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我正在阅读Bird、Klein 和 Loper的NLTK 书籍,但遇到了一个问题。我正在阅读这本书是为了丰富自己的个人生活,而不是为了上课。

我遇到的问题是 4.29:

编写一个递归函数,以字母排序的顺序漂亮地打印一个 trie,例如:

chair: 'flesh' ---t: 'cat' --ic: 'stylish' ---en: 'dog'

我正在使用书中的这段代码来创建 trie:

def insert(trie, key, value):
    if key:
        first, rest = key[0], key[1:]
        if first not in trie:
            trie[first] = {}
        insert(trie[first], rest, value)
    else:
        trie['value'] = value

trie = {}
insert(trie, 'chat', 'cat')
insert(trie, 'chien', 'dog')
insert(trie, 'chair', 'flesh')
insert(trie, 'chic', 'stylish')

我从这个讨论中修改了一个函数的答案,该函数递归地遍历 trie 并提取完整的键和值:

def trawl_trie(trie):
    unsorted = []
    for key, value in trie.items():
        if 'value' not in key:
            for item in trawl_trie(trie[key]):
                unsorted.append(key + item)
        else:
            unsorted.append(': ' + value)

    return unsorted

但是我无法使用递归来制作按字母顺序排列的列表,也无法弄清楚如何使用递归来替换键的重复部分。我能做的最好的事情是创建一个辅助函数来遍历上述函数的结果:

def print_trie(trie):

    # sort list alphabetically
    alphabetized = list(sorted(set(trawl_trie(trie))))


    print(alphabetized[0])

    # compare the 2nd item ~ to the previous one in the list.
    for k in range(1, len(alphabetized)):
        # separate words from value
        prev_w, prev_d = (re.findall(r'(\w+):', alphabetized[k - 1]), re.findall(r': (\w+)', alphabetized[k - 1]))
        curr_w, curr_d = (re.findall(r'(\w+):', alphabetized[k]), re.findall(r': (\w+)', alphabetized[k]))
        word = ''

        # find parts that match and replace them with dashes
        for i in range(min(len(prev_w[0]), len(curr_w[0]))):
            if prev_w[0][i] == curr_w[0][i]:
                word += prev_w[0][i]

        curr_w[0] = re.sub(word, '-' * len(word), curr_w[0])
        print(curr_w[0] + ": " + str(curr_d[0]))

这将是输出:

print_trie(trie)

chair: flesh
---t: cat
--ic: stylish
---en: dog

有谁知道是否可以使用一个递归函数获得相同的结果?或者我被困在使用递归函数来遍历 trie 时,使用第二个辅助函数来让一切看起来都很好?

干杯,

  • MC
4

2 回答 2

0
def insert(trie, key, value):
    """Insert into Trie"""
    if key:
        first, rest = key[0], key[1:]
        if first not in trie:
            trie[first] = {}
        insert(trie[first], rest, value)
    else:
        trie['value'] = value

def display(trie, s = ""):
  """Recursive function to Display Trie entries in alphabetical order"""
  first = True
  for k, v in sorted(trie.items(), key = lambda x: x[0]):
    # dictionary sorted based upon the keys
    if isinstance(v, dict):
      if first:
        prefix = s + k          # first to show common prefix
        first = False
      else:
        prefix = '-'*len(s) + k  # dashes for common prefix

      display(v, prefix)   # s+k is extending string s for display by appending current key k
    else:
      print(s, ":", v)  # not a dictionary, so print current   # not a dictionary, so print current string s and value

# Create Trie
trie = {}
insert(trie, 'chat', 'cat')
insert(trie, 'chien', 'dog')
insert(trie, 'chair', 'flesh')
insert(trie, 'chic', 'stylish')

#Display Use Recursive function (second argument will default to "" on call)
display(trie)

输出

chair : flesh
---t : cat
--ic : stylish
---en : dog
于 2019-11-01T02:48:50.147 回答
0

我修改了 DarrylG 的答案(再次感谢!),方法是添加一个通过递归传递的接受键列表和一对for循环遍历该列表以查看字符串开头是否有任何公共元素可以更换。

2019 年 11 月 2 日编辑:此修改有一个我没有注意到的错误。它会在第一次正确运行,但在随后的运行中,它将替换太多字符,即破折号:

----r: flesh
---t: cat
---c: stylish
----n: dog
def insert(trie, key, value):
    """Insert into Trie"""
    if key:
        first, rest = key[0], key[1:]
        if first not in trie:
            trie[first] = {}
        insert(trie[first], rest, value)
    else:
        trie['value'] = value

def display(trie, s = "", final = []):

    """Recursive function to Display Trie entries in alphabetical order"""

    for k, v in sorted(trie.items(), key = lambda x: x[0]):

        # dictionary sorted based upon the keys
        if isinstance(v, dict):
            display(v, s + k, final)   # s+k is extending string s for display by appending current key k
        else:
            # replace common elements at beginning of strings with dashes
            i = sum([any([f.startswith(string[:j]) for f in final]) for j in range(1, len(string))])
            string = '-' * i + s[i:]

            print(string + ":", v)  # not a dictionary, so print current edited s and value
            final.append(s)


# Create Trie
trie = {}
insert(trie, 'chat', 'cat')
insert(trie, 'chien', 'dog')
insert(trie, 'chair', 'flesh')
insert(trie, 'chic', 'stylish')

#Display Use Recursive function (second argument will default to "" on call)
display(trie)

于 2019-11-01T05:52:28.433 回答