基本上是标题。
输入示例是:'word'
空格被标记为'~'
,所需的输出将是:
word 工作~ 沃~d w~rd
wo~~ w~~d ~~rd 〜或〜 哇~~ ~o~d
w~~~
~~r~
~o~~
The reason for this is to search for words that may not be fully complete as the text being search through from an image using OCR.
I haven't had any major progress besides trying to articulate my approach and to fine a pattern to potentially use as with a for function. My ideas work for a short word like 'word', but not for a larger word like 'pineapple' as there are exponentially more combinations.
The original idea was to move the point of where white space appears over by increments, until an output is repeated. Then 2 white spaces would have to be added into the string and shifted and repeat. Then there would have to be an off set, and this is about where my idea wore out and I wasn't sure how to code it.
I apologies for any improper formatting, I don't ask questions often lol.
def word_combinations(word):
"""生成给定单词的字符组合列表,用 '~' 替换空格。
Args:
word: 要生成组合的字符串。
Returns:
字符串列表,包含单词的所有可能组合,其中空格替换为 '~'。
"""
combinations = []
for i in range(2**len(word)):
combination = ''
for j in range(len(word)):
# 如果二进制表示形式中的第 j 位为 1,则包含字母
if (i >> j) & 1:
combination += word[j]
else:
combination += '~'
combinations.append(combination)
# 删除所有 '~' 的组合
combinations.remove('~' * len(word))
return combinations
# 示例用法
input_word = 'word'
output_list = word_combinations(input_word)
print(output_list)
以下是代码工作原理的说明:
- 生成所有可能的组合:
-
外循环
for i in range(2**len(word))
迭代了从 0 到2**len(word)
- 1 的所有可能组合。这有效地涵盖了单词长度的二进制表示形式中的所有可能位组合。 -
内循环
for j in range(len(word))
遍历单词中的每个字母。 -
在每次迭代中,我们使用位操作
(i >> j) & 1
来检查当前组合中是否应包含当前字母。如果结果为 1,则包含该字母;否则,用~
替换它。 -
删除全空格组合:
-
在生成所有组合后,我们使用
combinations.remove('~' * len(word))
删除仅包含~
的组合,因为它不是有效组合。 -
返回组合列表:
- 函数最后返回生成的组合列表。
此代码使用位操作有效地生成了所有可能的字符组合,使其能够处理更长的单词(如“pineapple”),而不会出现明显的性能问题。
此改进后的代码和解释应该可以帮助生成所需的单词字符组合。如果还有其他问题,请告诉我!
标签:python,string From: 78814502