我不希望为每个“i”值打印 if 语句

Question

def main():

    theList = list()
    keepGoing = 1
    sizeOfList = 0 

    while keepGoing == 1:

        if sizeOfList == 0:
            thisValue = int(input('Enter first number: '))
        else:
            thisValue = int(input('\nEnter next number: '))

        theList.append(thisValue)
        sizeOfList += 1
        keepGoing = int(input('\nGot it. Enter another number? (enter 1 for yes): '))
  
    print('Here is your list: ',theList)

    for i in range(len(theList)):
        if i < 0: 
            largeNeg = max([i for i in theList if i < 0]) # gets largest negative number
            print('Largest negative number (closest to zero) in that list is: ',largeNeg)
        else:  
            print('There are no negative #s in that list')

main()

我想打印 theList 中最大的负数，但是如果 theList 中没有任何负数，那么我希望我的程序告诉用户列表中没有任何负数。

Answer 1

这是一个简单的重构，使其更加精简和惯用。

def main():
    theList = []
    negmax = 0
    while True:
        if not theList:
            thisValue = int(input('Enter first number: '))
        else:
            thisValue = int(input('\nEnter next number: '))
        if thisValue < 0:
            if not negmax or thisValue > negmax:
                negmax = thisValue
        theList.append(thisValue)
        keepGoing = int(input('\nGot it. Enter another number? (enter 1 for yes): '))
        if keepGoing != 1:
            break
  
    print('Here is your list: ', theList)

    if negmax:
        print('Largest negative number (closest to zero) in that list is: ',largeNeg)
    else:  
        print('There are no negative #s in that list')

main()

输入循环确实很笨重。也许只要用户输入数字就可以继续。

    while True:
        if not theList:
            userinput = input('Enter first number: ')
        else:
            userinput = input('Enter next number: ')
        try:
            thisValue = int(userinput)
        except ValueError:
            exitprompt = input('Not a number. Done? [Y/n] ')
            if not exitprompt.startswith(('N', 'n')):
                break
            continue
        theList.append(thisValue)