Python3.x-如何获取整数变量并将其转换为大量输入？

假设您有这样一个代码：

shape= int(input("How many sides are in your shape?: "))
print("So you have", shape,"number of sides.")

>>>
How many sides are in your shape?: 3
So you have 3 number of sides.
What is the 1st letter in your shape's name?: A 
# 'That is an input that I cannot figure out'
What is the 2nd letter in your shape's name?: B 
# 'That is an input that I cannot figure out'
What is the 3rd letter in your shape's name?: C
# 'That is an input that I cannot figure out'
So your shape's name is ABC. 

现在我想做的是获取用户输入的内容（当然是整数），并根据用户刚刚输入的内容创建大量输入。大概是这样的：

shape= int(input("How many sides are in your shape?: "))
print("So you have", shape,"number of sides.")

>>>
How many sides are in your shape?: 3
So you have 3 number of sides.
What is the 1st letter in your shape's name?: A 
# 'That is an input that I cannot figure out'
What is the 2nd letter in your shape's name?: B 
# 'That is an input that I cannot figure out'
What is the 3rd letter in your shape's name?: C
# 'That is an input that I cannot figure out'
So your shape's name is ABC. 

此外，如果改变输入数量的“形状”变量也改变了。因此：

>>>
How many sides are in your shape?: 5
So you have 3 number of sides.
What is the 1st letter in your shape's name?: E
# 'That is an input that I cannot figure out'
What is the 2nd letter in your shape's name?: F
# 'That is an input that I cannot figure out'
What is the 3rd letter in your shape's name?: H
# 'That is an input that I cannot figure out'
What is the 4th letter in your shape's name?: G
# 'That is an input that I cannot figure out'
So your shape's name is EFHG. 

---

如果序号对输出很重要（“第一个字母”、“第二个字母”、“第三个字母”等），那么我会：

def to_ordinal(n:int) -> str:
    """takes a number and returns its ordinal

    >>> to_ordinal(1)
    1st
    >>> to_ordinal(155)
    155th
    """

    endings = {1: "st", 2: "nd", 3: "rd"}

    if 11 <= n < 20:
        return str(n) + "th"
    else:
        return str(n) + endings.get(n % 10, "th")

---

使用shape迭代并附加到字符串

# Create an empty string.
name = ""
# For indices 0 through the length of shape minus 1,
for i in range(shape):
    # request user input for that index and add the letter inputted to the end of the string.
    name = name + input("Input letter %d: " % i)
# Then print the string.
print("\nYour shape's name is " + name + ".")

---

因此，我最终使用@cameronroytaylor和@AdamSmith的响应来创建此代码

shape= int(input("How many sides are in your shape?: "))
print("So you have", shape,"number of sides.")

然后加上这个,

num = 0

looper = shape+1

name = ""

for n in range(looper):
    num += 1
    def to_ordinal(n:int) -> str:

        endings = {1: "st", 2: "nd", 3: "rd"}

        if 11 <= n < 20:
            return str(n) + "th"
        else:
            return str(n) + endings.get(n % 10, "th")

    if num > 1:
        name = name + input(f"What is the {to_ordinal(n)} letter in your first shape's name?: ")
    # the f"{some_var}" syntax only works in py3.6+
    # otherwise use `inputs.append("Name your {} side: ".format(to_ordinal(n))`

print("\nYour first", shape+"'s name is " + name + ".")

num=0
活套=形状+1
name=“”
对于范围内的n（活套）：
num+=1
定义到顺序（n:int）->str:
结尾={1:“st”，2:“nd”，3:“rd”}
如果11 1：
name=name+input（f“第一个形状名称中的{to_序数（n）}字母是什么？：”）
#f“{some_var}”语法仅在py3.6中有效+
#否则使用`inputs.append（“命名{}端：”.format（to_序数（n））`
打印（“\n您的第一个”，shape+”的名称为“+name+”）

---

请检查@cameronroytaylor和@AdamSmith的回答，因为他们帮助得出了这些答案。他们值得表扬。

范围（X）内的构造

，

将在封闭的块

X

上循环。希望这会有帮助，因为我不理解你的例子（你的第二个代码块指出应该有5个提示，但只提示了4次…）出于好奇——为什么要使用字符串呢？@AdamSmith我将这个问题解释为#ObGamErOGE关心将用户输入附加到一个名称中。你的代码非常适合生成输入函数的输入（提示）。使用这两个代码片段#ObGamErOGE应该能够实现他们的目标。这很公平！明智地说：字符串连接效率非常低。如果

形状

变得非常大，这可能会导致问题。如果你积累到一个列表中，那么使用

'.join（该列表）

您以更便宜的价格获得了相同的输出。您是如何做到的。我的意思是它工作得很好，但我不明白为什么。请任何人向我解释它是如何工作的。我编辑了代码，以便有注释解释它是如何工作的。如果有任何不合理的地方，请告诉我，我将为您解释。此外，您可能希望更改将

input（）

参数中的

i

转换为

（i+1）

，如果您希望它从

1

开始，而不是从

0

@fuzzything44开始，感谢您的编辑。看起来我们都错过了一些边缘案例（您的代码捕捉到了“22”，但不正确地给出了“12”）我认为我当前的编辑是正确的。我得到这个错误的原因是：回溯（最近一次调用last）：文件“-------------------------------”，第19行，在范围（1，num_inputs+1）内的n中：TypeError:必须是str，而不是strint@ObGamErOGE哦，因为我很笨，没有将用户输入转换为int:）