十个绿瓶子程序python

Question

我正在用Python编写一个程序，该程序根据用户输入的瓶子数量打印十个绿色瓶子程序。这就是我到目前为止所拥有的。

x=input("How many bottles to start with?\n")                                                    
while x > 0:                                                    
  if(x==1):                                                 
    print (str(x) + " "+"green bottle, hanging on the wall")                                                    
    print (str(x) + " "+ "green bottle hanging on the wall")                                                    
    print ("and if one green bottle, should accidentally falls")                                                    
    x = x - 1                                                   
    print("there'd be no green bottle hanging on the wall")                                                 
    print("\n")                                                 
  elif(x==2):                                                   
    print (str(x) + " "+"green bottles, hanging on the wall")                                                   
    print (str(x) + " "+ "green bottles hanging on the wall")                                                   
    print ("and if one green bottle should accidentally fall")                                                  
    x = x - 1                                                   
    print ("there'd be "  + str (x)  + " green bottle, hanging on the wall")                                                    
    print("\n")                                                 
  else:                                                 
    print (str(x) + " "+"green bottles, hanging on the wall")                                                   
    print (str(x) + " "+ "green bottles hanging on the wall")                                                   
    print ("and if one green bottle should accidentally falls")                                                 
    x = x - 1                                                   
    print ("there'd be "  + str (x)  + " green bottles, hanging on the wall")                                                   
    print("\n")                                                 

问题1:我还需要将“1”改为“1”。我正在考虑写一个函数，因为我只需要把1改成10。请告诉我怎么做。

Answer 1

问题1这可以通过一条if语句简单地完成。由于您提到您只需要1到10，因此语句应如下所示

x = int(input("How many bottles to start with? "))
if (x > 10 or x < 1):
    print(f"Can't start with {x} bottles. Pick a number between 1 and 10.")
    exit()

问题2可以用字典来完成的。

map_to_str = {
    0: "NO",
    1: "ONE",
    2: "TWO",
    3: "THREE",
    4: "FOUR",
    5: "FIVE",
    6: "SIX",
    7: "SEVEN",
    8: "EIGHT",
    9: "NINE",
    10: "TEN"
}

代码的另一个问题是大量代码不必要地重复。首先，我将使用for循环。我还会写一个函数，以字符串的形式返回瓶子的数量。

def bottles_to_str(num_of_bottles):
    return f"{map_to_str[num_of_bottles]} green bottle" + "s"*(num_of_bottles!=1)

此函数将number作为参数，并使用字典将其转换为字符串。它还检查瓶子的数量是否不等于1 (False也可以解释为0，True也可以解释为1)。如果瓶子数不是1，它的计算结果是"s"*1，它是's'，否则它的计算结果是"s"*0，它是空字符串''。

主循环将如下所示

for i in range(x, 0, -1):
    print(f"{bottles_to_str(i)} hanging on the wall,")
    print(f"{bottles_to_str(i)} hanging on the wall,")
    print(f"And if one green bottle should accidentally fall,")
    print(f"There'll be {bottles_to_str(i-1)} hanging on the wall\n")

Answer 2

要将数字转换为相应的英语单词，建议使用字典：

x = int(input("How many bottles to start with?\n")) # Convert the input to an int
int_to_str = {
1: "one",
2: "two",
...
10: "ten"
}
print ("there'd be "  + int_to_str[x] + " green bottles, hanging on the wall")

Answer 3

toAlpha = {0 : "no",
           1 : "one",
           2 : "two",
           3 : "three",
           4 : "four",
           5 : "five",}  # And so on..

def bottleORbottles(x):
    if x > 1:
        return "bottles"
    return "bottle"

x=input("How many bottles to start with?\n")
if x == 0 :
    print("WE DONT HAVE ANY BOTTLES..")
    
while x > 0:
    print((toAlpha[x] + " green " + bottleORbottles(x) + " , hanging on the wall \n") *2)
    print ("and if one green bottle, should accidentally fall")
    x = x - 1
    print("there'd be " + toAlpha[x] + " green  " + bottleORbottles(x) + " hanging on the wall \n")

看看如何通过使用小函数来消除冗余。