numpy实现目标检测中的IOU和NMS

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交并比（Intersection-over-Union，IoU），目标检测中使用的一个概念，是产生的候选框（candidate bound）与原标记框（ground truth bound）的交叠率，即它们的交集与并集的比值。最理想情况是完全重叠，即比值为1。

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计算公式：

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代码实现1：

def calculateIoU(candidateBound, groundTruthBound):
    cx1 = candidateBound[0]
    cy1 = candidateBound[1]
    cx2 = candidateBound[2]
    cy2 = candidateBound[3]

    gx1 = groundTruthBound[0]
    gy1 = groundTruthBound[1]
    gx2 = groundTruthBound[2]
    gy2 = groundTruthBound[3]

    carea = (cx2 - cx1) * (cy2 - cy1) #C的面积
    garea = (gx2 - gx1) * (gy2 - gy1) #G的面积

    x1 = max(cx1, gx1)
    y1 = max(cy1, gy1)
    x2 = min(cx2, gx2)
    y2 = min(cy2, gy2)
    w = max(0, x2 - x1)
    h = max(0, y2 - y1)
    area = w * h #C∩G的面积

    iou = area / (carea + garea - area)

    return iou

代码实现2：

import numpy as np
def compute_iou(box1, box2, wh=False):
    """
    compute the iou of two boxes.
    Args:
        box1, box2: [xmin, ymin, xmax, ymax] (wh=False) or [xcenter, ycenter, w, h] (wh=True)
        wh: the format of coordinate.
    Return:
        iou: iou of box1 and box2.
    """
    if wh == False:
        xmin1, ymin1, xmax1, ymax1 = box1
        xmin2, ymin2, xmax2, ymax2 = box2
    else:
        xmin1, ymin1 = int(box1[0]-box1[2]/2.0), int(box1[1]-box1[3]/2.0)
        xmax1, ymax1 = int(box1[0]+box1[2]/2.0), int(box1[1]+box1[3]/2.0)
        xmin2, ymin2 = int(box2[0]-box2[2]/2.0), int(box2[1]-box2[3]/2.0)
        xmax2, ymax2 = int(box2[0]+box2[2]/2.0), int(box2[1]+box2[3]/2.0)

    ## 获取矩形框交集对应的左上角和右下角的坐标（intersection）
    xx1 = np.max([xmin1, xmin2])
    yy1 = np.max([ymin1, ymin2])
    xx2 = np.min([xmax1, xmax2])
    yy2 = np.min([ymax1, ymax2])

    ## 计算两个矩形框面积
    area1 = (xmax1-xmin1) * (ymax1-ymin1) 
    area2 = (xmax2-xmin2) * (ymax2-ymin2)

    inter_area = (np.max([0, xx2-xx1])) * (np.max([0, yy2-yy1]))　#计算交集面积
    iou = inter_area / (area1+area2-inter_area+1e-6)　＃计算交并比

    return iou

在目标检测中，常会利用非极大值抑制算法(NMS)对生成的大量候选框进行后处理，去除冗余的候选框，得到最具代表性的结果，以加快目标检测的效率。NMS其实是为了消除多余的候选框，找到最佳的bbox。

NMS的算法逻辑为：

for object in all objects:
    (1) 获取当前目标类别下所有bbx的信息
    (2) 将bbx按照confidence从高到低排序,并记录当前confidence最大的bbx
    (3) 计算最大confidence对应的bbx与剩下所有的bbx的IOU,移除所有大于IOU阈值的bbx
    (4) 对剩下的bbx，循环执行(2)和(3)直到所有的bbx均满足要求（即不能再移除bbx）

代码实现：

import numpy as np

class Bounding_box:
    def __init__(self, x1, y1, x2, y2, score):
        self.x1 = x1
        self.y1 = y1
        self.x2 = x2
        self.y2 = y2
        self.score = score

def get_iou(boxa, boxb):
    max_x = max(boxa.x1, boxb.x1)
    max_y = max(boxa.y1, boxb.y1)
    min_x = min(boxa.x2, boxb.x2)
    min_y = min(boxa.y2, boxb.y2)
    if min_x <= max_x or min_y <= max_y:
        return 0
    area_i = (min_x - max_x) * (min_y - max_y)
    area_a = (boxa.x2 - boxa.x1) * (boxa.y2 - boxa.y1)
    area_b = (boxb.x2 - boxb.x1) * (boxb.y2 - boxb.y1)
    area_u = area_a + area_b - area_i
    return float(area_i) / float(area_u)

def NMS(box_lists, k):
    box_lists = sorted(box_lists, key=lambda x: x.score, reverse=True)
    NMS_lists = [box_lists[0]]
    temp_lists = []
    for i in range(k):
        for j in range(1, len(box_lists)):
            iou = get_iou(NMS_lists[i], box_lists[j])
            if iou < 0.7:
                temp_lists.append(box_lists[j])
        if len(temp_lists) == 0:
            return NMS_lists
        box_lists = temp_lists
        temp_lists = []
        NMS_lists.append(box_lists[0])
    return NMS_lists

box1 = Bounding_box(13, 22, 268, 367, 0.124648176)
box2 = Bounding_box(18, 27, 294, 400, 0.35818103)
box3 = Bounding_box(234, 123, 466, 678, 0.13638769)
box_lists = [box1, box2, box3]
NMS_list = NMS(box_lists, 2)
print NMS_list
print NMS_list[0].x1

numpy实现目标检测中的IOU和NMS

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