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python - 如何使用 PIL 确定具有共享值的像素区域

转载 作者:行者123 更新时间:2023-11-28 22:01:57 25 4
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我需要将图像划分为RGB值通过特定测试的像素区域。
我可以扫描图像并检查每个像素的值,但是将它们聚类成区域然后获取这些区域坐标(x,y,宽度,高度)的部分让我完全黑暗:)
这是我到目前为止的代码

from PIL import Image

def detectRedRegions(PILImage):
image = PILImage.load()
width, height = PILImage.size
reds = []
h = 0
while h < height:
w = 0
while w < width:
px = image[w, h]
if is_red(px):
reds.append([w, h])
# Here's where I'm being clueless
w +=1
h +=1

我读了很多关于集群的书,但我无法完全理解这个主题,任何适合我需要的代码示例都会很棒(并希望能启发

谢谢!

最佳答案

[编辑]

虽然下面的解决方案有效,但可以做得更好。这是一个具有更好名称和更好性能的版本:

from itertools import product
from PIL import Image, ImageDraw


def closed_regions(image, test):
"""
Return all closed regions in image who's pixels satisfy test.
"""
pixel = image.load()
xs, ys = map(xrange, image.size)
neighbors = dict((xy, set([xy])) for xy in product(xs, ys) if test(pixel[xy]))
for a, b in neighbors:
for cd in (a + 1, b), (a, b + 1):
if cd in neighbors:
neighbors[a, b].add(cd)
neighbors[cd].add((a, b))
seen = set()
def component(node, neighbors=neighbors, seen=seen, see=seen.add):
todo = set([node])
next_todo = todo.pop
while todo:
node = next_todo()
see(node)
todo |= neighbors[node] - seen
yield node
return (set(component(node)) for node in neighbors if node not in seen)


def boundingbox(coordinates):
"""
Return the bounding box that contains all coordinates.
"""
xs, ys = zip(*coordinates)
return min(xs), min(ys), max(xs), max(ys)


def is_black_enough(pixel):
r, g, b = pixel
return r < 10 and g < 10 and b < 10


if __name__ == '__main__':

image = Image.open('some_image.jpg')
draw = ImageDraw.Draw(image)
for rect in disjoint_areas(image, is_black_enough):
draw.rectangle(boundingbox(region), outline=(255, 0, 0))
image.show()

与下面的 disjoint_areas() 不同,closed_regions() 返回像素坐标集而不是它们的边界框。

此外,如果我们使用 flooding代替连通分量算法,我们可以使它更简单,速度大约提高一倍:

from itertools import chain, product
from PIL import Image, ImageDraw


flatten = chain.from_iterable


def closed_regions(image, test):
"""
Return all closed regions in image who's pixel satisfy test.
"""
pixel = image.load()
xs, ys = map(xrange, image.size)
todo = set(xy for xy in product(xs, ys) if test(pixel[xy]))
while todo:
region = set()
edge = set([todo.pop()])
while edge:
region |= edge
todo -= edge
edge = todo.intersection(
flatten(((x - 1, y), (x, y - 1), (x + 1, y), (x, y + 1)) for x, y in edge))
yield region

# rest like above

它的灵感来自 Eric S. Raymond's version of floodfill .

[/编辑]

人们可能会使用 floodfill,但我喜欢这样:

from collections import defaultdict
from PIL import Image, ImageDraw


def connected_components(edges):
"""
Given a graph represented by edges (i.e. pairs of nodes), generate its
connected components as sets of nodes.

Time complexity is linear with respect to the number of edges.
"""
neighbors = defaultdict(set)
for a, b in edges:
neighbors[a].add(b)
neighbors[b].add(a)
seen = set()
def component(node, neighbors=neighbors, seen=seen, see=seen.add):
unseen = set([node])
next_unseen = unseen.pop
while unseen:
node = next_unseen()
see(node)
unseen |= neighbors[node] - seen
yield node
return (set(component(node)) for node in neighbors if node not in seen)


def matching_pixels(image, test):
"""
Generate all pixel coordinates where pixel satisfies test.
"""
width, height = image.size
pixels = image.load()
for x in xrange(width):
for y in xrange(height):
if test(pixels[x, y]):
yield x, y


def make_edges(coordinates):
"""
Generate all pairs of neighboring pixel coordinates.
"""
coordinates = set(coordinates)
for x, y in coordinates:
if (x - 1, y - 1) in coordinates:
yield (x, y), (x - 1, y - 1)
if (x, y - 1) in coordinates:
yield (x, y), (x, y - 1)
if (x + 1, y - 1) in coordinates:
yield (x, y), (x + 1, y - 1)
if (x - 1, y) in coordinates:
yield (x, y), (x - 1, y)
yield (x, y), (x, y)


def boundingbox(coordinates):
"""
Return the bounding box of all coordinates.
"""
xs, ys = zip(*coordinates)
return min(xs), min(ys), max(xs), max(ys)


def disjoint_areas(image, test):
"""
Return the bounding boxes of all non-consecutive areas
who's pixels satisfy test.
"""
for each in connected_components(make_edges(matching_pixels(image, test))):
yield boundingbox(each)


def is_black_enough(pixel):
r, g, b = pixel
return r < 10 and g < 10 and b < 10


if __name__ == '__main__':

image = Image.open('some_image.jpg')
draw = ImageDraw.Draw(image)
for rect in disjoint_areas(image, is_black_enough):
draw.rectangle(rect, outline=(255, 0, 0))
image.show()

此处,同时满足 is_black_enough() 的相邻像素对被解释为图中的边。此外,每个像素都被视为自己的邻居。由于这种重新解释,我们可以对图使用连通分量算法,这很容易实现。结果是像素满足is_black_enough()的所有区域的边界框序列。

关于python - 如何使用 PIL 确定具有共享值的像素区域,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12321899/

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