115 lines
4.0 KiB
Python
115 lines
4.0 KiB
Python
from .Config import *
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def LimitStringLength(data, max_length:int=50) -> str:
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s:str = data if isinstance(data, str) else str(data)
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if len(s) <= max_length:
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return s
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else:
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inside_str = "\n...\n...\n"
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# 计算头尾部分的长度
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head_length = max_length // 2
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tail_length = max_length - head_length - len(inside_str) # 3 是省略号的长度
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# 截取头尾部分并连接
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return s[:head_length] + inside_str + s[-tail_length:]
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def FillString(data:Any,
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max_length: int = 50,
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fill_char: str = " ",
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side: Literal["left", "right", "center"] = "right"
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) -> str:
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s:str = data if isinstance(data, str) else str(data)
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char = fill_char[0]
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if len(s) >= max_length:
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return s
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else:
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if side == "left":
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return s + char * (max_length - len(s))
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elif side == "right":
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return char * (max_length - len(s)) + s
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elif side == "center":
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left = (max_length - len(s)) // 2
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right = max_length - len(s) - left
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return char * left + s + char * right
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else:
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raise ValueError(f"Unsupported side: {side}")
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def Bytes2Strings(lines:List[bytes], encoding='utf-8') -> List[str]:
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return [line.decode(encoding) for line in lines]
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def Bytes2String(lines:List[bytes], encoding='utf-8') -> str:
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return "".join(Bytes2Strings(lines, encoding))
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def word_segmentation(
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sentence,
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cut_all: bool = False,
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HMM: bool = True,
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use_paddle: bool = False
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) -> Sequence[Optional[Union[Any, str]]]:
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try:
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import jieba
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return jieba.dt.cut(str(sentence), cut_all=cut_all, HMM=HMM, use_paddle=use_paddle)
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except ImportError:
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raise ValueError("jieba is not install")
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def GetEditorDistanceAndOperations(s1:str, s2:str) -> Tuple[int, List[Tuple[str, int, int, str]]]:
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"""
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计算两个字符串的编辑距离和操作序列
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操作格式: (操作类型, 开始位置, 结束位置, 内容)
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位置基于源字符串s1
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"""
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m, n = len(s1), len(s2)
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# 使用简单的LCS算法来找到最长公共子序列
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# 然后基于LCS生成操作序列
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lcs = [[0] * (n + 1) for _ in range(m + 1)]
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# 构建LCS表
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if s1[i - 1] == s2[j - 1]:
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lcs[i][j] = lcs[i - 1][j - 1] + 1
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else:
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lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1])
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# 基于LCS生成操作序列
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operations = []
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i, j = m, n
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while i > 0 or j > 0:
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if i > 0 and j > 0 and s1[i - 1] == s2[j - 1]:
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# 字符匹配，不需要操作
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i -= 1
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j -= 1
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elif j > 0 and (i == 0 or lcs[i][j - 1] >= lcs[i - 1][j]):
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# 需要插入s2[j-1]
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# 找到插入位置（在s1中的位置）
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insert_pos = i
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operations.insert(0, ("add", insert_pos, insert_pos, s2[j - 1]))
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j -= 1
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else:
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# 需要删除s1[i-1]
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operations.insert(0, ("delete", i - 1, i, s1[i - 1]))
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i -= 1
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# 合并连续的操作
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merged_operations = []
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for op in operations:
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if merged_operations and merged_operations[-1][0] == op[0]:
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last_op = merged_operations[-1]
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if op[0] == "add" and last_op[2] == op[1]:
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# 合并连续的添加操作
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merged_operations[-1] = (op[0], last_op[1], op[2], last_op[3] + op[3])
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elif op[0] == "delete" and last_op[2] == op[1]:
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# 合并连续的删除操作
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merged_operations[-1] = (op[0], last_op[1], op[2], last_op[3] + op[3])
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else:
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merged_operations.append(op)
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else:
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merged_operations.append(op)
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# 计算编辑距离
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edit_distance = m + n - 2 * lcs[m][n]
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return edit_distance, merged_operations
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