Python实现自定义排除规则的孤立索引识别

现有基于Python编写的孤立索引识别代码，基础判定规则为：索引需至少拥有2个相邻邻居才不属于孤立索引，坐标(i,j)的合法邻居仅包含正交相邻的四个坐标：(i+1,j),(i-1,j),(i,j+1),(i,j-1)。现需在现有逻辑基础上新增两项排除规则：

• 始终排除索引(0,0)，不将其判定为孤立索引
• 若某索引是最后一行的唯一索引，则不对其做孤立判定

预期测试结果 #

• 输入indices = [(0, 0),(0,1),(1,0),(2,0)]时，预期返回的孤立索引仅为(0,1)
• 输入indices = [(0, 0),(1,0),(2,0)]时，预期无孤立索引返回

修正方案 #

原有代码的邻居计数逻辑本身符合正交相邻判定规则，仅需要在孤立判定环节前新增两个排除规则的判断分支即可，修正后完整代码如下：

import numpy as np

def get_isolated_indices(indices):
    indices = np.array(indices)
    indices = indices[np.argsort(indices[:, 0])]
    
    # 约定坐标对第一个值为行号，第二个值为列号
    max_vals = np.max(indices, axis=0)
    max_row = max_vals[0]

    rows = indices[:, 0]
    cols = indices[:, 1]
    # 提前统计最后一行的索引总数，用于第二条排除规则判定
    last_row_count = np.sum(rows == max_row)
    
    isolated = []
    
    for i in range(len(indices)):
        this_row = indices[i, 0]
        this_col = indices[i, 1]
        
        # 排除规则1：直接跳过(0,0)的孤立判定
        if (this_row, this_col) == (0, 0):
            continue
        # 排除规则2：直接跳过最后一行唯一索引的孤立判定
        if this_row == max_row and last_row_count == 1:
            continue
        
        # 计算合法相邻坐标范围
        nearby_row_vals = list(range(max(0, this_row - 1), min(this_row + 1, max_vals[0]) + 1))
        nearby_col_vals = list(range(max(0, this_col - 1), min(this_col + 1, max_vals[1]) + 1))
        
        same_row_mask = rows == this_row
        same_col_mask = cols == this_col
        
        diff_rows_mask = (rows >= np.min(nearby_row_vals)) & (rows <= np.max(nearby_row_vals)) & (rows != this_row)
        diff_cols_mask = (cols >= np.min(nearby_col_vals)) & (cols <= np.max(nearby_col_vals)) & (cols != this_col)
        
        intersection1 = np.logical_and(same_row_mask, same_col_mask)
        intersection2 = np.logical_and(same_row_mask, diff_cols_mask)
        intersection3 = np.logical_and(diff_rows_mask, same_col_mask)
        
        final_intersection = np.logical_or(np.logical_or(intersection1, intersection2), intersection3)
        # 减去自身重复计数，得到真实正交邻居数量
        num_neighbors = np.sum(final_intersection) - 1
        
        if num_neighbors <= 1:
            isolated.append((this_row, this_col))
            
    return isolated

# 测试用例验证
if __name__ == "__main__":
    indices1 =  [(0, 0),(0,1),(1,0),(2,0)]
    print("测试用例1结果：", get_isolated_indices(indices1)) # 输出 [(0, 1)]
    indices2 =  [(0, 0),(1,0),(2,0)]
    print("测试用例2结果：", get_isolated_indices(indices2)) # 输出 []

验证说明 #

运行上述代码后两个测试用例输出完全符合预期：

• 测试用例1中(0,0)被规则排除，(2,0)作为最后一行唯一索引被规则排除，(1,0)拥有(0,0)、(2,0)两个邻居不满足孤立条件，仅(0,1)只有1个邻居被判定为孤立
• 测试用例2中(0,0)被规则排除，(2,0)作为最后一行唯一索引被规则排除，(1,0)拥有(0,0)、(2,0)两个邻居不满足孤立条件，最终无孤立索引返回

内容的提问来源于stack exchange，提问作者Wiz123