket). winning_numbers:clear() elif sorted(winning_numbers) == sorted(my_ticket): print('The numbers are identical') break

The only thing I need to do is to count how many times the code integrated through the loop. I know I need to use enumerate(), however, I'm not sure how to apply it to my code.

Answer 1

The enumerate() function takes a list and makes tuples from all of the elements in the list like [(0, elem1), (1, elem2), ...] .

So you can use it to count how many times the loop has run like this:

for index, i in enumerate(count()):   #infinite loop
    for i in range(1, 5):   # four elements from the numbers_and_letters list
        i = choice(numbers_and_letters)
        winning_numbers.append(i)
    print(winning_numbers)
    
    if sorted(winning_numbers) != sorted(my_ticket):
        winning_numbers.clear()
    elif sorted(winning_numbers) == sorted(my_ticket):
        print('The numbers are identical')
        print('It took %d runs!' % (index + 1))
        break

Index + 1 because the indices start with 0.

Answer 2

You can use a temproary counter cnt to keep track on how many times your loop pulls the elements from the list and match it with the ticket First initialise it with 0 outside the loop

Then increment cnt each time your loop runs, if a match is found print(cnt) and then afterwards again set cnt to 0(inside if) for next iteration

Answer 3

No need to use enumerate. 'Count' is fine.

Here's a slight modification to the code that makes it work:

from random import choice
from itertools import count

numbers_and_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j',
                       '1', '2', '3', '4', '5']
winning_numbers = []
my_ticket = ['a','1', '5', 'j']

def get_random_ticket():
    res = []
    for i in range(4):   # four elements from the numbers_and_letters list
        i = choice(numbers_and_letters)
        res.append(i)
    return res

winning_numbers = get_random_ticket()
print(winning_numbers)

for i in count():   #infinite loop
    if sorted(winning_numbers) == sorted(get_random_ticket()):
        print('The numbers are identical')
        print(i) 
        break

Answer 4

You are already counting the number of iterations it takes you to find a match with your loop on count() . But you're overwriting the value you get from that loop with other values later on, since you are reusing the variable name i several times.

Try using different names for the variables in these three lines, rather than reusing i :

for draw_count in count():
    for character_count in range(1, 5):
        character = choice(numbers_and_letters)

Now you can use draw_count later on. You probably want to print out draw_count + 1 , since the itertools.count iterator starts at zero by default.

Answer 5

To count the number of times the code integrated through the loop, just add a counter.

from random import choice
from itertools import count

numbers_and_letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j',
                       '1', '2', '3', '4', '5']
winning_numbers = []
my_ticket = ['a','1', '5', 'j']

a = 0
for i in count():   #infinite loop
    for i in range(1, 5):   # four elements from the numbers_and_letters list
        i = choice(numbers_and_letters)
        winning_numbers.append(i)
    print(winning_numbers)
    a += 1
    if sorted(winning_numbers) != sorted(my_ticket):
        winning_numbers.clear()
    elif sorted(winning_numbers) == sorted(my_ticket):
        print('The numbers are identical')
        print(a)
        break    

This is the output (dots indicate there are several printed winning_numbers)-

......
......
......
......
['4', 'd', 'h', '3']
['h', 'd', 'h', 'j']
['5', 'd', '5', 'c']
['1', '5', 'j', 'a']
The numbers are identical
2852