Question

我刚刚开始学习Python，并对我正在阅读的教科书中出现的练习提出疑问。

我发现了一个功能正常的解决方案，但我想知道是否有更简单/推荐的解决方案？

问题：

numXs = int(input('How many times should I print the letter X? '))
toPrint = ' '
#concatenate X to print numXs times print(toPrint)

我的解决方案：

numXs = int(input('How many times should I print the letter X? '))
toPrint = ''
while (toPrint == ''):
    if numXs == 0:
        numXs = int(input('Enter an integer != 0 '))
    else:
        toPrint = abs(numXs) * 'X'
print(toPrint)

Answer 1

我建议您事先处理所有数据检查和纠正，这样您的实际算法就会简单得多：

numXs = 0

while numXs <= 0:
    numXs = int(input('How many times should I print the letter X? '))
    if numXs <= 0:
        print('Enter an integer > 0')

print('X' * numXs)

Answer 2

一个简单的解决方案是

try:
    print("x" * int(input("how many times?")))
except:
    print("you entered an invalid number")

如果您想要零或负面检查

try:
    num = int(input("how many times?"))
    if num > 0:
        print("x" * num)
    else:
        print("need a positive integer")
except:
    print("not a number")

如果您希望永远发生这种情况，只需将其包裹在while循环中

while True:
    #code above

Answer 3

您没有处理错误的转换（输入“Eight”） - 因此您可以将其缩短为：

print(abs(int(input("How many?")))*"X") #loosing the ability to "requery" on 0

相同的功能：

numXs = abs(int(input('How many times should I print the letter X? ')))
while not numXs: # 0 is considered FALSE
    numXs = abs(int(input('Enter an integer != 0 ')))

print(numXs * 'X')

更安全的：

def getInt(msg, errorMsg):
    '''Function asks with _msg_ for an input, if convertible to int returs
        the number as int. Else keeps repeating to ask with _errorMsg_ until
        an int can be returned'''
    i = input(msg)
    while not i.isdigit():
        i = input(errorMsg)

    return int(i)

我正在使用isdigit()来确定input是否是我们可以使用的数字。您还可以在try:周围执行except:和int(yourInput)，这样您就可以捕获不可以转换为整数的输入：

def getIntWithTryExcept(msg, errorMsg):
    '''Function asks with _msg_ for an input, if convertible to int returs
        the number as int. Else keeps repeating to ask with _errorMsg_ until
        an int can be returned'''
    i = input(msg)  # use 1st input message
    num = 0
    try:
        num = int(i)     # try convert, if not possible except: will happen
    except:
        while not num:   # repeat with 2nd message till ok
            i = input(errorMsg)
            try:
                num = int(i)
            except:      # do nothing, while is still True so it repeats
                pass

    return int(i) # return the number

循环连接练习

3 个答案: