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python - 检测图像上的对象边界太慢

问题描述

我有一个label_image, 作为一个看起来像这样的数据框。请注意，一些对象，那些标记为1并被3其他对象完全吞没/包围的对象，没有背景将对象彼此分开。在label_image这里作为输入给出。我自己没有制作它，因为我没有实际图像（如 tif、jpg 等）

dummy_img = pd.DataFrame(np.array([
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
]))

我想得到物体的轮廓。

我最初写的是这样的：

import pandas as pd
import numpy as np
from scipy.sparse import coo_matrix
from scipy.ndimage import binary_erosion

def outline(label_image):
    res_list = []
    
    mask = binary_erosion(label_image.values)
    label_image[mask] = 0
    c = coo_matrix(label_image)
    if c.data.size > 0:
        df = pd.DataFrame({'coords': list(zip(c.col, c.row)), 'label': c.data})
        df = df.groupby('label')['coords'].apply(lambda group_series: group_series.tolist()).reset_index()
        df = df.astype({"label": int})
    else:
        df = pd.DataFrame()
    return df

这给出了一个输出：

label   coords
2       [(3, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3...
4       [(12, 3), (13, 3), (14, 3), (15, 3), (16, 3), ...

这是错误的。它不仅错过了内部对象，而且检测到的对象的派生坐标也是错误的。例如label 4，它返回之间的轮廓labels 4 and 0并忽略之间的轮廓labels 4 and 3。

我做了这个修复

def outline_fix(label_image):
    res_list = []
    coo = coo_matrix(label_image)
    labels = np.unique(coo.data)
    for label in sorted(set(labels)):
        #print('label: %d' % label)
        c = coo.copy()
        c.data[c.data != label] = 0
        c = c.toarray()
        mask = binary_erosion(c)
        c[mask] = 0
        c = coo_matrix(c)
        if c.data.size > 0:
            df = pd.DataFrame({'coords': list(zip(c.col, c.row)), 'label': c.data})
            df = df.groupby('label')['coords'].apply(lambda group_series: group_series.tolist()).reset_index()
            df = df.astype({"label": int})
        else:
            df = pd.DataFrame()
        res_list.append(df)
        
    if res_list:
        out = pd.concat(res_list).astype({"label": int})
    else:
        out = pd.DataFrame()
        
    return out

返回：

label   coords
1       [(5, 6), (6, 6), (7, 6), (8, 6), (9, 6), (5, 7...
2       [(3, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3...
3       [(12, 5), (13, 5), (14, 5), (10, 6), (11, 6), ...
4       [(12, 3), (13, 3), (14, 3), (15, 3), (16, 3), ...

效果很好。不是 100% 完美，例如，因为label 4它错过了两对坐标。坐标列表应该有 length=32 而不是我得到的 30，但这是我可以忍受的，它不是很重要。

修正后的功能的问题在于它非常缓慢。在实际情况下，我有一个2000-by-2000包含多个2800对象的数组。并非所有对象都是“嵌套的”，但它可能很多，取决于传入的图像。

代码在大约 4 分钟内完成（对于真实的真实案例），这太长了。请问有没有其他选择，或者有什么想法可以加快速度？

非常感谢任何帮助

标签： pythonpandasimage-processingimage-segmentationscikit-image

解决方案

.apply根据我的经验，当你必须使用函数时，pandas 非常慢。出于这个原因，我更喜欢对其进行操作.values然后重新分配。话虽如此，熊猫可以制作得非常快，但它本身就是一门艺术。检查此代码，对于大图像，它肯定会比 pandas 实现更快：

import numpy as np
import pandas as pd
from collections import Counter

dummy_img = pd.DataFrame(np.array([
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,1,1,1,1,1,3,3,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,3,3,3,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,2,2,2,2,2,2,2,2,2,4,4,4,4,4,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
]))

# Array of boundary points
# row - label, col - point coordinates

img_np = dummy_img.values
b_pts = [[] for i in range(4)] 

# idea is simple - run a 3x3 window and check whenever the mean value
# inside the window differs from element at the center. 

img_padded = np.pad(img_np, ((1, 1), (1, 1)))
    
for r, row in enumerate(img_np):
  for c, elem in enumerate(row):
    if Flase in (img_padded[r:r+3, c:c+3] == elem):
      b_pts[elem-1].append((r, c))

for elem in b_pts:     
  print(elem)

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