我有一個 PizzaPriceCalculator 函式，它接受 3 個引數：醬汁（字串）、澆頭（最初是字串）和忠誠客戶（布林值）。
我有 2 個字典：這樣的 sauceMenu 和 toppingsMenu：
sauceMenu = {
"regular": 4.35,
"alfredo": 4.6,
"marinara": 5.85,
"mix": 5.47
}
toppingsMenu = {
"cheese": 0.5,
"pepperoni": 0.75,
"salami": 0.65,
"olives": 1.0,
"mushroom": 0.35
}
函式是這樣呼叫的：
pizzaPriceCalculator("marinara", "pepperoni", True)
我只是遍歷兩個字典，如果有匹配和總和醬的回傳成本以及列印最終價格的澆頭成本。
這是完整的代碼：
def pizzaPriceCalculator (sauce, toppings, loyalCustomer):
sauceMenu = {
"regular": 4.35,
"alfredo": 4.6,
"marinara": 5.85,
"mix": 5.47
}
toppingsMenu = {
"cheese": 0.5,
"pepperoni": 0.75,
"salami": 0.65,
"olives": 1.0,
"mushroom": 0.35
}
for i in sauceMenu:
if sauce is i:
print(sauceMenu[i])
baseTeaAmount = sauceMenu[i]
for j in toppingsMenu:
if toppings is j:
print(toppingsMenu[j])
toppingsAmount = toppingsMenu[j]
priceBeforeDiscount = baseTeaAmount toppingsAmount
print(priceBeforeDiscount)
if loyalCustomer:
priceAfterDiscount = priceBeforeDiscount - 1
print(priceAfterDiscount)
# return priceAfterDiscount;
pizzaPriceCalculator("marinara", "pepperoni", True)
問題是我被告知要重構代碼并使第二個引數（澆頭）成為字串串列，以便您可以選擇許多澆頭。
pizzaPriceCalculator("marinara", ["pepperoni", "cheese", "olives"], True)
如何匹配陣列迭代器和物件迭代器，以便將醬汁和所有配料相加？
我嘗試遍歷澆頭串列（來自輸入）并檢查它是否與我們的 toppingsMenu 字典迭代器匹配，但它不起作用：
for j in toppingsMenu:
for k in toppings:
if toppings[k] is j:
print(toppingsMenu[j])
toppingsAmount = toppingsMenu[j]
另外，我如何將所有挑選的頂部價格相互添加，最后添加到醬汁價格以獲得最終成本？
任何建議表示贊賞。
uj5u.com熱心網友回復：
按照下面的代碼。假設，對于醬汁，輸入只是字串中的一項，而澆頭將是許多專案的串列......
def pizzaPriceCalculator (sauce, toppings, loyalCustomer):
sauceMenu = {"regular": 4.35, "alfredo": 4.6, "marinara": 5.85, "mix": 5.47}
toppingsMenu = {"cheese": 0.5, "pepperoni": 0.75, "salami": 0.65, "olives": 1.0, "mushroom": 0.35}
baseTeaAmount = sauceMenu[sauce]
toppingsAmount = 0
for item in toppings:
toppingsAmount = toppingsAmount toppingsMenu[item]
priceBeforeDiscount = baseTeaAmount toppingsAmount
if loyalCustomer:
priceAfterDiscount = priceBeforeDiscount - 1
print(priceAfterDiscount)
else:
print(priceBeforeDiscount)
pizzaPriceCalculator("marinara", ["pepperoni", "cheese", "olives"], True)
# 7.1 (output)
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