时间:2023-07-05 22:54:01 | 来源:网站运营
时间:2023-07-05 22:54:01 来源:网站运营
用Python做一个游戏辅助脚本:转载!1 # -*- coding:utf-8 -*- 2 3 import win32gui 4 import time 5 from PIL import ImageGrab, Image 6 import numpy as np 7 import operator 8 from pymouse import PyMouse 9 10 11 class GameAssist: 12 13 def __init__(self, wdname): 14 """初始化""" 15 16 # 取得窗口句柄 17 self.hwnd = win32gui.FindWindow(0, wdname) 18 if not self.hwnd: 19 print("窗口找不到,请确认窗口句柄名称:【%s】" % wdname ) 20 exit() 21 22 # 窗口显示最前面 23 win32gui.SetForegroundWindow(self.hwnd) 24 25 # 小图标编号矩阵 26 self.im2num_arr = [] 27 28 # 主截图的左上角坐标和右下角坐标 29 self.scree_left_and_right_point = (299, 251, 768, 564) 30 # 小图标宽高 31 self.im_width = 39 32 33 # PyMouse对象,鼠标点击 34 self.mouse = PyMouse() 35 36 def screenshot(self): 37 """屏幕截图""" 38 39 # 1、用grab函数截图,参数为左上角和右下角左标 40 # image = ImageGrab.grab((417, 257, 885, 569)) 41 image = ImageGrab.grab(self.scree_left_and_right_point) 42 43 # 2、分切小图 44 # exit() 45 image_list = {} 46 offset = self.im_width # 39 47 48 # 8行12列 49 for x in range(8): 50 image_list[x] = {} 51 for y in range(12): 52 # print("show",x, y) 53 # exit() 54 top = x * offset 55 left = y * offset 56 right = (y + 1) * offset 57 bottom = (x + 1) * offset 58 59 # 用crop函数切割成小图标,参数为图标的左上角和右下角左边 60 im = image.crop((left, top, right, bottom)) 61 # 将切割好的图标存入对应的位置 62 image_list[x][y] = im 63 64 return image_list 65 66 def image2num(self, image_list): 67 """将图标矩阵转换成数字矩阵""" 68 69 # 1、创建全零矩阵和空的一维数组 70 arr = np.zeros((10, 14), dtype=np.int32) # 以数字代替图片 71 image_type_list = [] 72 73 # 2、识别出不同的图片,将图片矩阵转换成数字矩阵 74 for i in range(len(image_list)): 75 for j in range(len(image_list[0])): 76 im = image_list[i][j] 77 78 # 验证当前图标是否已存入 79 index = self.getIndex(im, image_type_list) 80 81 # 不存在image_type_list 82 if index < 0: 83 image_type_list.append(im) 84 arr[i + 1][j + 1] = len(image_type_list) 85 else: 86 arr[i + 1][j + 1] = index + 1 87 88 print("图标数:", len(image_type_list)) 89 90 self.im2num_arr = arr 91 return arr 92 93 # 检查数组中是否有图标,如果有则返回索引下表 94 def getIndex(self,im, im_list): 95 for i in range(len(im_list)): 96 if self.isMatch(im, im_list[i]): 97 return i 98 99 return -1100101 # 汉明距离判断两个图标是否一样102 def isMatch(self, im1, im2):103104 # 缩小图标,转成灰度105 image1 = im1.resize((20, 20), Image.ANTIALIAS).convert("L")106 image2 = im2.resize((20, 20), Image.ANTIALIAS).convert("L")107108 # 将灰度图标转成01串,即系二进制数据109 pixels1 = list(image1.getdata())110 pixels2 = list(image2.getdata())111112 avg1 = sum(pixels1) / len(pixels1)113 avg2 = sum(pixels2) / len(pixels2)114 hash1 = "".join(map(lambda p: "1" if p > avg1 else "0", pixels1))115 hash2 = "".join(map(lambda p: "1" if p > avg2 else "0", pixels2))116117 # 统计两个01串不同数字的个数118 match = sum(map(operator.ne, hash1, hash2))119120 # 阀值设为10121 return match < 10122123 # 判断矩阵是否全为0124 def isAllZero(self, arr):125 for i in range(1, 9):126 for j in range(1, 13):127 if arr[i][j] != 0:128 return False129 return True130131 # 是否为同行或同列且可连132 def isReachable(self, x1, y1, x2, y2):133 # 1、先判断值是否相同134 if self.im2num_arr[x1][y1] != self.im2num_arr[x2][y2]:135 return False136137 # 1、分别获取两个坐标同行或同列可连的坐标数组138 list1 = self.getDirectConnectList(x1, y1)139 list2 = self.getDirectConnectList(x2, y2)140 # print(x1, y1, list1)141 # print(x2, y2, list2)142143 # exit()144145 # 2、比较坐标数组中是否可连146 for x1, y1 in list1:147 for x2, y2 in list2:148 if self.isDirectConnect(x1, y1, x2, y2):149 return True150 return False151152 # 获取同行或同列可连的坐标数组153 def getDirectConnectList(self, x, y):154155 plist = []156 for px in range(0, 10):157 for py in range(0, 14):158 # 获取同行或同列且为0的坐标159 if self.im2num_arr[px][py] == 0 and self.isDirectConnect(x, y, px, py):160 plist.append([px, py])161162 return plist163164 # 是否为同行或同列且可连165 def isDirectConnect(self, x1, y1, x2, y2):166 # 1、位置完全相同167 if x1 == x2 and y1 == y2:168 return False169170 # 2、行列都不同的171 if x1 != x2 and y1 != y2:172 return False173174 # 3、同行175 if x1 == x2 and self.isRowConnect(x1, y1, y2):176 return True177178 # 4、同列179 if y1 == y2 and self.isColConnect(y1, x1, x2):180 return True181182 return False183184 # 判断同行是否可连185 def isRowConnect(self, x, y1, y2):186 minY = min(y1, y2)187 maxY = max(y1, y2)188189 # 相邻直接可连190 if maxY - minY == 1:191 return True192193 # 判断两个坐标之间是否全为0194 for y0 in range(minY + 1, maxY):195 if self.im2num_arr[x][y0] != 0:196 return False197 return True198199 # 判断同列是否可连200 def isColConnect(self, y, x1, x2):201 minX = min(x1, x2)202 maxX = max(x1, x2)203204 # 相邻直接可连205 if maxX - minX == 1:206 return True207208 # 判断两个坐标之间是否全为0209 for x0 in range(minX + 1, maxX):210 if self.im2num_arr[x0][y] != 0:211 return False212 return True213214 # 点击事件并设置数组为0215 def clickAndSetZero(self, x1, y1, x2, y2):216 # print("click", x1, y1, x2, y2)217218 # (299, 251, 768, 564)219 # 原理:左上角图标中点 + 偏移量220 p1_x = int(self.scree_left_and_right_point[0] + (y1 - 1)*self.im_width + (self.im_width / 2))221 p1_y = int(self.scree_left_and_right_point[1] + (x1 - 1)*self.im_width + (self.im_width / 2))222223 p2_x = int(self.scree_left_and_right_point[0] + (y2 - 1)*self.im_width + (self.im_width / 2))224 p2_y = int(self.scree_left_and_right_point[1] + (x2 - 1)*self.im_width + (self.im_width / 2))225226 time.sleep(0.2)227 self.mouse.click(p1_x, p1_y)228 time.sleep(0.2)229 self.mouse.click(p2_x, p2_y)230231 # 设置矩阵值为0232 self.im2num_arr[x1][y1] = 0233 self.im2num_arr[x2][y2] = 0234235 print("消除:(%d, %d) (%d, %d)" % (x1, y1, x2, y2))236 # exit()237238 # 程序入口、控制中心239 def start(self):240241 # 1、先截取游戏区域大图,然后分切每个小图242 image_list = self.screenshot()243244 # 2、识别小图标,收集编号245 self.image2num(image_list)246247 print(self.im2num_arr)248249 # 3、遍历查找可以相连的坐标250 while not self.isAllZero(self.im2num_arr):251 for x1 in range(1, 9):252 for y1 in range(1, 13):253 if self.im2num_arr[x1][y1] == 0:254 continue255256 for x2 in range(1, 9):257 for y2 in range(1, 13):258 # 跳过为0 或者同一个259 if self.im2num_arr[x2][y2] == 0 or (x1 == x2 and y1 == y2):260 continue261 if self.isReachable(x1, y1, x2, y2):262 self.clickAndSetZero(x1, y1, x2, y2)263264265 if __name__ == "__main__":266 # wdname 为连连看窗口的名称,必须写完整267 wdname = u'宠物连连看经典版2,宠物连连看经典版2小游戏,4399小游戏 www.4399.com - Google Chrome'268269 demo = GameAssist(wdname)270 demo.start()GameAssist.py
关键词:游戏,辅助,脚本