蟒蛇抛硬币

Question

我对 Python 非常陌生，我必须创建一个模拟掷硬币的游戏，并要求用户输入应该掷硬币的次数。基于该响应，程序必须选择一个随机数，该数是 0 或 1（并决定哪个代表“正面”，哪个代表“反面”）指定的次数。计算产生的“正面”数量和“反面”数量，并向用户呈现以下信息：包含模拟抛硬币的列表，以及产生的正面数量和反面数量的摘要。例如，如果用户输入 5，则抛硬币模拟可能会导致 [‘heads’, ‘tails’, ‘tails’, ‘heads’, ‘heads’]。程序应该打印如下内容：“ [‘heads’, ‘tails’, ‘tails’, ‘heads’, ‘heads’]

这是我目前所拥有的，但它根本不起作用......

import random

def coinToss():
    number = input("Number of times to flip coin: ")
    recordList = []
    heads = 0
    tails = 0
    flip = random.randint(0, 1)
    if (flip == 0):
        print("Heads")
        recordList.append("Heads")
    else:
        print("Tails")
        recordList.append("Tails")
    print(str(recordList))
    print(str(recordList.count("Heads")) + str(recordList.count("Tails")))

Answer 1

您需要loop 来执行此操作。我建议使用for 循环：

import random
def coinToss():
    number = input("Number of times to flip coin: ")
    recordList = []
    heads = 0
    tails = 0
    for amount in range(number):
         flip = random.randint(0, 1)
         if (flip == 0):
              print("Heads")
              recordList.append("Heads")
         else:
              print("Tails")
              recordList.append("Tails")
    print(str(recordList))
    print(str(recordList.count("Heads")) + str(recordList.count("Tails")))

我建议你阅读this on for loops。

另外，您可以将number 作为parameter to the function 传递：

import random
def coinToss(number):
    recordList, heads, tails = [], 0, 0 # multiple assignment
    for i in range(number): # do this 'number' amount of times
         flip = random.randint(0, 1)
         if (flip == 0):
              print("Heads")
              recordList.append("Heads")
         else:
              print("Tails")
              recordList.append("Tails")
    print(str(recordList))
    print(str(recordList.count("Heads")) + str(recordList.count("Tails")))

那么，最后需要调用函数：coinToss()。

Answer 2

你快到了：

1) 你需要调用函数：

coinToss()

2) 你需要设置一个循环来重复调用random.randint()。

Answer 3

我会选择以下内容：

from random import randint
num = input('Number of times to flip coin: ')
flips = [randint(0,1) for r in range(num)]
results = []
for object in flips:
        if object == 0:
            results.append('Heads')
        elif object == 1:
            results.append('Tails')
print results

Answer 4

这可能更符合 Python 风格，尽管不是每个人都喜欢列表推导式。

import random

def tossCoin(numFlips):      
    flips= ['Heads' if x==1 else 'Tails' for x in [random.randint(0,1) for x in range(numflips)]]
    heads=sum([x=='Heads' for x in flips])
    tails=numFlips-heads

Answer 5

import random
import time



flips = 0
heads = "Heads"
tails = "Tails"

heads_and_tails = [(heads),
                   (tails)]
while input("Do you want to coin flip? [y|n]") == 'y':
    print(random.choice(heads_and_tails))
    time.sleep(.5)
    flips += 1


else:
    print("You flipped the coin",flips,"times")
    print("Good bye")

你可以试试这个，我有它，所以它会询问你是否要掷硬币，然后当你说“不”或 n 时，它会告诉你掷硬币的次数。 （这是在 python 3.5 中）

Answer 6

创建一个包含head和tail两个元素的列表，并使用choice() from random得到掷硬币的结果。要获取 head 或 tail 出现的次数，请将计数附加到列表中，然后使用集合中的 Counter(list_name) 。使用 uin() 调用

##coin flip
import random
import collections
def tos():
    a=['head','tail']
    return(random.choice(a))
def uin():
    y=[]
    x=input("how many times you want to flip the coin: ")
    for i in range(int(x)):
        y.append(tos())
    print(collections.Counter(y))

Answer 7

您可以这样做：

import random
options = ['Heads' , 'Tails']
number = int(input('no.of times to flip a coin : ')
for amount in range(number):
   heads_or_tails = random.choice(options)
   print(f" it's {heads_or_tails}")
print()
print('end')

Answer 8

我是这样做的。可能不是最好和最有效的方法，但是现在您有不同的选择可供选择。我做了 10000 次循环，因为在练习中已经说明了这一点。

#Coinflip program
import random

numberOfStreaks = 0
emptyArray = []
for experimentNumber in range(100):
#Code here that creates a list of 100 heads or tails values
headsCount = 0
tailsCount = 0
#print(experimentNumber)
for i in range(100):
    if random.randint(0, 1) == 0:
        emptyArray.append('H')
        headsCount +=1
    else:
        emptyArray.append('T')
        tailsCount += 1   

#Code here that checks if the list contains a streak of either heads or tails of 6 in a row
heads = 0
tails = 0
headsStreakOfSix = 0
tailsStreakofSix = 0

for i in emptyArray:
    if i == 'H':
        heads +=1
        tails = 0
        if heads == 6:
            headsStreakOfSix += 1
            numberOfStreaks +=1
            
    if i == 'T':
        tails +=1
        heads = 0
        if tails == 6:
            tailsStreakofSix += 1
            numberOfStreaks +=1
            
#print('\n' + str(headsStreakOfSix))
#print('\n' + str(tailsStreakofSix))
#print('\n' + str(numberOfStreaks))
print('\nChance of streak: %s%%' % (numberOfStreaks / 10000))

Answer 9

#program to toss the coin as per user wish and count number of heads and tails
import random
toss=int(input("Enter number of times you want to toss the coin"))
tail=0
head=0
for i in range(toss):
    val=random.randint(0,1)
    if(val==0):
        print("Tails")
        tail=tail+1
    else:
        print("Heads")
        head=head+1
print("The total number of tails is {} and head is {} while tossing the coin {} times".format(tail,head,toss))
    

Answer 10

解决眼前的问题

最高投票的答案实际上并没有运行，因为它将一个字符串传递给range()（而不是一个 int）。

这是一个解决两个问题的解决方案：刚刚提到的range() 问题，以及在最后两行的print() 语句中对str() 的调用可以变得多余。编写这个 sn-p 是为了尽可能少地修改原始代码。

def coinToss():
    number = int(input("Number of times to flip coin: "))
    recordList = []
    heads = 0
    tails = 0
    for _ in range(number):
        flip = random.randint(0, 1)
        if (flip == 0):
            recordList.append("Heads")
        else:
            recordList.append("Tails")
    print(recordList)
    print(recordList.count("Tails"), recordList.count("Heads"))

---

更简洁的方法

但是，如果您正在寻找更简洁的解决方案，您可以使用列表推导式。只有一个其他答案具有列表理解，但您可以使用一个而不是两个列表理解嵌入从 {0, 1} 到 {"Heads", "Tails"} 的映射：

def coinToss():
    number = int(input("Number of times to flip coin: "))
    recordList = ["Heads" if random.randint(0, 1) else "Tails" for _ in range(number)]
    print(recordList)
    print(recordList.count("Tails"), recordList.count("Heads"))

Answer 11

import random

def coinToss(number):
    heads = 0
    tails = 0
    for flip in range(number):
        coinFlip = random.choice([1, 2])

        if coinFlip == 1:
            print("Heads")
            recordList.append("Heads")
        else:
            print("Tails")
            recordList.append("Tails")

number = input("Number of times to flip coin: ")
recordList = []
if type(number) == str and len(number)>0:
    coinToss(int(number))
    print("Heads:", str(recordList.count("Heads")) , "Tails:",str(recordList.count("Tails")))

Answer 12

投掷 N 枚硬币的所有可能性

def Possible(n, a):
    if n >= 1:
        Possible(n // 2, a)
    z = n % 2
    z = "H" if z == 0 else "T"
    a.append(z)
    return a


def Comb(val):
    for b in range(2 ** N):
        A = Possible(b, [])
        R = N - len(A)
        c = []
        for x in range(R):
            c.append("H")
        Temp = (c + A)
        if len(Temp) > N:
            val.append(Temp[abs(R):])
        else:
            val.append(Temp)
    return val


N = int(input())
for c in Comb([]):
    print(c)

Answer 13

heads = 1
tails = 0 

input("choose 'heads' or 'tails'. ").upper()

random_side = random.randint(0, 1)

if random_side == 1:
    print("heads you win")
else:
    print("sorry you lose ")