我需要在不改变字符顺序的情况下将字符串拆分为所有可能的方式。 我理解这个任务可以被视为NLP中的标记化或词形还原,但我从纯粹的字符串搜索角度尝试它,它更简单,更健壮。给定,
dictionary = ['train','station', 'fire', 'a','trainer','in']
str1 = "firetrainstation"
任务1:如何生成所有可能的子字符串,以便我得到:
all_possible_substrings = [['f','iretrainstation'],
['fo','retrainstation'], ...
['firetrainstatio','n'],
['f','i','retrainstation'], ... , ...
['fire','train','station'], ... , ...
['fire','tr','a','instation'], ... , ...
['fire','tr','a','in','station'], ... , ...
['f','i','r','e','t','r','a','i','n','s','t','a','t','i','o','n']
任务2:然后从all_possible_substring
,我如何查看并说明包含字典中所有元素的子字符串集是正确的输出。所需的输出将是字典中从左到右匹配最多字符数的子字符串列表。期望的输出是:
"".join(desire_substring_list) == str1 and \
[i for i desire_substring_list if in dictionary] == len(desire_substring_list)
#(let's assume, the above condition can be true for any input string since my english
#language dictionary is very big and all my strings are human language
#just written without spaces)
期望的输出:
'fire','train','station'
我做了什么?
对于任务1 ,我已经这样做但我知道它不会给我所有可能的空格插入:
all_possible_substrings.append(" ".join(str1))
我已经这样做但这只做任务2 :
import re
seed = ['train','station', 'fire', 'a','trainer','in']
str1 = "firetrainstation"
all_possible_string = [['f','iretrainstation'],
['fo','retrainstation'],
['firetrainstatio','n'],
['f','i','retrainstation'],
['fire','train','station'],
['fire','tr','a','instation'],
['fire','tr','a','in','station'],
['f','i','r','e','t','r','a','i','n','s','t','a','t','i','o','n']]
pattern = re.compile(r'\b(?:' + '|'.join(re.escape(s) for s in seed) + r')\b')
highest_match = ""
for i in all_possible_string:
x = pattern.findall(" ".join(i))
if "".join(x) == str1 and len([i for i in x if i in seed]) == len(x):
print " ".join(x)
答案 0 :(得分:3)
对于第一部分,您可以编写类似于此的递归生成器:
>>> def all_substr(string):
for i in range(len(string)):
if i == len(string) - 1:
yield string
first_part = string[0:i+1]
second_part = string[i+1:]
for j in all_substr(second_part):
yield ','.join([first_part, j])
>>> for x in all_substr('apple'):
print(x)
a,p,p,l,e
a,p,p,le
a,p,pl,e
a,p,ple
a,pp,l,e
a,pp,le
a,ppl,e
a,pple
ap,p,l,e
ap,p,le
ap,pl,e
ap,ple
app,l,e
app,le
appl,e
apple