Question

我正在用Python为Burger shop开发一个应用程序网络，该应用程序将从数据库中获取一些数据，我需要管理这些数据

从数据库中，我将得到一个像这样的元组列表 check_table = [（1，'Classic'，'Cheddar'，'medium'），（2，'Big'，'Cheddar'，'rare'），（3，'Classic'，'Cheddar'，'rare'） ]

在这里，我试图根据烹饪水平来获取适当数量的牛排，例如稀有，中等，做得很好

我花了很多时间，但是我得不到正确计数的代码，我什至试图制作类似

的语句

如果check_table [x] [y] ==“经典”和check_table [x] [y] ==“稀有”：

牛排+ = 1 稀有+ = 1

代码如下：

check_table = [(1,'Classic', 'Cheddar', 'Medium'), (2,'Big', 'Cheddar', 'Rare'), (3,'Classic', 'Cheddar', 'Rare')]

# Variables
steaks = 0
rare = 0
medium = 0
w_done = 0

# Getting all the data from a nested for
for x in range(len(check_table)):
    for y in range(len(check_table[0])):
        # Getting the number of steaks
        if check_table[x][y] == "Classic":
            steaks += 1
            # Getting the style of the steak
            if check_table[x][y] == "Rare":
                rare += 1
            elif check_table[x][y] == 'Medium':
                medium += 1
            elif check_table[x][y] == 'Well-done':
                w_done += 1

        elif check_table[x][y] == "Big":
            steaks += 2
            # Getting steak style
            if check_table[x][y] == 'Rare':
                rare += 2
            elif check_table[x][y] == 'Medium':
                medium += 2
            elif check_table[x][y] == 'Well-done':
                w_done += 2

print("# Steaks ", steaks)
print("# Rare ", rare)
print("# Medium ", medium )
print("# Well-done ", w_done)

我希望得到这样的东西牛排4 稀有3 中1 做得好0

因为一个经典汉堡包含一个牛排，而一个大汉堡包含两个牛排，所以我得到了它，就像if语句不在那里一样

牛排4 稀有0 中0 做得好0

希望有人可以帮我解决这个问题，谢谢大家

Answer 1

check_table = [(1,'Classic', 'Cheddar', 'Medium'), (2,'Big', 'Cheddar', 'Rare'), (3,'Classic', 'Cheddar', 'Rare')]

# Variables
steaks = 0
rare = 0
medium = 0
w_done = 0

# Getting all the data from a nested for
for x in check_table:
    print(x)
      # Getting the number of steaks
    if x[1] == "Classic":
        steaks += 1
          # Getting the style of the steak
        if x[3] == "Rare":
            rare += 1
        elif x[3] == 'Medium':
            medium += 1
        elif x[3] == 'Well-done':
            w_done += 1

    elif x[1] == "Big":
        steaks += 2
          # Getting steak style
        if x[3] == 'Rare':
            rare += 2
        elif x[3] == 'Medium':
            medium += 2
        elif x[3] == 'Well-done':
            w_done += 2

print("# Steaks ", steaks)
print("# Rare ", rare)
print("# Medium ", medium )
print("# Well-done ", w_done)

Answer 2

您应该尝试将所有这些if/elif语句转换为某种数据结构的查找。这将使所有内容更易于推理，更易于编辑，并使您的代码更加简洁。

例如，与其检查if / elif多少牛排，不如查看经典牛排与大牛排的区别：

steak_count = {
    'Classic': 1,
    'Big': 2
}
count = steak_count['Big'] # 2 steaks for 'Big'

现在，如果您添加其他类型的汉堡，则无需再次编写整套if/elif语句。您可以通过以下方式将此想法传达给其他人：

check_table = [(1,'Classic', 'Cheddar', 'Medium'), (2,'Big', 'Cheddar', 'Rare'), (3,'Classic', 'Cheddar', 'Rare')]

steak_count = {
    'classic': 1,
    'big': 2
}
# Variables
counts = {
    'steaks': 0,
    'rare': 0,
    'medium': 0,
    'w_done': 0  
}

for order_num, steak, cheese, cooked in check_table:
    count = steak_count[steak.lower()]
    counts['steaks'] += count
    counts[cooked.lower()] += count

print(counts)
# {'steaks': 4, 'rare': 3, 'medium': 1, 'w_done': 0}

如果在python中嵌套则嵌套

2 个答案: