Question

我在视频中进行了物体检测，并在一帧中找到了5到6个坐标。在每个帧中，坐标之间的距离因移动物体而变化。我的文本文件有13列，第一列是帧号和6个坐标，如下所示。我必须找到坐标之间最大空间的框架编号。每个坐标之间的最小任意距离不应该相互交叉，因为我想根据坐标之间的间距来分割视频。

是否有任何预先编写的算法用于执行此类分段，还是应该循环？

enter image description here

def func1(value):                   
    return df['column_A'][df['column_A'] < value].count()

for name, df in df.groupby(['id_2']):
    for j in df.index:
       y = func1(df['column_A'].ix[j])   
       print y

预期输出类似于

视频片段中的第一个最大空间301.467 0.713235 0.852451 0.752451 0.753922 0.380392 0.373039 0.282843 0.291667 0.262255 0.283824 0.236765 0.218627 301.6 0.717647 0.864706 0.752941 0.761275 0.753431 0.743627 0.388235 0.383333 0.286765 0.294608 0.237745 0.207353 301.667 0.722059 0.876471 0.756863 0.763726 0.395098 0.391667 0.284314 0.289216 0.238725 0.198529 nan nan 301.867 0.72451 0.884314 0.762745 0.771078 0.40098 0.401961 0.290686 0.292647 0.240196 0.207843 0.241176 0.190686 302 0.731373 0.897059 0.760784 0.771569 0.413725 0.422549 0.302941 0.292157 0.25098 0.179412 nan nan 302.067 0.743627 0.917157 0.765686 0.777941 0.430392 0.444608 0.305392 0.293137 0.257843 0.17402 nan nan 302.2 0.754902 0.928431 0.768627 0.782843 0.440686 0.458333 0.316667 0.294608 0.266176 0.17451 nan nan 302.267 0.768137 0.937745 0.770098 0.789216 0.454902 0.476961 0.323039 0.296569 0.276471 0.186275 nan nan 302.467 0.783824 0.939706 0.769608 0.794608 0.47598 0.498039 0.328431 0.29951 0.281863 0.194118 nan nan 302.8 0.792647 0.940686 0.768627 0.8 0.486765 0.510294 0.337745 0.305392 0.285294 0.2 nan nan

视频片段中的第二个最大空间: 301.467 to 301.867

Answer 1

以下脚本首先将您的数据转换为行列表。对于每一行，丢弃初始值，然后创建包含所有对的排列列表。然后使用maths.hypot来计算距离。对于按距离排序的每个集合，将显示max_distance。

import math, itertools

data = """301.467 0.713235 0.852451 0.752451 0.753922 0.380392 0.373039 0.282843          0.291667 0.262255 0.283824 0.236765 0.218627
301.6 0.717647 0.864706 0.752941 0.761275 0.753431 0.743627 0.388235 0.383333 0.286765 0.294608 0.237745 0.207353
301.667 0.722059 0.876471 0.756863 0.763726 0.395098 0.391667 0.284314 0.289216 0.238725 0.198529 nan nan
301.867 0.72451 0.884314 0.762745 0.771078 0.40098 0.401961 0.290686 0.292647 0.240196 0.207843 0.241176 0.190686
302 0.731373 0.897059 0.760784 0.771569 0.413725 0.422549 0.302941 0.292157 0.25098 0.179412 nan nan
302.067 0.743627 0.917157 0.765686 0.777941 0.430392 0.444608 0.305392 0.293137 0.257843 0.17402 nan nan
302.2 0.754902 0.928431 0.768627 0.782843 0.440686 0.458333 0.316667 0.294608 0.266176 0.17451 nan nan
302.267 0.768137 0.937745 0.770098 0.789216 0.454902 0.476961 0.323039 0.296569 0.276471 0.186275 nan nan
302.467 0.783824 0.939706 0.769608 0.794608 0.47598 0.498039 0.328431 0.29951 0.281863 0.194118 nan nan
302.8 0.792647 0.940686 0.768627 0.8 0.486765 0.510294 0.337745 0.305392 0.285294 0.2 nan nan"""

def calc_distance(c1, c2):
    return math.hypot(c2[0] - c1[0], c2[1] - c1[1])

results = []

for row in data.split("\n"):
    cols = [float(col) for col in row.split(" ") if len(col)]
    max_distance = 0
    for c1,c2 in itertools.permutations(zip(cols[1:], cols[2:]), 2):
        max_distance = max(max_distance, calc_distance(c1,c2))

    results.append((max_distance, cols[0]))

results.sort()

for distance, frame in results:
    print "%-15s %f" % (frame, distance)

这将产生以下结果，显示按最大距离顺序排序的帧：

301.467         0.814885
301.6           0.831112
301.667         0.847618
301.867         0.860743
302.0           0.869144
302.067         0.887828
302.8           0.897788
302.267         0.898022
302.2           0.898471
302.467         0.898814

这是使用Python 2.7.9完成的。我没有手动验证结果。

根据您的评论进行更新

我仍然不完全确定我理解这个问题，但是下面的解决方案计算了所有帧的最小和最大坐标，并显示了找到每个外围坐标的帧。这将为您提供适合自动裁剪的信息。

# Hold the frame number and the value for each side
# Where 100 is greater than the expected height and width of the frame
# Assuming top left is (0,0)

frame_top = (0.0, 100.0)    
frame_left = (0.0, 100.0)
frame_bottom = (0.0, 0.0)
frame_right = (0.0, 0.0)

for row in data.split("\n"):
    cols = [float(col) for col in row.split(" ") if len(col)]
    icols = iter(cols[1:])

    for x in icols:
        y = next(icols)

        if x and y: # Skip any nan values
            if x < frame_left[1]:
                frame_left = (cols[0], x)
            if x > frame_right[1]:
                frame_right = (cols[0], x)
            if y < frame_top[1]:
                frame_top = (cols[0], y)
            if y > frame_bottom[1]:
                frame_bottom = (cols[0], y)

print "Frame %f has top %f" % frame_top
print "Frame %f has left %f" % frame_left
print "Frame %f has bottom %f" % frame_bottom
print "Frame %f has right %f" % frame_right

这给出了以下输出：

Frame 302.067000 has top 0.174020
Frame 301.467000 has left 0.236765
Frame 302.800000 has bottom 0.940686
Frame 302.800000 has right 0.792647

实际上它正在计算所有点的最小边界矩形。你当然可以提交较小范围的帧。

如何找到五点之间的最大距离？

1 个答案: