视频分割中的IOU计算问题(一维向量的IOU) |
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视频分割中的IOU计算问题(一维向量的IOU)
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对视频的帧进行分割得到向量，举例说明 p = [1,1,0,0,1,1,0,0,0,1,2,2,2,3,3,3,4,4,5,5] y = [1,1,1,1,0,0,0,2,2,2,2,2,3,3,3,4,4,5,5,5]其中，p为预测帧的分类，y为真实数据帧的分类。 计算p和y的IOU，也就是一维向量的IOU。首先明白IOU的概念，IOU的计算公式如下。 看如下简图，
很容易看出交集与并集的计算公式如下， intersection = minimum(p_end, y_end) - maximum(p_start, y_start) union = maximum(p_end, y_end) - minimum(p_start[j], y_start)因此，可以求出IOU. 完整代码如下。 import numpy as np p = [1,1,0,0,1,1,0,0,0,1,2,2,2,3,3,3,4,4,5,5] y = [1,1,1,1,0,0,0,2,2,2,2,2,3,3,3,4,4,5,5,5] def get_labels_start_end_time(frame_wise_labels, bg_class=["background"]): labels = [] starts = [] ends = [] last_label = frame_wise_labels[0] if frame_wise_labels[0] not in bg_class: labels.append(frame_wise_labels[0]) #如果标签第一帧不是"background"，labels[0]=第一帧的动作 starts.append(0) #starts[0]=0 for i in range(len(frame_wise_labels)): if frame_wise_labels[i] != last_label: if frame_wise_labels[i] not in bg_class: labels.append(frame_wise_labels[i]) starts.append(i) if last_label not in bg_class: ends.append(i) last_label = frame_wise_labels[i] #labels是动作标签 #starts是动作开始的位置 #end是动作结束的位置 if last_label not in bg_class: ends.append(i + 1) return labels, starts, ends p_label, p_start,p_end = get_labels_start_end_time(p, bg_class=["background"]) y_label, y_start,y_end = get_labels_start_end_time(y, bg_class=["background"]) tp = 0 fp = 0 hits = np.zeros(len(y_label)) overlap = 0.1 for j in range(len(p_label)): min_end_py = np.minimum(p_end[j], y_end) max_start_py = np.maximum(p_start[j], y_start) max_end_py = np.maximum(p_end[j], y_end) min_start_py = np.minimum(p_start[j], y_start) intersection = np.minimum(p_end[j], y_end) - np.maximum(p_start[j], y_start) # 交叉 union = np.maximum(p_end[j], y_end) - np.minimum(p_start[j], y_start) IoU = (1.0 * intersection / union) * ([p_label[j] == y_label[x] for x in range(len(y_label))])
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