Question

我想限制用户输入1 - 223，不包括＆＃34; 3,7,10,12和＃34; 当用户输入其中一个号码时，我想打印错误并要求重新提升用户的条目..我在列出排除的号码和重新编码代码时遇到了问题。

这是我停止的地方

for i in range(numRan):
    ranNums.append(int(raw_input( "Range %d Number ? (1-223)\n" % (i+1) )))
    if i in []:
        print "Rang is not allowed!"

Answer 1

这样做 - 包括在输入无效时重新输入：

allowed_nums = set(range(1,224)) - {3,7,10,12}
for i in range(numRan):
    while True:
        inpt = int(raw_input("Range %d Number? (1-223)\n" % (i+1)))
        if inpt in allowed_nums:
            ranNums.append(inpt)
            break
        print "Range is not allowed!"

所以这基本上起作用的方式是，对于每次迭代或输入，我们创建一个无限循环，只有当输入有效时才会中断，从而进入下一个输入迭代。

Answer 2

试试这个：

not_allowed = {3,7,10,12}
for i in range(numRan):
    while True: # to repeat raw_input if the input is not_allowed
        val = int(raw_input( "Range %d Number ? (1-223)\n" % (i+1) )))
        if val in not_allowed or val < 1 or val > 223:
            print "Rang is not allowed!"
        else:
            ranNums.append(val)
            break

Answer 3

我通常会为必须重新提示用户的输入编写一个函数

def limited_input(prompt="",include_only=None,not_allowed=None,not_allowed_err="",cast=None,wrongcasterr=""):
    if not_allowed is None:
        not_allowed = set()
    in_ = None
    while in_ is None:
        in_ = raw_input(prompt)
        if cast:
            try: in_ = cast(in_)
            except ValueError:
                print wrongcasterr
                in_ = None
                continue
        if (include_only and in_ not in include_only) or in_ in not_allowed:
            print not_allowed_err
            in_ = None
    return in_

然后你可以像往常一样使用它：

for i in range(numRan):
    ranNums.append(limited_input(prompt="Range %d Number ? (1-223)\n" % (i+1),
                                 include_only = range(1,224),
                                 not_allowed = {3,7,10,12},
                                 not_allowed_err = "You cannot use 3,7,10,12"
                                 cast = int
                                 wrongcasterr = "You must enter an integer"))

这个解决方案非常便于携带，我通常将它包含在我的python库中的某个模块或其他模块中。在这个系统上它位于utils.inpututils但是YMMV

Answer 4

您可能更容易分离要求函数的函数和创建数字列表层，如下所示：

restrict = [3,7,10,12]

# This function returns an input number with the right 
# characteristics. 
def strictInput():
    v = int(raw_input('Number?'))
    if v >0 and v<224 and v not in restrict: return v
    print 'Illegal number ..., try again'
    return strictInput()

# Now call this function however many times you need ...
inputNumbers = [strictInput() for i in range(numRan)]

Python输入限制和重新定位

4 个答案: