怎样通过数字提示，输入答案检索正确与否？

zhang1497 · 发表于 2021-11-23 21:59

本帖最后由 zhang1497 于 2021-11-25 19:30 编辑

先上图

最新在练习数字编码表，于是就想着能不能用python做一个程序。

经过指导，代码这样写了：
1。’新的问题来了，怎么实现一直出题，然后答对以后删题，继续出题。答错了不删题，继续下一题。直到字典里的元素为空时，提示闯关成功?
2。经过自己摸索，翻书，查百度。完成了。
3.还要再加上一个提前结束的指令，有什么没那么多时间。还没研究。下午再研究下。

[Python] 纯文本查看 复制代码

import random
A={"00":"望远镜","01":"小树","02":"铃儿","03":"凳子","04":"轿车","05":"手套","06":"手枪","07":"锄头","08":"溜冰鞋","09":"猫",
    "0":"呼啦圈","1":"蜡烛","2":"鹅","3":"耳朵","4":"帆船","5":"钩子","6":"勺子","7":"镰刀","8":"眼睛","9":"哨子","10":"棒球",
    "11":"梯子","12":"椅儿","13":"医生","14":"钥匙","15":"鹦鹉","16":"石榴","17":"仪器","18":"糖葫芦","19":"衣钩","20":"香烟",
    "21":"鳄鱼","22":"双胞胎","23":"和尚","24":"闹钟","25":"二胡","26":"河流","27":"耳机","28":"恶霸","29":"饿囚","30":"三轮车",
    "31":"鲨鱼","32":"扇儿","33":"星星","34":"三丝","35":"山虎","36":"山鹿","37":"山鸡","38":"妇女","39":"山丘","40":"司令",
    "41":"蜥蜴","42":"柿儿","43":"石山","44":"蛇","45":"师父","46":"饲料","47":"司机","48":"石板","49":"湿狗","50":"武林",
    "51":"工人","52":"鼓儿","53":"乌纱帽","54":"青年","55":"火车","56":"蜗牛","57":"武器","58":"尾巴","59":"蜈蚣","60":"榴莲",
    "61":"儿童","62":"牛儿","63":"流沙","64":"螺丝","65":"绿壶","66":"溜溜球","67":"绿漆","68":"喇叭","69":"太极","70":"麒麟",
    "71":"鸡翼","72":"企鹅","73":"花旗参","74":"骑士","75":"西服","76":"汽油","77":"机器","78":"青蛙","79":"气球","80":"巴黎",
    "81":"白蚁","82":"靶儿","83":"芭蕉扇","84":"巴士","85":"保姆","86":"八路","87":"白旗","88":"爸爸","89":"芭蕉","90":"酒瓶",
    "91":"球衣","92":"球儿","93":"旧伞","94":"首饰","95":"酒壶","96":"蝴蝶","97":"旧旗","98":"酒杯","99":"舅舅"}
while len(A) != 0:                              #建立一个判断，当列表不为空时，执行以下代码。
    qustion = random.choice(list(A.keys()))     #用字典A的key建立一个list(列表)，通过random函数随机抽一个key，定义question变量
    answer = input('%s-' % (qustion))           #输入key键所对应的value值，例如给的是“00”这个键，那输入“望远镜”这个值。
    if answer == A[qustion]:                    #判断输入的这个值是不是对应了A列表里的key键。
        print("恭喜你，答对了")                    #如果对了，打印“恭喜你答对了”
        del A[qustion]                          #删除这个key,value。不断做对，列表A会越爱越少，直到变成空列表。
        continue                                #答对了题目，继续出题。
    else:                                       #如果输入的这个值不是key键所对应的。
        print('不好意思，你答错了，正确答案是',A[qustion])         #打印“不好意思，你错了”，并输入正确答案。不删题处理。
print("恭喜你过关了")                             #如果列表为空了，说明全部做对了，打印“恭喜过关”

4.晚上给它加上了分数和次数。接下来要再完善下多少分才算通关，否则失败。以及如何提前退出循环系统。注释如有不对还望提醒。

[Python] 纯文本查看 复制代码

import random
A={"00":"望远镜","01":"小树","02":"铃儿","03":"凳子","04":"轿车","05":"手套","06":"手枪","07":"锄头","08":"溜冰鞋","09":"猫",
    "0":"呼啦圈","1":"蜡烛","2":"鹅","3":"耳朵","4":"帆船","5":"钩子","6":"勺子","7":"镰刀","8":"眼睛","9":"哨子","10":"棒球",
    "11":"梯子","12":"椅儿","13":"医生","14":"钥匙","15":"鹦鹉","16":"石榴","17":"仪器","18":"糖葫芦","19":"衣钩","20":"香烟",
    "21":"鳄鱼","22":"双胞胎","23":"和尚","24":"闹钟","25":"二胡","26":"河流","27":"耳机","28":"恶霸","29":"饿囚","30":"三轮车",
    "31":"鲨鱼","32":"扇儿","33":"星星","34":"三丝","35":"山虎","36":"山鹿","37":"山鸡","38":"妇女","39":"山丘","40":"司令",
    "41":"蜥蜴","42":"柿儿","43":"石山","44":"蛇","45":"师父","46":"饲料","47":"司机","48":"石板","49":"湿狗","50":"武林",
    "51":"工人","52":"鼓儿","53":"乌纱帽","54":"青年","55":"火车","56":"蜗牛","57":"武器","58":"尾巴","59":"蜈蚣","60":"榴莲",
    "61":"儿童","62":"牛儿","63":"流沙","64":"螺丝","65":"绿壶","66":"溜溜球","67":"绿漆","68":"喇叭","69":"太极","70":"麒麟",
    "71":"鸡翼","72":"企鹅","73":"花旗参","74":"骑士","75":"西服","76":"汽油","77":"机器","78":"青蛙","79":"气球","80":"巴黎",
    "81":"白蚁","82":"靶儿","83":"芭蕉扇","84":"巴士","85":"保姆","86":"八路","87":"白旗","88":"爸爸","89":"芭蕉","90":"酒瓶",
    "91":"球衣","92":"球儿","93":"旧伞","94":"首饰","95":"酒壶","96":"蝴蝶","97":"旧旗","98":"酒杯","99":"舅舅"}
sum = 0                                         #定义一个变量初始值为0,收集做对题目数量
times = 20                                      #20次机会
while len(A) != 0 and times > 0 :               #建立一个判断，当列表不为空，而且输入次数不为0，执行以下代码。
    qustion = random.choice(list(A.keys()))     #用字典A的key建立一个list(列表)，通过random函数随机抽一个key，定义question变量
    answer = input('%s--' % (qustion))          #输入key键所对应的value值，例如给的是“00”这个键，那输入“望远镜”这个值。
    times = times - 1                           #次数减1
    if answer == A[qustion]:                    #判断输入的这个值是不是对应了A列表里的key键。
        print("恭喜你，答对了")                    #如果对了，打印“恭喜你答对了”
        del A[qustion]                          #删除这个key,value。不再出现这个题，避免重复做题。
        sum = sum + 1                           #答对一题记一分
        continue                                #答对了题目，继续出题。
    else:                                       #如果输入的这个值不是key键所对应的。
        print('不好意思，你答错了，正确答案是',A[qustion])         #打印“不好意思，你错了”，并输入正确答案。不删题处理。
print("恭喜你过关了,做对了",sum,"道题，共计",sum*5,"分")           #20次机会用完了以后看看自己得了多少分

5，加了一个错误次数，以及错误时要多做5题。还有提前退出指令。

[Python] 纯文本查看 复制代码

import random
A={"00":"望远镜","01":"小树","02":"铃儿","03":"凳子","04":"轿车","05":"手套","06":"手枪","07":"锄头","08":"溜冰鞋","09":"猫",
    "0":"呼啦圈","1":"蜡烛","2":"鹅","3":"耳朵","4":"帆船","5":"钩子","6":"勺子","7":"镰刀","8":"眼睛","9":"哨子","10":"棒球",
    "11":"梯子","12":"椅儿","13":"医生","14":"钥匙","15":"鹦鹉","16":"石榴","17":"仪器","18":"糖葫芦","19":"衣钩","20":"香烟",
    "21":"鳄鱼","22":"双胞胎","23":"和尚","24":"闹钟","25":"二胡","26":"河流","27":"耳机","28":"恶霸","29":"饿囚","30":"三轮车",
    "31":"鲨鱼","32":"扇儿","33":"星星","34":"三丝","35":"山虎","36":"山鹿","37":"山鸡","38":"妇女","39":"山丘","40":"司令",
    "41":"蜥蜴","42":"柿儿","43":"石山","44":"蛇","45":"师父","46":"饲料","47":"司机","48":"石板","49":"湿狗","50":"武林",
    "51":"工人","52":"鼓儿","53":"乌纱帽","54":"青年","55":"火车","56":"蜗牛","57":"武器","58":"尾巴","59":"蜈蚣","60":"榴莲",
    "61":"儿童","62":"牛儿","63":"流沙","64":"螺丝","65":"绿壶","66":"溜溜球","67":"绿漆","68":"喇叭","69":"太极","70":"麒麟",
    "71":"鸡翼","72":"企鹅","73":"花旗参","74":"骑士","75":"西服","76":"汽油","77":"机器","78":"青蛙","79":"气球","80":"巴黎",
    "81":"白蚁","82":"靶儿","83":"芭蕉扇","84":"巴士","85":"保姆","86":"八路","87":"白旗","88":"爸爸","89":"芭蕉","90":"酒瓶",
    "91":"球衣","92":"球儿","93":"旧伞","94":"首饰","95":"酒壶","96":"蝴蝶","97":"旧旗","98":"酒杯","99":"舅舅"}
sum = 0                                         #定义一个变量初始值为0,收集做对题目数量
false = 0                                       #错误次数
times = 20                                      #20次机会
while times > 0 :                               #建立一个判断，当列表不为空，而且输入次数不为0，执行以下代码。
    question = random.choice(list(A.keys()))    #用字典A的key建立一个list(列表)，通过random函数随机抽一个key，定义question变量
    answer = input('%s--' % (question))         #输入key键所对应的value值，例如给的是“00”这个键，那输入“望远镜”这个值。
    times = times - 1                           #次数减1
    if answer != 'x':                           #输入X提前退出练习
        if answer == A[question]:                    #判断输入的这个值是不是对应了A列表里的key键。
            #print("恭喜你，答对了")                    #如果对了，打印“恭喜你答对了”
            del A[question]                          #删除这个key,value。不再出现这个题，避免重复做题。
            sum = sum + 1                           #答对一题记一分
            continue                                #答对了题目，继续出题。
        else:                                       #如果输入的这个值不是key键所对应的。
            times = times + 5                       #错了再继续练习5次
            false = false + 1                       #错误次数累计加1
            print('不好意思，你答错了，正确答案是',question,'--',A[question])     #打印“不好意思，你错了”，并输入正确答案。不删题处理。
    else:
        print('已退出')                                #输入x就直接退出了
        break                                         #退出循环
print('----------测试结束----------\n正确：', sum, '\n错误：', false)   #正确错误情况
print('--------恭喜你挑战成功-------' if sum > 15 else '------挑战失败，再接再厉------')

qianshang666 · 发表于 2021-11-23 22:01

用列表一一对应不就行了

zhang1497 · 发表于 2021-11-23 22:03

qianshang666 发表于 2021-11-23 22:01
用列表一一对应不就行了

问题是如何解决00,01,02,03？

zhang1497 · 发表于 2021-11-23 22:29

想到一个思路，建立一个序号列表，用来电脑出题的，然后再建立一个序号和答案为字符串的列表。这样用电脑输出的元素+自己输入的答案组成新的字符串，然后判断这个新的字符串是不是第二个列表里的元素，是就正确，不是就错误。我去试试。

qizhi45 · 发表于 2021-11-23 23:41

Python有个数据类型是字典即key-value对应
思路：先按照上面那个图构建字典，将数字作为key，文字作为value;
循环出题时先判断字典是不是为空，为空直接输出回答正确;
在范围中random一个key值访问其value，判断key对应的value存不存在，不存在则结束当前循环进入下一轮即再random一个；
然后跟你输入的答案对应，然后再判断你的对错逻辑，对的时候删除字典中的这个键值对。错的答案也结束当前循环。

yamika · 发表于 2021-11-23 23:50

用列表存储一个key-value对，存储结构大致像这样[('0','呼啦圈‘),('1','蜡烛')。。。]，00.01这些按字符串处理，然后顺序遍历列表即可

zhang1497 · 发表于 2021-11-23 23:54

我是通过import random和字符串相加的原理来做的。

zhang1497 · 发表于 2021-11-23 23:55

yamika 发表于 2021-11-23 23:50
用列表存储一个key-value对，存储结构大致像这样[('0','呼啦圈‘),('1','蜡烛')。。。]，00.01这些按字符串 ...

我没有试过这类，能简单写个代码我学习下吗？

楼主你好萌 · 发表于 2021-11-24 08:54

这玩意，用字典很轻松就实现了

zhang1497 · 发表于 2021-11-24 09:57

楼主你好萌发表于 2021-11-24 08:54
这玩意，用字典很轻松就实现了

还没接触过字典，书才翻到几十页。哈哈。能编写给框架看看吗？

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