Question

我试图写一个简单的if语句，它取决于一组条件（列表中出现的是0）但是当我的两个条件都是0时，我遇到了错误，只有一个显示语句。

if let1 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
elif let2 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
elif let3 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")

我已尝试使用或声明，但无法产生所需的输出，通知用户他们的输入中的一个，两个或三个可能不在我程序的其余部分中使用的文本正文中。

对于缺乏succint术语而道歉，因为我对编程很陌生。

由于

编辑：感谢您的回复，但我应该给予全部功能......

def segment_sequence(mycorpus, letter1, letter2, letter3):
    corpus_string = "".join(mycorpus)
    a = letter1
    b = letter2
    c = letter3
    d = a + b
    e = b + c
    let1 = corpus_string.count(a)
    let2 = corpus_string.count(b)
    let3 = corpus_string.count(c)
    let1let2 = corpus_string.count(d)
    let2let3 = corpus_string.count(e)
    print(let1,let2,let3)
    if let1 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
    if let2 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
    if let3 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")
    break
    else:
        trans_prob1 = let1let2/let1
        trans_prob2 = let2let3/let2
        if trans_prob1 > trans_prob2:
            print('Here is the proposed boundary given the training corpus:\n')
            print('Proposed end of one word:' ,d,'\n')
            print('Proposed beginning of new word:' ,c,'\n')
        elif trans_prob2 > trans_prob1:
            print('Here is the proposed boundary given the training corpus:\n')
            print('Proposed end of one word: ',e,'\n')
            print('Proposed beginning of new word: ',a,'\n')
        elif trans_prob1 == trans_prob2:
            print('Both ',d,' and ',e,' are equally possible word endings')
        return()

使用if而不是elif的问题是它没有中断，然后我得到除以0的错误。我没有找到任何过于复杂的简单而准确的用户错误消息。

再次感谢您

Answer 1

使用elif尝试使用if，如下所示：

if let1 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
if let2 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
if let3 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")

Answer 2

如果您希望针对每种情况测试条件，请不要使用elif：

此外，为了确保您的功能在触发案例时中止，这会导致除数为零，我在测试电池后添加了一个标志和一个测试以退出。

def segment_sequence(mycorpus, letter1, letter2, letter3):
    corpus_string = "".join(mycorpus)
    a = letter1
    b = letter2
    c = letter3
    d = a + b
    e = b + c
    let1 = corpus_string.count(a)
    let2 = corpus_string.count(b)
    let3 = corpus_string.count(c)
    let1let2 = corpus_string.count(d)
    let2let3 = corpus_string.count(e)
    print(let1,let2,let3)

    must_abort = False

    if let1 == 0:
       print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
        must_abort = True
    if let2 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
        must_abort = True
    if let3 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")
        must_abort = True

    if must_abort:
        return

    # no need for else any longer here.
    trans_prob1 = let1let2/let1
    trans_prob2 = let2let3/let2
    if trans_prob1 > trans_prob2:
        print('Here is the proposed boundary given the training corpus:\n')
        print('Proposed end of one word:' ,d,'\n')
        print('Proposed beginning of new word:' ,c,'\n')
    elif trans_prob2 > trans_prob1:
        print('Here is the proposed boundary given the training corpus:\n')
        print('Proposed end of one word: ',e,'\n')
        print('Proposed beginning of new word: ',a,'\n')
    elif trans_prob1 == trans_prob2:
        print('Both ',d,' and ',e,' are equally possible word endings')
    return

现在，无论先前的条件是否得到满足，每个案例都会被评估并采取行动。

重构主函数之外的中止案例：

def segment_sequence(mycorpus, letter1, letter2, letter3):
    corpus_string = "".join(mycorpus)
    a = letter1
    b = letter2
    c = letter3
    d = a + b
    e = b + c
    let1 = corpus_string.count(a)
    let2 = corpus_string.count(b)
    let3 = corpus_string.count(c)
    let1let2 = corpus_string.count(d)
    let2let3 = corpus_string.count(e)
    print(let1,let2,let3)

    if must_abort(let1, let2, let3):
        return

    trans_prob1 = let1let2/let1
    trans_prob2 = let2let3/let2
    if trans_prob1 > trans_prob2:
        print('Here is the proposed boundary given the training corpus:\n')
        print('Proposed end of one word:' ,d,'\n')
        print('Proposed beginning of new word:' ,c,'\n')
    elif trans_prob2 > trans_prob1:
        print('Here is the proposed boundary given the training corpus:\n')
        print('Proposed end of one word: ',e,'\n')
        print('Proposed beginning of new word: ',a,'\n')
    elif trans_prob1 == trans_prob2:
        print('Both ',d,' and ',e,' are equally possible word endings')
    return


def must_abort(let1, let2, let3):
    abort = False
    if let1 == 0:
       print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
        abort = True
    if let2 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
        abort = True
    if let3 == 0:
        print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")
        abort = True
    return abort

Answer 3

如果elif的工作方式如此，它会检查条件会在哪里获得真实状态，之后就会破坏。所以你必须使用if。

if let1 == 0:
   print("Segmentation unavailable as corpus does not contain the following letter ",a, "\n")
if let2 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",b, "\n")
if let3 == 0:
    print("Segmentation unavailable as corpus does not contain the following letter ",c, "\n")

压缩if语句以提供更准确的输出

3 个答案:

重构主函数之外的中止案例：