python 如何實(shí)現(xiàn)非極大值抑制算法?相信很多沒有經(jīng)驗(yàn)的人對(duì)此束手無策,為此本文總結(jié)了問題出現(xiàn)的原因和解決方法,通過這篇文章希望你能解決這個(gè)問題。
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算法原理
非極大值抑制算法(Non-maximum suppression, NMS)的本質(zhì)是搜索局部極大值,抑制非極大值元素。
算法的作用
當(dāng)算法對(duì)一個(gè)目標(biāo)產(chǎn)生了多個(gè)候選框的時(shí)候,選擇 score
最高的框,并抑制其他對(duì)于改目標(biāo)的候選框
適用場(chǎng)景
一幅圖中有多個(gè)目標(biāo)(如果只有一個(gè)目標(biāo),那么直接取 score
最高的候選框即可)。
算法的輸入
算法對(duì)一幅圖產(chǎn)生的所有的候選框,以及每個(gè)框?qū)?yīng)的 score
(可以用一個(gè) 5 維數(shù)組 dets
表示,前 4 維表示四個(gè)角的坐標(biāo),第 5 維表示分?jǐn)?shù)),閾值 thresh
。
算法的輸出
正確的候選框組(dets
的一個(gè)子集)。
細(xì)節(jié)
score
從大到小排序。IoU
大于 thresh
的框設(shè)為抑制。參考代碼
# -------------------------------------------------------- # Fast R-CNN # Copyright (c) 2015 Microsoft # Licensed under The MIT License [see LICENSE for details] # Written by Ross Girshick # -------------------------------------------------------- import numpy as np cimport numpy as np cdef inline np.float32_t max(np.float32_t a, np.float32_t b): return a if a >= b else b cdef inline np.float32_t min(np.float32_t a, np.float32_t b): return a if a <= b else b def cpu_nms(np.ndarray[np.float32_t, ndim=2] dets, np.float thresh): cdef np.ndarray[np.float32_t, ndim=1] x1 = dets[:, 0] cdef np.ndarray[np.float32_t, ndim=1] y1 = dets[:, 1] cdef np.ndarray[np.float32_t, ndim=1] x2 = dets[:, 2] cdef np.ndarray[np.float32_t, ndim=1] y2 = dets[:, 3] cdef np.ndarray[np.float32_t, ndim=1] scores = dets[:, 4] cdef np.ndarray[np.float32_t, ndim=1] areas = (x2 - x1 + 1) * (y2 - y1 + 1) cdef np.ndarray[np.int_t, ndim=1] order = scores.argsort()[::-1] cdef int ndets = dets.shape[0] cdef np.ndarray[np.int_t, ndim=1] suppressed = \ np.zeros((ndets), dtype=np.int) # nominal indices cdef int _i, _j # sorted indices cdef int i, j # temp variables for box i's (the box currently under consideration) cdef np.float32_t ix1, iy1, ix2, iy2, iarea # variables for computing overlap with box j (lower scoring box) cdef np.float32_t xx1, yy1, xx2, yy2 cdef np.float32_t w, h cdef np.float32_t inter, ovr keep = [] for _i in range(ndets): i = order[_i] if suppressed[i] == 1: continue keep.append(i) ix1 = x1[i] iy1 = y1[i] ix2 = x2[i] iy2 = y2[i] iarea = areas[i] for _j in range(_i + 1, ndets): j = order[_j] if suppressed[j] == 1: continue xx1 = max(ix1, x1[j]) yy1 = max(iy1, y1[j]) xx2 = min(ix2, x2[j]) yy2 = min(iy2, y2[j]) w = max(0.0, xx2 - xx1 + 1) h = max(0.0, yy2 - yy1 + 1) inter = w * h ovr = inter / (iarea + areas[j] - inter) if ovr >= thresh: suppressed[j] = 1 return keep