Python - 使用递归函数按字母顺序打印出 trie

Question

我正在通过 Bird、Klein 和 Loper 的 NLTK book 工作，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，以按字母顺序漂亮地打印 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')

我修改了来自 this discussion 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

有谁知道用一个递归函数是否可以得到相同的结果？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

Answer 1

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

Answer 2

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已接受键的列表和一对 for-loops 遍历此列表以查看是否存在是可以替换的字符串开头的任何常见元素。

编辑 11/2/19：这个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会替换太多字符将破折号，即：

----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)