基于重叠和预测的lable的框架框

Question

此处输入映像我有一个数据框

image     label    bb0         bb1                 bb2                 bb3
im1       L1    0.16602577  0.20398015          0.6750162              1.0
im1       L2    0.38657066  0.5192988000000001  0.45647129999999997    0.93410575
im1       L2    0.019637141 0.66581047          0.1369443              0.9160644
im1       L2    0.3700079   0.32554457          0.45773253             0.581625

在 ，BB3 是按顺序Y1，X1，Y2，X2的预测边界框的坐标。我想根据重叠区域（IOU）合并每个标签的边界框，如果IOU大于某个阈值，则将框合并，否则将其保留为原样。在最终数据框架中，我只需要更新的边界框。 可以在附件图像中看到场景。中间的两个橙色盒子将组合起来制作黄色盒子，所有其他盒子都将保持不变。最终数据将包含三行，一个用于L1，一个不是重叠的橙色盒，一个新的黄色框。我想以这种方式更新数据框。

我正在使用延迟代码 -

def iou(box1, box2):
    assert box1['bb1'] < box1['bb3']
    assert box1['bb0'] < box1['bb2']
    assert box2['bb1'] < box2['bb3']
    assert box2['bb0'] < box2['bb2']

    # determine the coordinates of the intersection rectangle
    x_left   = max(box1['bb1'], box2['bb1'])
    y_top    = max(box1['bb0'], box2['bb0'])
    x_right  = min(box1['bb3'], box2['bb3'])
    y_bottom = min(box1['bb2'], box2['bb2'])

    if x_right < x_left or y_bottom < y_top:
        iou1 = 0
    else:
        intersection_area = (x_right - x_left) * (y_bottom - y_top)
        # compute the area of both AABBs
        bb1_area = (box1["bb3"] - box1["bb1"]) * (box1["bb2"] - box1['bb0'])
        bb2_area = (box2["bb3"] - box2["bb1"]) * (box2["bb2"] - box2['bb0'])
        #intersection over union
        iou1 = intersection_area / float(bb1_area + bb2_area - intersection_area)
    return iou1


def get_iou(df):
    df_out = pd.DataFrame(columns = df.columns)
    for i in (df["image"].unique()):
        print(i)
        #df_out = pd.DataFrame(columns = df.columns)
        k1 = df[df["image"] == str(i)]
        k2 = k1["class"].value_counts().reset_index()
        k3 = k2[k2["class"] > 1]
        k8 = k2[k2["class"] == 1]
        k9 = df[(df["image"] == str(i)) & (df["class"].isin(k8["index"]))]
        df_out = df_out.append(k9, ignore_index = True)
        
        new_df = pd.DataFrame(columns = ["class","Combinations", "IOU"])
        for j in k3["index"].unique():
            k4 = df[(df["image"] == str(i)) & (df["class"] == str(j))]
            k4["unique_id"] = list(range(len(k4)))
            #k5 = k4
            #k6 = k5.to_dict(orient='records')
            #k7 = list(itertools.combinations(k6, 2))
            gg = list(itertools.combinations(k4["unique_id"], 2))
            #new_df = pd.DataFrame(columns = ["Combinations", "IOU"])
            #print(k7, "\n")
            df_out2 = pd.DataFrame(columns = df.columns)
            l = (0,1)
            while l in gg:
                #print(l)
                box1 = k4[k4["unique_id"] == l[0]].to_dict('records')[0]
                #print(box1)
                box2 = k4[k4["unique_id"] == l[1]].to_dict('records')[0]
                iou1 = iou(box1, box2)
                assert iou1 >= 0.0
                assert iou1 <= 1.0
                print(iou1)
                if iou1 <= 0.05:
                    print("No Overlap")
                else:
                    x_left_new   = min(box1["bb1"], box2["bb1"])
                    x_right_new  = max(box2["bb3"], box1["bb3"])
                    y_top_new    = min(box2["bb0"], box2["bb0"])
                    y_bottom_new = max(box1["bb2"], box2["bb2"])
                    new_bb = {'image' : box1["image"], 
                              'class' : box1['class'], 
                              'score' : np.nan, 
                              'bb0' : y_top_new, 
                              'bb1' : x_left_new, 
                              'bb2' : y_bottom_new, 
                              'bb3' : x_right_new,
                              'key' : box1["key"]}
                    k4 = k4.append(new_bb, ignore_index = True)
                    k4 = k4[~k4["unique_id"].isin(l)]
                    k4["unique_id"] = list(range(len(k4)))
                    gg = list(itertools.combinations(k4["unique_id"], 2))
                    df_out2 = df_out2.append(k4)