Question

给出以下几行文字

TOKYO-BLING.1 H02-AVAILABLE
TOKYO-BLING.1 H02-MIDDLING
TOKYO-BLING.1 H02-TOP
TOKYO-BLING.2 H04-USED
TOKYO-BLING.2 H04-AVAILABLE
TOKYO-BLING.2 H04-CANCELLED
WAY-VERING.1 H03-TOP
WAY-VERING.2 H03-USED
WAY-VERING.2 H03-AVAILABLE
WAY-VERING.1 H03-CANCELLED

我想做一些解析来产生一些合理的分组。上面的列表可以分组如下

TOKYO-BLING.1 H02-AVAILABLE
TOKYO-BLING.1 H02-MIDDLING
TOKYO-BLING.1 H02-TOP

TOKYO-BLING.2 H04-USED
TOKYO-BLING.2 H04-AVAILABLE
TOKYO-BLING.2 H04-CANCELLED

WAY-VERING.2 H03-USED
WAY-VERING.2 H03-AVAILABLE

WAY-VERING.1 H03-TOP
WAY-VERING.1 H03-CANCELLED

任何人都可以建议一种算法（或某种方法）可以扫描给定数量的文本，并确定文本可以如上所述进行分组。显然每个小组都可以进一步。我想我正在寻找一个很好的解决方案来查看一个短语列表，并找出如何最好地通过一些常见的字符串序列对它们进行分组。

Answer 1

这是一种方式：

对条目进行排序
确定每个条目之间的公共前缀的长度
通过在公共前缀短于上一个条目

示例实施：

def common_count(t0, t1):
  "returns the length of the longest common prefix"
  for i, pair in enumerate(zip(t0, t1)):
    if pair[0] != pair[1]:
      return i
  return i

def group_by_longest_prefix(iterable):
  "given a sorted list of strings, group by longest common prefix"
  longest = 0
  out = []

  for t in iterable:
    if out: # if there are previous entries 

      # determine length of prefix in common with previous line
      common = common_count(t, out[-1])

      # if the current entry has a shorted prefix, output previous 
      # entries as a group then start a new group
      if common < longest:
        yield out
        longest = 0
        out = []
      # otherwise, just update the target prefix length
      else:
        longest = common

    # add the current entry to the group
    out.append(t)

  # return remaining entries as the last group
  if out:
    yield out

使用示例：

text = """
TOKYO-BLING.1 H02-AVAILABLE
TOKYO-BLING.1 H02-MIDDLING
TOKYO-BLING.1 H02-TOP
TOKYO-BLING.2 H04-USED
TOKYO-BLING.2 H04-AVAILABLE
TOKYO-BLING.2 H04-CANCELLED
WAY-VERING.1 H03-TOP
WAY-VERING.2 H03-USED
WAY-VERING.2 H03-AVAILABLE
WAY-VERING.1 H03-CANCELLED
"""

T = sorted(t.strip() for t in text.split("\n") if t)

for L in group_by_longest_prefix(T):
  print L

这会产生：

['TOKYO-BLING.1 H02-AVAILABLE', 'TOKYO-BLING.1 H02-MIDDLING', 'TOKYO-BLING.1 H02-TOP']
['TOKYO-BLING.2 H04-AVAILABLE', 'TOKYO-BLING.2 H04-CANCELLED', 'TOKYO-BLING.2 H04-USED']
['WAY-VERING.1 H03-CANCELLED', 'WAY-VERING.1 H03-TOP']
['WAY-VERING.2 H03-AVAILABLE', 'WAY-VERING.2 H03-USED']

在此处查看此行动：http://ideone.com/1Da0S

Answer 2

您可以按空格分割每个字符串，然后创建dict。

我就这样做了：

f = open( 'hotels.txt', 'r' )   # read the data
f = f.readlines()               # convert to a list of strings (with newlines)
f = [ i.strip() for i in f ]    # take off the newlines
h = [ i.split(' ') for i in f ] # split using whitespace
                                # now h is a list of lists of strings

keys = [ i[0] for i in h ]      # keys = ['TOKYO-BLING.1','TOKYO-BLING.1',...]
keys = list( set( keys ) )      # take out redundant elements

d = dict()                      # start a dict
for i in keys:                  # initialize dict with empty lists
    d[i] = list()               # (one for each key)

for i in h:                     # for each list in h, append a suffix
    d[i[0]].append(i[1])        # to the appropriate prefix (or key)

这会产生：

{'TOKYO-BLING.1': ['H02-AVAILABLE', 'H02-MIDDLING', 'H02-TOP'],\
 'TOKYO-BLING.2': ['H04-USED', 'H04-AVAILABLE', 'H04-CANCELLED'],\
 'WAY-VERING.1': ['H03-TOP', 'H03-CANCELLED'],\
 'WAY-VERING.2': ['H03-USED', 'H03-AVAILABLE']}

Answer 3

Generalized Suffix tree可行或suffix array

Answer 4

这是我的，它开始时更短：

import os

def prefix_groups(data):
    """Return a dictionary of {prefix:[items]}."""
    lines = data[:]
    groups = dict()
    while lines:
        longest = None
        first = lines.pop()
        for line in lines:
            prefix = os.path.commonprefix([first, line])
            if not longest:
                longest = prefix
            elif len(prefix) > len(longest):
                longest = prefix
        if longest:
            group = [first]
            rest = [item for item in lines if longest in item]
            [lines.remove(item) for item in rest]
            group.extend(rest)
            groups[longest] = group
        else:
            # Singletons raise an exception
            raise IndexError("No prefix match for {}!".format(first))
    return groups

if __name__ == "__main__":
    from pprint import pprint
    data = """
    TOKYO-BLING.1 H02-AVAILABLE
    TOKYO-BLING.1 H02-MIDDLING
    TOKYO-BLING.1 H02-TOP
    TOKYO-BLING.2 H04-USED
    TOKYO-BLING.2 H04-AVAILABLE
    TOKYO-BLING.2 H04-CANCELLED
    WAY-VERING.1 H03-TOP
    WAY-VERING.2 H03-USED
    WAY-VERING.2 H03-AVAILABLE
    WAY-VERING.1 H03-CANCELLED
    """
    data = [line.strip() for line in data.split('\n') if line.strip()]
    groups = prefix_groups(data)
    pprint(groups)

输出：

{'TOKYO-BLING.1 H02-': ['TOKYO-BLING.1 H02-AVAILABLE',
                        'TOKYO-BLING.1 H02-MIDDLING',
                        'TOKYO-BLING.1 H02-TOP'],
 'TOKYO-BLING.2 H04-': ['TOKYO-BLING.2 H04-USED',
                        'TOKYO-BLING.2 H04-AVAILABLE',
                        'TOKYO-BLING.2 H04-CANCELLED'],
 'WAY-VERING.1 H03-': ['WAY-VERING.1 H03-TOP', 'WAY-VERING.1 H03-CANCELLED'],
 'WAY-VERING.2 H03-': ['WAY-VERING.2 H03-USED', 'WAY-VERING.2 H03-AVAILABLE']}

最长的共同序列组

4 个答案: