Python：AssertionError

Question

首先，那是我需要测试的代码：

class ParserError(Exception):
    pass

class Sentence(object):

    def __init__(self, subject, verb, object):
self.subject = subject [1]
        self.verb = verb[1]
        self.object = object[1]

def peek(word_list):
    if word_list:
        word = word_list[0]
        return word[0]
    else:
        return None

def match(word_list, expecting):
if word_list:
        word = word_list.pop(0)

        if word[0] == expecting:
            return word
        else:
            return None
    else:
        return None

def skip (word_list, word_type):
    while peek(word_list) == word_type:
        match(word_list, word_type)

def parse_verb(word_list):
    skip(word_list, 'stop')

    if peek(word_list) == 'verb':
        return match(word_list, 'verb')
    else: 
        raise ParserError("Expected a verb next.")


def parse_object(word_list):
    skip(word_list,'stop')
    next = peek(word_list)

    if next == 'noun':
        return match(word_list, 'noun')
    if next == 'direction':
        return match(word_list, 'direction')
    else:
        raise ParserError('Expected a noun or direction next.')

def parse_subject(word_list, subj):
    verb = parse_verb(word_list)
    obj = parse_object(word_list)

    return Sentence(subj, verb, obj) # 执行 class Sentence

def parse_sentence(word_list):
    skip(word_list, 'stop')

    start = peek(word_list)

    if start == 'noun':
        subj = match(word_list, 'noun')
        return parse_sentence(word_list, subj)
    elif start == 'verb':
        # assume the subject is the player then
        return parse_subject(word_list, ('noun', 'player'))
    else:
        raise ParserError("Must start with subject, object, or verb not: %s" % start)

我输入的列表是[（＆＃39;动词＆＃39;，＆＃39;品味＆＃39;），（＆＃39;方向＆＃39;，＆＃39;好＆＃39;）， （＆＃39;名词＆＃39;＆＃39;比萨＆＃39）] 这个列表中的第一个值是（＆＃39;动词＆＃39;，＃39;味道＆＃39;），它的word_type是＆＃39;动词＆＃39;，它适合我在编码中给出的word_type但是我收到了错误 断言错误：无！=＆＃39;动词＆＃39; 我认为这是不合理的。

def test_skip():
    skipA = skip([('verb','taste'),('direction','good'),('noun','pizza')], 'verb')
    assert_equal(skipA, 'verb')
#    assert_equal(skipA, None)

Answer 1

函数skip不返回任何值（这意味着它返回None）。

所以skipA是None，断言失败。

Answer 2

可能存在问题＆＃39;跳过＆＃39;方法，因为＆＃39;跳过＆＃39;没有返回任何意味着跳过＆＃39;返回无

Answer 3

您需要在跳过功能中返回。你在偷看和匹配中返回，但是你调用skip并从那里想要一个值的范围意味着你也需要返回那里。