Python – 有一个变量既是 int 又是 str

Question

这是我的代码：

def retest2():
    print "Type in another chapter title! Or type \"Next\" to move on."
    primenumbers2()
def primenumbers1():
    print "--------------------------------------------------\nChapters in books are usually given the cardinal numbers 1, 2, 3, 4, 5, 6 and so on.\nBut I have decided to give my chapters prime numbers 2, 3, 5, 7, 11, 13 and so on because I like prime numbers.\n\nType in the chapter title of my book (a prime number) and I will tell you what cardinal number the chapter is."
def primenumbers2():
    prime = (str(input("\n")))
    chapter = (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233)

    if "Next" in prime or "next" in prime:
        print "--------------------------------------------------\nOnto the next thing."
    elif int(prime) in chapter:
        print "Chapter ",chapter.index(int(prime)) + 1
        retest2()
    elif prime.isalpha:
        print "That is not one of my chapter numbers because {0} is not a prime number. Try again.".format(prime)
        primenumbers2()

primenumbers1()
primenumbers2()

所以我要做的是让用户输入一个素数，输出是与该素数相关的基数。但是，我希望用户可以通过输入"Next" 或"next" 来选择进入下一个功能。因此，我的变量输入prime 需要同时适应字符串和整数输入。所以我将其设置为(str(input("\n")))，然后在需要时将其转换为int(prime)。

除了"Next" 或"next" 以外的字符串输入时，一切正常。例如，如果我输入“okay”，我会收到错误消息：

File "prime.py", line x, in primenumbers2
    prime = (str(input("\n")))
  File "<string>", line 1, in <module>
NameError: name 'okay' is not defined

但是如果我输入“4”，它不是一个素数，程序就会运行，我得到：

That is not one of my chapter numbers because 4 is not a prime number. Try again.

然后程序循环回到primenumbers2() 以重新启动该功能。

请帮我完成这项工作！

Answer 1

python2 需要使用raw_input，python2 中的input 基本上是eval(raw_input())，并且由于您没有定义变量okay，所以会出现错误。

In [10]: prime = (str(input("\n")))

foo
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-10-15ff4b8b32ce> in <module>()
----> 1 prime = (str(input("\n")))

<string> in <module>()

NameError: name 'foo' is not defined    
In [11]: foo = 1 
In [12]: prime = (str(input("\n")))
foo
In [13]: prime = (raw_input("\n"))
next
In [14]: print prime
next

你应该很少使用input。

您还应该使用 while 循环，如下所示：

 def primenumbers2():
    chapter = (
        2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107,
        109,
        113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233)
    while True:
        prime = input("Type in another chapter title! Or type \"Next\" to move on.")
        if "next"  == prime.lower():
            print("--------------------------------------------------\nOnto the next thing.")
            break
        try:
            p = int(prime)
            if p in chapter:
                print("Chapter {}".format(chapter.index(p) + 1))
            else:
                 print("That is not one of my chapter numbers because {0}"
                  " is not a prime number. Try again.".format(prime))
        except ValueError:
            print("Invalid input")

不完全确定你想做什么，但使用一段时间，检查用户输入是否等于next，如果不尝试使用 try/except 强制转换为 int 是一个更好的主意。

将章节设为字典也是一个更好的主意，通过按键访问章节号：

def primenumbers2():
    chaps = (
        2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107,
        109,
        113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233)

    chapter = dict(zip(chaps, range(1, len(chaps) + 1)))
    while True:
        prime = input("Type in another chapter title! Or type \"Next\" to move on.")
        if "next" == prime.lower():
            print("--------------------------------------------------\nOnto the next thing.")
            break
        try:
            p = int(prime)
            if p in chapter:
                print("Chapter {}".format(chapter[p]))
            else:
                print("That is not one of my chapter numbers because {}"
                      " is not a prime number. Try again.".format(prime))
        except ValueError:
            print("Invalid input")

range(1, len(chaps) + 1)) 表示章节中的每个p 都有一个与其在元组中的索引相对应的值。

Answer 2

您要做的是将用户输入的值转换为字符串或整数。如果您使用 input，正如 Padraic 所说，它等于 eval(raw_input())，您将评估输入，因为它是一个 python 命令：如果您编写任何不存在的名称，这将引发错误。要解决此问题，请使用 raw_input 函数。它将返回一个str 对象，即输入文本。

然后，您要查找此文本是数字还是字符串。一种解决方案是使用异常：

try:
    prime = int(prime)

    # Here the code assuming prime is a number, an `int`
except ValueError:
    # here the code assuming prime is a string, `str`, for example
    # 'Next', 'next' or 'okay'

Answer 3

尝试使用 isdigit 来避免异常：

def primenumbers2():
prime = (str(input("\n")))
chapter = (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233)

if "Next" in prime or "next" in prime:
    print "--------------------------------------------------\nOnto the next thing."
elif prime.isdigit() and int(prime) in chapter:
    print "Chapter ",chapter.index(int(prime)) + 1
    retest2()
else:
    print "That is not one of my chapter numbers because {0} is not a prime number. Try again.".format(prime)
    primenumbers2()