203 lines
7.1 KiB
Python
203 lines
7.1 KiB
Python
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#!/usr/bin/env python
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# -*- coding: utf-8 -*-
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"""
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@project:
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@File : 从合成里边提取剩余屋顶并计算潜力
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@Author : qiqq
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@create_time : 2023/11/6 15:29
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"""
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import os
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from tqdm import tqdm
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from collections import namedtuple
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from PIL import Image
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import numpy as np
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import cv2
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import random
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def generate_random_colors(n):
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Cls = namedtuple('cls', ['name', 'id', 'color'])
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colors = [
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Cls('backgroud', 0, (0, 0, 0))
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]
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while len(colors) < n:
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# 随机生成一个元组，元组中每个值在 0-255 之间
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color = Cls("roof"+str(len(colors)),len(colors),tuple(random.randint(0, 255) for _ in range(3)))
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colors.append(color)
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return list(colors)
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def get_putpalette(Clss, color_other=[0, 0, 0]):
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'''
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灰度图转8bit彩色图
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:param Clss:颜色映射表
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:param color_other:其余颜色设置
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:return:
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'''
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putpalette = []
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for cls in Clss:
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putpalette += list(cls.color)
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putpalette += color_other * (255 - len(Clss))
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return putpalette
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##chatgpt的代码二值图像连通区域分析与标记算法
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def connected_component_analysis(binary_image):
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# 使用findContours函数查找连通区域
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contours, _ = cv2.findContours(binary_image, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE) #最合适的一个
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# 对每个连通区域进行标记
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labeled_image = np.zeros_like(binary_image, dtype=np.uint16)
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for i, contour in enumerate(contours):
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cv2.drawContours(labeled_image, [contour], -1, i + 1, -1)
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# 统计连通区域数量
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num_labels = len(contours)
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return labeled_image, num_labels
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###############################################
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#1.从合成的多目标图中里边单独提取剩余屋顶形成二值图并保存
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def extract_remainroof():
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savename="remainroof"
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savepath_root=""
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path1=r""#多目标结果图的路径
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savepath=os.path.join(savepath_root,savename)
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if not os.path.exists(savepath):
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os.makedirs(savepath)
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imgss=[i for i in os.listdir(path1) if i.endswith("png")]
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for i in tqdm(imgss):
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im=os.path.join(path1,i)
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imm=np.array(Image.open(im))
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imm=(imm==1) * 1 #剩余屋顶在多目标结果图中的索引是1
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result = np.uint8(imm)
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#可视化保存
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palette = [0, 0, 0, 255, 0, 0, 0, 255, 0, 0, 0, 255]
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result = Image.fromarray(result).convert('P') #
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result.putpalette(palette)
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result.save(os.path.join(savepath, i))
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#对1提取的剩余屋顶进行实例化
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def remainroof_instanced():
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pathh = r"" #剩余屋顶保存的文件夹
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savepath_root=""
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savename = "r"
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savepath = os.path.join(savepath_root, savename)
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if not os.path.exists(savepath):
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os.makedirs(savepath)
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for i in tqdm(os.listdir(pathh)):
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imgp = os.path.join(pathh, i)
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binary_image = np.array(Image.open(imgp))
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labeled_image, num_labels = connected_component_analysis(binary_image)
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#实例化后只对那些图中小于256个实例的图像进行可视化保存，大于256的可视化不了
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if num_labels<256:
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palette = generate_random_colors(num_labels)
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result = np.uint8(labeled_image)
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result = Image.fromarray(result).convert('P')
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# 原来的
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l = get_putpalette(palette)
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result.putpalette(l)
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result.save(os.path.join(savepath,i))
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else:
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result = np.uint8(labeled_image)
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result = Image.fromarray(result).convert('P')
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result.save(os.path.join(savepath, i))
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'''
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3.
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取交集
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2步骤的实例化步骤存在缺陷比如 一个圆环，实例化之后会将这个圆环填补成完整的原 导致实例化后的屋顶偏大
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用没实例化之前的 * 实例化后的
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'''
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def yu():
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path1=r""#没实例化之前的 #也就是过程1的结果
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path2=r""#实例化之后的 #也就是过程2的结果
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savepath_root=""
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savename = ""
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savepath = os.path.join(savepath_root, savename)
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if not os.path.exists(savepath):
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os.makedirs(savepath)
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for i in tqdm(os.listdir(path1)):
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img1 = os.path.join(path1, i)
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img2 = os.path.join(path2, i)
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binary_image = np.array(Image.open(img1))
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instanceimage = np.array(Image.open(img2))
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result=binary_image*instanceimage
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num_labels=np.unique(result)
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t=len(num_labels)
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if t< 256:
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palette = generate_random_colors(t)
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result = np.uint8(result)
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result = Image.fromarray(result).convert('P')
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# 原来的
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l = get_putpalette(palette)
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result.putpalette(l)
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result.save(os.path.join(savepath, i))
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# print("1ok")
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else:
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result = np.uint8(result)
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result = Image.fromarray(result).convert('P')
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result.save(os.path.join(savepath, i))
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# print("2ok")
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#4.计算潜力
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def potentialmain():
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pathh = r""#过程3的结果
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gsd=0.26*0.26 #每个像素实际所占的面积
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singlepvare = 1.28 #每块光伏板的面积
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th=296 #阈值，小于这个阈值的屋顶就不计算他的潜力了 296是像素。
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total_pv=0
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total_roof=0
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avalibale=1
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for index,i in enumerate(os.listdir(pathh)):
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imgp = os.path.join(pathh, i)
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_image = np.array(Image.open(imgp))
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liss=np.unique(_image)
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labeled_image=_image
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image_deployedpv=0
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image_roof=0
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for n in liss:
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nnumbelr = np.sum(labeled_image == n)
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if nnumbelr <th:
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labeled_image[labeled_image==n]=0 ##!!
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newnum_labels = np.unique(labeled_image)
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for k in newnum_labels[1:]: # 注意0是背景
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current_instance_roofpiex = np.sum(labeled_image == k) # 统计当前实例的像素个数
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current_instance_roofare = current_instance_roofpiex * gsd * avalibale ##计算当前实例的面积
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current_instance_deployedpv = current_instance_roofare // singlepvare # 当前实例的可部署的光伏组件的数量
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print(f"当前图{index+1}的处理后的实例{k}的像素为{current_instance_roofpiex}，面积为{current_instance_roofare}，可安装的光伏组件块数为{current_instance_deployedpv}")
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# pertotal_pv = pertotal_pv + deployedpv
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# pertotal_roof = pertotal_roof + roofare
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image_deployedpv += current_instance_deployedpv
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image_roof += current_instance_roofare
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total_pv += image_deployedpv # 总计可以部署的光伏组件数量
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total_roof += image_roof
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print(f"当前图{i}的，可安装的光伏组件块数为{image_deployedpv}")
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print("++++++++++++++++++++++++++")
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print("总计的图像可安装的光伏面板的块数为：", total_pv)
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print("总计可用的屋顶的面积为：", total_roof)
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print("200ww总计光伏组件容量：", total_pv * 200)
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return total_pv
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potentialmain()
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