The work involves the retrieval of this function, the following is what I have made to try and resolve this assignment:
Figure, file Exercises.txt file is generated at the title, but Answers.txt was generated at the answer. Grade.txt generated at retrieval after final answer:
Wherein the code generating portion is subject to the following:
Its code is:
#小学四则运算题目问题
#需要在生成题目的同时生成答案
import random
doc1 = open('Exercises.txt','w') #打开Exercises.txt文件编写
doc2 = open('Answers.txt','w') #同上
#随机生成四种运算（加减乘除）并生成对应答案
op = ['+','-','x','÷'] #题目生成语言列表
num1 = num2 = 0
for i in range(100): #题目循环
num = i + 1
num1 =random.randint(0,99)
num2 =random.randint(0,99)
opnum = random.randint(0,3) #随机生成0到3的整数以生成不同运算
if opnum == 3 and num2 == 0: #被除数不能等于0
opnum = random.randint(0,2)
doc1.write('第{:0>3}题 {:0>2}{}{:0>2} = \n'.format(num,num1,op[opnum],num2))
#下面四步是分组判断写入答案
if opnum == 0: #加法
doc2.write('第{:0>3}题 {} \n'.format(num,num1+num2))
elif opnum == 1: #减法
doc2.write('第{:0>3}题 {} \n'.format(num,num1-num2))
elif opnum == 2: #乘法
doc2.write('第{:0>3}题 {} \n'.format(num,num1*num2))
else : #除法
if isinstance(num1/num2,int): #判断答案是否为分数
doc2.write('第{:0>3}题 {} \n'.format(num,num1/num2))
else:
if num1 > num2:
doc2.write('第{:0>3}题 {}又{}/{} \n'.format(num,num1//num2,num1%num2,num2))
else:
doc2.write('第{:0>3}题 {}/{} \n'.format(num,num1,num2))
doc1.close #关闭文件并保存Can work to change the code, the corresponding output txt document will show as follows:
Exercises.txt run once after the file:
Answers.txt file after the first run:
These are the topic of the code.
The search part, I encountered difficulties, made the first attempt the following:
Its code is:
data =[]
for line in open("Exercises.txt","r"): #设置文件对象并读取每一行文件
data.append(line) #将每一行文件加入到list中
num1 =0 #大循环数
num2 =1 #小循环数
ex1 =data[num1] #expression表达式
ex2 =data[num2]
tp1 =ex1[1:4]#topic题目数
tp2 =ex2[1:4]
num11 =ex1[7:9] #第一操作数
num12 =ex1[10:12] #第二操作数
num21 =ex2[7:9]
num22 =ex2[7:9]
op1 =ex1[9] #操作符
op2 =ex2[9]
list_repeat =[] #重复题目
exnum =len(data) #读取题目数
while num1 <=len(data):
while num2 <=len(data)-num1-1:
if op1 ==op2 and op1 in ['+','x']:
if num11 ==num21 or num11 ==num22:
list_repeat =list_repeat +tp1 +tp2
num2 =num2 +1
elif op1 ==op2 and op1 in ['-','÷']:
if num11 ==num21 and num12 ==num22:
list_repeat =list_repeat +tp1 +tp2
num2 =num2 +1
else:
num2 =num2 +1
num1 =num1 +1
num2 =num1 +1
print(list_repeat)Visible code computation is extremely large, need to calculate the minimum number if the topic is 10.0001 trillion times , seen in this way is extremely unscientific (when testing my computer have been crying), which I continued for the second attempts.
Then retrieve different from the first, the second time I first narrowed the range, the first for the first time in narrowing the answer, find the same answer when the subject, you can reduce a lot of time in this case the subject computer workload to reduce computation time.
second attempt code is as follows:
#读取txt文件2
data_answer =[] #Answers.txt文件语句
data_Exercises =[] #Exercises.txt文件语句
for line in open("Answers.txt","r"): #设置文件对象并读取每一行文件
data_answer.append(line) #将每一行文件加入到list中
for line in open("Exercises.txt","r"): #同上
data_Exercises.append(line)
#生成一个只存有答案的列表
loopnum1 =0 #循环数
answers ="" #答案字符串
while loopnum1 <= len(data_answer) -1:
answer =data_answer[loopnum1] #答案语句
awnum =answer[7:-2] #答案数据
answers = answers +awnum +" "
loopnum1 =loopnum1 +1
awlist =answers.split(sep=" ") #答案列表
del awlist[-1] #删除最后一个空字符串
#索引相同答案的列表
loopnum2 =0 #循环数
repeat =""
while loopnum2 <= len(data_answer) -1:
item =awlist[loopnum2] #每个元素
itnum =str(awlist.count(item)) #每个元素重复个数
repeat =repeat +itnum +" "
loopnum2 =loopnum2 +1
rep =repeat.split(sep=" ") #所有元素出现次数的字符串
del rep[-1] #删除最后一个空字符串
#生成重复题目的题号
repnum =rep.count('2') #重复题目总数
loopnum3 =0
aimnum =""
repaimnum = 0
while loopnum3 <=repnum -1:
repaimnum =rep.index('2') +1 +loopnum3 #题号
aimnum =aimnum +str(repaimnum) +" "
rep.remove('2')
loopnum3 =loopnum3 +1
print("一共有{}道题目错误，他们的题号是{}".format(repnum,aimnum))
# 文本进度条,纯属好玩才加的
import time
scale = 50
print("检查重复题目开始".center(scale//2,"_"))
start = time.perf_counter()
for i in range(scale + 1):
a = "■" * i
b = "□" * (scale - i)
c = (i / scale) * 100
dur = time.perf_counter() - start
print("\r{:3^.0f}%[{}{}]{:.1f}秒".format(c,a,b,dur),end="")
# └百分数 └进度条 └秒数
time.sleep(0.1)
print("\n"+"检查结束".center((scale +8)//2,"_"))
Remove the progress bar is purely for fun, a total of 42 lines of code, and is not yet completed, the following is the result of the current runtime:
