pythonversion3的ammend配方程序解决方案

Question

我想知道是否有人能解决这个问题。 我不断地出现错误，我的程序无法相应地为OCR GCSE计算的配方任务工作。我已经设法在一个外部文本文件中创建添加配方，并搜索现有配方。在

我遇到的问题是重新计算新的数量（所有的数字数据）为不同的人数的食谱。我使用的是pythonversion3。在

我的修正程序部分如下。（我默认了一个人的所有配方），这样更容易重新计算新的数量。在

我的代码如下。**

我提前感谢你的帮助。在

import os

def modify():
    #create a boolean variable to use as a flag
    found = False

    #Get the search value and the new recipe information.
    search = input ("Enter a recipe name to search for: ")
    new_number_of_people =input("Enter the new number of people to serve (default is 1):")


    #open the original recipeList.txt file.
    recipeList_file=open("recipeList.txt", "r")

    #open the temporary file.
    temp_file = open("temp.txt", "w")

    #Read the first record's recipe name field
    recipe = recipeList_file.readline()


    #Read the rest of the file.
    while recipe != "":
         #Read the recipe item,quantity, units and number of people.
        ingredient1 = str(recipeList_file.readline())
        quantity1 =float(recipeList_file.readline())
        units1 = str (recipeList_file.readline())
        ingredient2 = str(recipeList_file.readline())
        quantity2 =float(recipeList_file.readline())
        units2 = str (recipeList_file.readline())        
        ingredient3 = str(recipeList_file.readline())
        quantity3 =float(recipeList_file.readline())
        units3 = str (recipeList_file.readline())        
        number_of_people = float(recipeList_file.readline())

        recipe = recipe.rstrip("\n")


         #write a new record with the temp file
        if recipe == search:
             #write the modified record to the temp file.
             temp_file.write(recipe + "\n")
             temp_file.write(ingredient1+ "\n")
             temp_file.write((quantity1*input(new_number_of_people)) + "\n")
             temp_file.write(units1 + "\n")
             temp_file.write(ingredient2+ "\n")
             temp_file.write((quantity2*input(new_number_of_people)) + "\n")
             temp_file.write(units2 + "\n")
             temp_file.write(ingredient3+ "\n")
             temp_file.write((quantity3*input(new_number_of_people)) + "\n")
             temp_file.write(units3 + "\n")
             temp_file.write((new_number_of_people) + "\n")

             #Set the found flag to True.
             found = True
        else:
                 #write the original record to the temp file.
                 #write the modified record to the temp file.
             temp_file.write(recipe + "\n")
             temp_file.write(ingredient1+ "\n")
             temp_file.write((quantity1*input(new_number_of_people)) + "\n")
             temp_file.write(units1 + "\n")
             temp_file.write(ingredient2+ "\n")
             temp_file.write((quantity2*input(new_number_of_people)) + "\n")
             temp_file.write(units2 + "\n")
             temp_file.write(ingredient3+ "\n")
             temp_file.write((quantity3*input(new_number_of_people)) + "\n")
             temp_file.write(units3 + "\n")
             temp_file.write((new_number_of_people) + "\n")


         #Read the next recipe
        ingredient1 = str(recipeList_file.readline())
        quantity1 = float(recipeList_file.readline())
        units1 = str (recipeList_file.readline())
        ingredient2 = str(recipeList_file.readline())
        quantity2 = float(recipeList_file.readline())
        units2 = str (recipeList_file.readline())
        ingredient3 = str(recipeList_file.readline())
        quantity3 = float(recipeList_file.readline())
        units3 = str (recipeList_file.readline())
        number_of_people = float(recipeList_file.readline())

        #Close the Recipe file and the temporary file.
    recipeList_file.close()
    temp_file.close()

        #Delete the original recipeList.txt file.
    os.remove ("recipeList.txt")

        #Rename the temporary file.
    os.rename("temp.txt", "recipeList.txt")

        #if the search was not found in the file display message
    if found:
        print ("The file has been updated.")

    else:
        print ("That recipe was not found in the file")
        #call the main function.
modify()

在收件人列表.txt格式：

^{pr2}$

Answer 1

输入文件的格式不是最好的。我将更好的每一个配方在一行和每一个数据用分隔符（逗号），一个csv格式。在

但由于输入是以这种格式给出的。。。在

首先，是一个好习惯，为每一项任务做一个功能。因此，我认为有三个任务：

1. 来自用户的输入：问题（）
2. 读取数据并进行处理：modify（）
3. 将处理后的数据保存到文件中：Save（）

question()函数。说明：

向用户请求输入数据，我以元组形式返回。在

modify()函数。说明：

打开文件输入后，我读取（）所有文件，并使用“\n”分隔符拆分每一行。这将生成一个包含文件每行内容的列表。在

由于配方的名称在文件中是唯一的，并且我想在调用函数时访问已确定的配方，我认为最好使用字典，其中 配方是键，其他数据（我把它放在一个列表中）是值。对， dict的值可以是Python中的任何对象，因此我使用列表。我定义了一个空白dict来添加配方。在

一个配方有12个数据（包括末尾的空行），因此使用zip(*[iter(recipeList)]*12)我创建了一个由12个元素组成的元组，可以用for语句进行迭代。在

元组的0索引是我用作键的配方的名称。元组的其余部分 （使用从第二个元素到末尾的片段[1:]）在列表中转换的是字典对的值。为什么要单子？因为元组在Python中是不可变的。在

然后我迭代列表（从第二个元素到第三个元素；数量）按食客的数量用乘积更新值。最后，我用人数更新了第十项。在

此函数返回更新字典。在

save()函数。 试着自己编写代码。您只需迭代字典（嵌套一个列表）并在一行中将每个数据写入配方的新文件。（别忘了写下每个字典对的键，就是菜谱的名字）。在

import os

def modify(recipe_name, number):
    recipeList_file = open('recipeList.txt', 'r', newline= None)
    recipeList = recipeList_file.read().split('\n')
    recipeDict = {}
    for i in zip(*[iter(recipeList)]*12):
        print(i)    # For debugging purpose
        recipeDict[i[0]] = list(i[1:])
        print(recipeDict.items())    # For debugging purpose
    for j in range(1, 10, 3):
        recipeDict[recipe_name][j] = str(int(recipeDict[recipe_name][j]) * number)
    recipeDict[recipe_name][9] = str(number)
    print(recipeDict.items())    # For debugging purpose
    recipeList_file.close()
    return recipeDict

def question():
    search = input("Enter a recipe name to search for: ")
    new_number_of_people = int(input("Enter the new number of people to serve (default is 1):"))
    return search, new_number_of_people


def save(dictionarySave):
    pass
    # Think yourself.

recipe_to_modify, number_to_modify = question()
new_recipe_dictionary = modify(recipe_to_modify, number_to_modify)
save(new_recipe_dictionary)