手写实现卷积和NMS

手写实现卷积python

import numpy as np
#k是卷积核假设n*n，feature是特征图假设是w*h
def convolution(feature,k):
    w,h=feature.shape
    n=len(k)
    new_feature=[]
    for i in range(w-n):
        line=[]
        for j in range(h-n):
            a=feature[i:i+n,j:j+n]
            #np.multiply()求两个矩阵的内积
            line.append(np.sum(np.multiply(k,a)))
        new_feature.append(line)
    return np.array(new_feature)

手写NMS

pytorch版本

import torch
#坐标原点在左上角，水平x，竖直y
#计算一个框与一堆框的iou
def iou(box,boxes):
    #先计算box的面积[x1,y1,x2,y2]
    box_area=(box[2]-box[0])*(box[3]-box[1])
    boxes_area=(boxes[:,2]-boxes[:,0])*(boxes[:,3]-boxes[:,1])
    #求交集，左上角最大，右下角最小
    xx1=torch.maximum(box[0],boxes[:,0])
    yy1=torch.maximum(box[1],boxes[:,1])
    xx2=torch.minimum(box[2],boxes[:,2])
    yy2=torch.minimum(box[3],boxes[:,3])
    w,h=torch.maximum(torch.Tensor([0]),xx2-xx1),torch.maximum(torch.Tensor([0]),yy2-yy1)
    over_area=w*h
    return over_area/(box_area+boxes_area-over_area)
 
def NMS(boxes,thresh=0.3):
    #根据boxes的置信度进行排序，假设置信度在第一列
    #argsort将numpy数组进行排序，返回索引,descending表示从大到小返回
    new_boxes=boxes[boxes[:,0].argsort(descending=True)]
    keep_boxes=[]
    while len(new_boxes)>0:
        max_box=new_boxes[0]
        keep_boxes.append(max_box)
        if len(new_boxes)>1:#除了评分最高的外，还有其他框
            other_boxes=new_boxes[1:]
            #torch.where返回满足条件的索引
            new_boxes=other_boxes[torch.where(iou(max_box[1:],other_boxes[:,1:])
        else:
            break
    #将一个列表里面的张量数据进行扩维拼接，本来1行5列，现在n行5列
    return torch.stack(keep_boxes)
boxes=torch.Tensor([[0.8,0,0,2,2],[0.6,1,1,4,4],[0.5,-1,-1,2,2],[0.75,-0.5,-0.5,2.5,2.5]])
print(NMS(boxes,0.3))

numpy版本

def nms(boxes, scores, threshold):
    # 根据得分降序排序
    indices = sorted(range(len(scores)), key=lambda i: scores[i], reverse=True)
    
    # 初始化选中的边界框列表
    selected_boxes = []
    
    while indices:
        # 选择得分最高的边界框
        i = indices[0]
        selected_boxes.append(boxes[i])
        
        # 计算当前边界框与其他边界框的IoU
        ious = [calculate_iou(boxes[i], boxes[j]) for j in indices[1:]]
        
        # 仅保留IoU低于阈值的边界框
        indices = [indices[j + 1] for j, iou in enumerate(ious) if iou < threshold]
    
    return selected_boxes
 
 
def calculate_iou(box1, box2):
    # 计算两个边界框的相交区域
    x1 = max(box1[0], box2[0])
    y1 = max(box1[1], box2[1])
    x2 = min(box1[2], box2[2])
    y2 = min(box1[3], box2[3])
    
    intersection = max(0, x2 - x1) * max(0, y2 - y1)
    
    # 计算两个边界框的联合区域
    area1 = (box1[2] - box1[0]) * (box1[3] - box1[1])
    area2 = (box2[2] - box2[0]) * (box2[3] - box2[1])
    
    union = area1 + area2 - intersection
    
    # 计算IoU（交并比）
    iou = intersection / union
    
    return iou