Question

我目前正在经历LPTHW而且我已经达到了excercise 48，这是我第一次碰到砖墙。

这是我给出的测试用例的第一部分

from nose.tools import *
from ex48 import lexicon

def test_direction():
    assert_equal(lexicon.scan("north"), [('direction', 'north')])
    result = lexicon.scan("north south east")
    assert_equal(result, [('direction', 'north'),
                          ('direction', 'south'),
                          ('direction', 'east')])

This question has been asked here before，我注意到目前为止我的解决方案与answer provided by robbyt完全相同。但它仍然无效。

def scan(thewords):

    directions = [('direction', 'north'), ('direction', 'south'), ('direction', 'east')]

    thewords = thewords.split()
    sentence = []

    for i in thewords:
        if i in directions:
            sentence.append(('direction', i))

        else:
            sentence.append(('error', i))


    return sentence

所以问题是：在输入（词）之后，如何正确搜索元组列表然后返回它所属的特定元组？

提前感谢任何答案和建议，真的坚持这个。

Answer 1

受@ Evee的第一个解决方案的启发（谢谢），这是我通过所有测试的解决方案。也许它使用的代码多于第二种解决方案，但它消除了定义的唯一方法之外的循环。

class Lexicon(object):
    def __init__(self):
        self.mapping = {
              'direction':  ['north', 'south', 'east', 'west'],
              'verb':       ['go', 'kill', 'eat'],
              'stop':       ['the', 'in', 'of'],
              'noun':       ['door', 'bear', 'princess', 'cabinet']
              }
        self.mapping_categories = self.mapping.keys()

    def scan(self, input):
        self.result = []

        for word in input.split():
            try:
                self.result.append(('number', int(word)))
            except ValueError:
                for category, item in self.mapping.items():
                    if word.lower() in item:
                        found_category = category
                        break
                    else:
                        found_category = 'error'
                self.result.append((found_category, word))

        return self.result

lexicon = Lexicon()

Answer 2

感谢Thomas K＆amp; amp;我设法完成了练习。我现在非常生气，现在看起来很简单......

directions = ['north', 'south', 'east', 'west', 'down', 'up', 'down', 'right']
verbs = ['go', 'stop', 'kill', 'eat']
stops = ['the', 'in', 'at', 'of', 'from', 'at', 'it']
nouns = ['door', 'bear', 'princess', 'cabinet']


def scan(thewords):

    thewords = thewords.split()
    sentence = []

    for i in thewords:
        if i in directions:
            sentence.append(('direction', i))

        elif i in verbs:
            sentence.append(('verb', i))

        elif i in stops:
            sentence.append(('stop', i))

        elif i in nouns:
            sentence.append(('noun', i))

        elif i.isdigit():
            sentence.append(('number', convert_number(i)))

        else:            
            sentence.append(('error', i))

    return sentence

def convert_number(s):
    try:
        return int(s)

    except ValueError:
        return None

Answer 3

我不知道练习48引入了多少，但我对你的解决方案有一些评论。

首先，对于每个单词列表使用单独的变量，维护此代码有点困难。其次，i通常仅在从0计算自然数时用作变量。

考虑：

_LEXICON = dict(
    direction = ['north', 'south', 'east', 'west', 'down', 'up', 'down', 'right'],
    verb = ['go', 'stop', 'kill', 'eat'],
    stop = ['the', 'in', 'at', 'of', 'from', 'at', 'it'],
    noun = ['door', 'bear', 'princess', 'cabinet'],
    number = ['0','1','2','3','4','5','6','7','8','9'],
)

def scan(words):
    result = []

    for word in words.split():
        found_category = 'error'
        for category, category_lexicon in _LEXICON.items():
            if word in category_lexicon:
                found_category = category
                break

        result.append((found_category, word))

    return result

但我们可以做得更好;寻找列表中的项目很慢。当你想要查找某些内容时，你需要一本字典：

_LEXICON = dict(...)
_LEXICON_INDEX = dict()
for category, words in _LEXICON:
    for word in words:
        _LEXICON_INDEX[word] = category

def scan(words):
    result = []

    for word in words.split():
        result.append((_LEXICON_INDEX.get(word, 'error'), word))

    return result

当然，这实际上并未通过练习中的所有测试。我会留给你修改我的代码。 ;）

Answer 4

据推测，你的“指示”列表最终将包含除“方向”之外的其他元素作为第一个条目（继@Thomas K的评论之后）。

如果列出元组的第二个元素：

valid_words = [x[1] for x in directions]

然后修改您的代码：

for i in thewords:
    if i in valid_words:
        sentence.append(directions[valid_words.index(i)])

会给你相关的元组。

LPTHW练习48帮助 - 在列表中使用元组

4 个答案: