Python 如何使用PIL确定具有共享值的像素区域
我需要将图像分割为RGB值通过特定测试的像素区域。Python 如何使用PIL确定具有共享值的像素区域,python,numpy,scipy,python-imaging-library,cluster-analysis,Python,Numpy,Scipy,Python Imaging Library,Cluster Analysis,我需要将图像分割为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,但是将它们聚集到区域中,然后获取这些区域的坐标(x、y、宽度、高度)的部分会让我完全陷入黑暗:) 这是我目前掌握的代码 from PIL import Image def detectRedRegions(PILImage): image = PILImage.load() width, height = PILImage.size reds = [] h =
我可以扫描图像并检查每个像素的值,但是将它们聚集到区域中,然后获取这些区域的坐标(x、y、宽度、高度)的部分会让我完全陷入黑暗:)
这是我目前掌握的代码
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
从PIL导入图像
def检测区域(PILImage):
image=PILImage.load()
宽度,高度=PILImage.size
红色=[]
h=0
当h<高度:
w=0
当w<宽度时:
px=图像[w,h]
如果是红色(px):
reds.append([w,h])
#这就是我无知的地方
w+=1
h+=1
我读了很多关于集群的书,但是我不能完全理解这个主题。任何适合我需要的代码示例都会很好(并且希望能给我启发)
谢谢![编辑]
虽然下面的解决方案有效,但可以改进。以下是一个名称更好、性能更好的版本:
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()
它的灵感来自于
[/编辑]
人们可能会使用洪水填充,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()
从集合导入defaultdict
从PIL导入图像,ImageDraw
def连接的_部件(边缘):
"""
给定由边(即节点对)表示的图,生成其
作为节点集连接的组件。
时间复杂度与边数成线性关系。
"""
邻居=默认DICT(设置)
对于边中的a、b:
邻居[a]。添加(b)
邻居[b]。添加(a)
seen=set()
def组件(节点,邻居=邻居,看见=看见,看见=看见。添加):
不可见=设置([节点])
next_unseen=unseen.pop
虽然看不见:
node=next_unseen()
请参阅(节点)
看不见|=邻居[节点]-看到
屈服点
返回(为邻居中的节点设置(组件(节点)),如果节点不在视图中)
def匹配_像素(图像,测试):
"""
生成像素满足测试要求的所有像素坐标。
"""
宽度,高度=image.size
像素=image.load()
对于x范围内的x(宽度):
对于X范围内的y(高度):
如果测试(像素[x,y]):
产量x,y
def make_边(坐标):
"""
生成所有相邻像素坐标对。
"""
坐标=设置(坐标)
对于坐标中的x,y:
如果(x-1,y-1)在坐标中:
收益率(x,y),(x-1,y-1)
如果(x,y-1)在坐标系中:
收益率(x,y),(x,y-1)
如果(x+1,y-1)在坐标中:
收益率(x,y),(x+1,y-1)
如果(x-1,y)在坐标中:
收益率(x,y),(x-1,y)
收益率(x,y),(x,y)
def边界框(坐标):
"""
返回所有坐标的边界框。
"""
xs,ys=zip(*坐标)
返回最小值(xs)、最小值(ys)、最大值(xs)、最大值(ys)
def不相交区域(图像,测试):
"""
返回所有非连续区域的边界框
谁在测试。
"""
对于每个in-connected_组件(生成_边(匹配_像素(图像,测试)):
屈服边界框(每个)
def是否足够黑(像素):
r、 g,b=像素
返回r<10,g<10,b<10
如果uuuu name uuuuuu='\uuuuuuu main\uuuuuuu':
image=image.open('some_image.jpg'))
draw=ImageDraw.draw(图像)
对于不相交区域中的rect(图像,黑色是否足够):
矩形(矩形,轮廓=(255,0,0))
image.show()
这里,两个都满足
的相邻像素对是足够黑的()
被解释为图形中的边。此外,每个像素都被视为其自己的邻居。由于这种重新解释,我们可以对图形使用连接组件算法,这非常容易实现。结果是所有区域的边界框序列,这些区域的像素满足是否足够黑()
在图像处理中,您需要的是区域标记或连接组件检测。
包中提供了一个实现。
因此,如果您安装了numpy+scipy,那么以下内容应该可以工作
import numpy as np
import scipy.ndimage as ndi
import Image
image = Image.load()
# convert to numpy array (no data copy done since both use buffer protocol)
image = np.asarray(image)
# generate a black and white image marking red pixels as 1
bw = is_red(image)
# labeling : each region is associated with an int
labels, n = ndi.label(bw)
# provide bounding box for each region in the form of tuples of slices
objects = ndi.find_objects(labels)
你到底想做什么?一般来说,具有特定颜色的区域不会是正方形,而是任意形状(尽管以某种方式连接),因此用简单的元组(如x、y、宽度、高度)来查找和定义区域将是一个挑战。我知道区域不会是正方形:)我所想的(很抱歉没有弄清楚这一点)是获取区域周围的边界框(应该足够精确,以满足我的需要)
import numpy as np
import scipy.ndimage as ndi
import Image
image = Image.load()
# convert to numpy array (no data copy done since both use buffer protocol)
image = np.asarray(image)
# generate a black and white image marking red pixels as 1
bw = is_red(image)
# labeling : each region is associated with an int
labels, n = ndi.label(bw)
# provide bounding box for each region in the form of tuples of slices
objects = ndi.find_objects(labels)