DDR爱好者之家

Python识别处理照片中的条形码
站长资源 2024/11/24 佚名
7 1538
DDR爱好者之家 Design By 杰米

最近一直在玩数独，突发奇想实现图像识别求解数独，输入到输出平均需要0.5s。

整体思路大概就是识别出图中数字生成list，然后求解。

输入输出demo

数独采用的是微软自带的Microsoft sudoku软件随便截取的图像，如下图所示：

Python识别处理照片中的条形码

经过程序求解后，得到的结果如下图所示：

Python识别处理照片中的条形码

def getFollow(varset, terminalset, first_dic, production_list):
    follow_dic = {}
    done = {}
    for var in varset:
        follow_dic[var] = set()
        done[var] = 0
    follow_dic["A1"].add("#")
    # for var in terminalset:
    #     follow_dic[var]=set()
    #     done[var] = 0
    for var in follow_dic:
        getFollowForVar(var, varset, terminalset, first_dic, production_list, follow_dic, done)
    return follow_dic
  
  
def getFollowForVar(var, varset, terminalset, first_dic, production_list, follow_dic, done):
    if done[var] == 1:
        return
    for production in production_list:
        if var in production.right:
            ##index这里在某些极端情况下有bug，比如多次出现var，index只会返回最左侧的
            if production.right.index(var) != len(production.right) - 1:
                follow_dic[var] = first_dic[production.right[production.right.index(var) + 1]] | follow_dic[var]
            # 没有考虑右边有非终结符但是为null的情况
            if production.right[len(production.right) - 1] == var:
                if var != production.left[0]:
                    # print(var, "吸纳", production.left[0])
                    getFollowForVar(production.left[0], varset, terminalset, first_dic, production_list, follow_dic,
                                    done)
                    follow_dic[var] = follow_dic[var] | follow_dic[production.left[0]]
  
    done[var] = 1

程序具体流程

程序整体流程如下图所示：

Python识别处理照片中的条形码

读入图像后，根据求解轮廓信息找到数字所在位置，以及不包含数字的空白位置，提取数字信息通过KNN识别，识别出数字；无数字信息的在list中置0；生成未求解数独list，之后求解数独，将信息在原图中显示出来。

def initProduction():
    production_list = []
    production = Production(["A1"], ["A"], 0)
    production_list.append(production)
    production = Production(["A"], ["E", "I", "(", ")", "{", "D", "}"], 1)
    production_list.append(production)
    production = Production(["E"], ["int"], 2)
    production_list.append(production)
    production = Production(["E"], ["float"], 3)
    production_list.append(production)
    production = Production(["D"], ["D", ";", "B"], 4)
    production_list.append(production)
    production = Production(["B"], ["F"], 5)
    production_list.append(production)
    production = Production(["B"], ["G"], 6)
    production_list.append(production)
    production = Production(["B"], ["M"], 7)
    production_list.append(production)
    production = Production(["F"], ["E", "I"], 8)
    production_list.append(production)
    production = Production(["G"], ["I", "=", "P"], 9)
    production_list.append(production)
    production = Production(["P"], ["K"], 10)
    production_list.append(production)
    production = Production(["P"], ["K", "+", "P"], 11)
    production_list.append(production)
    production = Production(["P"], ["K", "-", "P"], 12)
    production_list.append(production)
    production = Production(["I"], ["id"], 13)
    production_list.append(production)
    production = Production(["K"], ["I"], 14)
    production_list.append(production)
    production = Production(["K"], ["number"], 15)
    production_list.append(production)
    production = Production(["K"], ["floating"], 16)
    production_list.append(production)
    production = Production(["M"], ["while", "(", "T", ")", "{", "D", ";", "}"], 18)
    production_list.append(production)
    production = Production(["N"], ["if", "(", "T", ")", "{", "D",";", "}", "else", "{", "D", ";","}"], 19)
    production_list.append(production)
    production = Production(["T"], ["K", "L", "K"], 20)
    production_list.append(production)
    production = Production(["L"], [">"], 21)
    production_list.append(production)
    production = Production(["L"], ["<"], 22)
    production_list.append(production)
    production = Production(["L"], [">="], 23)
    production_list.append(production)
    production = Production(["L"], ["<="], 24)
    production_list.append(production)
    production = Production(["L"], ["=="], 25)
    production_list.append(production)
    production = Production(["D"], ["B"], 26)
    production_list.append(production)
    production = Production(["B"], ["N"], 27)
    production_list.append(production)
    return production_list
 
 
source = [[5, "int", " 关键字"], [1, "lexicalanalysis", " 标识符"], [13, "(", " 左括号"], [14, ")", " 右括号"], [20, "{", " 左大括号"],
          [4, "float", " 关键字"], [1, "a", " 标识符"], [15, ";", " 分号"], [5, "int", " 关键字"], [1, "b", " 标识符"],
          [15, ";", " 分号"], [1, "a", " 标识符"], [12, "=", " 赋值号"], [3, "1.1", " 浮点数"], [15, ";", " 分号"], [1, "b", " 标识符"],
          [12, "=", " 赋值号"], [2, "2", " 整数"], [15, ";", " 分号"], [8, "while", "  关键字"], [13, "(", " 左括号"],
          [1, "b", " 标识符"], [17, "<", " 小于号"], [2, "100", " 整数"], [14, ")", " 右括号"], [20, "{", " 左大括号"],
          [1, "b", " 标识符"], [12, "=", " 赋值号"], [1, "b", " 标识符"], [9, "+", " 加 号"], [2, "1", " 整数"], [15, ";", " 分号"],
          [1, "a", " 标识符"], [12, "=", " 赋值号"], [1, "a", " 标识符"], [9, "+", " 加号"], [2, "3", " 整数"], [15, ";", " 分号"],
          [21, "}", " 右大括号"], [15, ";", " 分号"], [6, "if", " 关键字"], [13, "(", " 左括号"], [1, "a", " 标识符"],
          [16, ">", " 大于号"], [2, "5", " 整数"], [14, ")", " 右括号"], [20, "{", " 左大括号"], [1, "b", " 标识符"],
          [12, "=", " 赋值号"], [1, "b", " 标识符"], [10, "-", " 减号"], [2, "1", " 整数"], [15, ";", " 分号"], [21, "}", " 右大括号"],
          [7, "else", " 关键字"], [20, "{", " 左大括号"], [1, "b", " 标识符"], [12, "=", " 赋值号"], [1, "b", " 标识符"],
          [9, "+", " 加号"], [2, "1", " 整数"], [15, ";", " 分号"], [21, "}", " 右大括号"], [21, "}", " 右大括号"]]

以上就是Python识别处理照片中的条形码的详细内容，更多关于python 识别条形码的资料请关注其它相关文章！

DDR爱好者之家 Design By 杰米
广告合作：本站广告合作请联系QQ：858582 申请时备注：广告合作（否则不回）
免责声明：本站资源来自互联网收集,仅供用于学习和交流,请遵循相关法律法规,本站一切资源不代表本站立场,如有侵权、后门、不妥请联系本站删除！
DDR爱好者之家 Design By 杰米
ddrfans.com DDR爱好者之家
8,675无损音乐
1,324高清电影
213破解软件
70,141收录资讯

《魔兽世界》大逃杀！60人新游玩模式《强袭风暴》3月21日上线

暴雪近日发布了《魔兽世界》10.2.6 更新内容，新游玩模式《强袭风暴》即将于3月21 日在亚服上线，届时玩家将前往阿拉希高地展开一场 60 人大逃杀对战。

艾泽拉斯的冒险者已经征服了艾泽拉斯的大地及遥远的彼岸。他们在对抗世界上最致命的敌人时展现出过人的手腕，并且成功阻止终结宇宙等级的威胁。当他们在为即将于《魔兽世界》资料片《地心之战》中来袭的萨拉塔斯势力做战斗准备时，他们还需要在熟悉的阿拉希高地面对一个全新的敌人──那就是彼此。在《巨龙崛起》10.2.6 更新的《强袭风暴》中，玩家将会进入一个全新的海盗主题大逃杀式限时活动，其中包含极高的风险和史诗级的奖励。

《强袭风暴》不是普通的战场，作为一个独立于主游戏之外的活动，玩家可以用大逃杀的风格来体验《魔兽世界》，不分职业、不分装备（除了你在赛局中捡到的），光是技巧和战略的强弱之分就能决定出谁才是能坚持到最后的赢家。本次活动将会开放单人和双人模式，玩家在加入海盗主题的预赛大厅区域前，可以从强袭风暴角色画面新增好友。游玩游戏将可以累计名望轨迹，《巨龙崛起》和《魔兽世界：巫妖王之怒 经典版》的玩家都可以获得奖励。

更新日志

友情链接

杰晶网络 DDR爱好者之家 南强小屋 黑松山资源网 白云城资源网
DDR爱好者之家 Design By 杰米