Question

我已经调查了这一点，但我找不到任何帮助我的事情（道歉，如果类似的东西的回答可以帮助我）。我正在编写一个货币转换器，它受到大量if的影响，但它看起来并不高效，我也无法想象它具有很好的可读性，所以我想知道在这种情况下我如何编写更有效的代码：

prompt = input("Input") #For currency, inputs should be written like "C(NUMBER)(CURRENCY TO CONVERT FROM)(CURRENCY TO CONVERT TO)" example "C1CPSP"

if prompt[0] == "C": #Looks at first letter and sees if it's "C". C = Currency Conversion
    #CP = Copper Piece, SP = Silver Piece, EP = Electrum Piece, GP = Gold Piece, PP = Platinum Piece
    ccint = int(''.join(list(filter(str.isdigit, prompt)))) # Converts Prompt to integer(Return string joined by str.(Filters out parameter(Gets digits (?), from prompt))))
    ccalpha = str(''.join(list(filter(str.isalpha, prompt)))) #Does the same thing as above expect with letters

    if ccalpha[1] == "C": #C as in start of CP
        acp = [ccint, ccint/10, ccint/50, ccint/100, ccint/1000] #Array of conversions. CP, SP, EP, GP, PP
        if ccalpha[3] == "C": #C as in start of CP
            print(acp[0]) #Prints out corresponding array conversion
        if ccalpha[3] == "S": #S as in start of SP, ETC. ETC.
            print(acp[1])
        if ccalpha[3] == "E":
            print(acp[2])
        if ccalpha[3] == "G":
            print(acp[3])
        if ccalpha[3] == "P":
            print(acp[4])
    if ccalpha[1] == "S":
        asp = [ccint*10, ccint, ccint/10, ccint/10, ccint/100]
        if ccalpha[3] == "C":
            print(asp[0])
        if ccalpha[3] == "S":
            print(asp[1])
        if ccalpha[3] == "E":
            print(asp[2])
        if ccalpha[3] == "G":
            print(asp[3])
        if ccalpha[3] == "P":
            print(asp[4])
    if ccalpha[1] == "E":
        aep = [ccint*50, ccint*5 ,ccint , ccint/2, ccint/20]
        if ccalpha[3] == "C":
            print(aep[0])
        if ccalpha[3] == "S":
            print(aep[1])
        if ccalpha[3] == "E":
            print(aep[2])
        if ccalpha[3] == "G":
            print(aep[3])
        if ccalpha[3] == "P":
            print(aep[4])
    if ccalpha[1] == "G":
        agp = [ccint*100, ccint*10, ccint*2, ccint, ccint/10]
        if ccalpha[3] == "C":
            print(agp[0])
        if ccalpha[3] == "S":
            print(agp[1])
        if ccalpha[3] == "E":
            print(agp[2])
        if ccalpha[3] == "G":
            print(agp[3])
        if ccalpha[3] == "P":
            print(agp[4])
    if ccalpha[1] == "P":
        app = [ccint*1000, ccint*100, ccint*20, ccint*10, ccint]
        if ccalpha[3] == "C":
            print(app[0])
        if ccalpha[3] == "S":
            print(app[1])
        if ccalpha[3] == "E":
            print(app[2])
        if ccalpha[3] == "G":
            print(app[3])
        if ccalpha[3] == "P":
            print(app[4])

Answer 1

您始终可以使用词典进行查找：

lookup = {'C': {'C': ccint, 'S': ccint/10, 'E': ccint/50, 'G': ccint/100, 'P': ccint/1000},
          'S': {'C': ccint*10, 'S': ccint, 'E': ccint/10, 'G': ccint/10, 'P': ccint/100},
          'E': {'C': ccint*50, 'S': ccint*5, 'E': ccint, 'G': ccint/2, 'P': ccint/20},
          'G': {'C': ccint*100, 'S': ccint*10, 'E': ccint*2, 'G': ccint, 'P': ccint/10},
          'P': {'C': ccint*1000, 'S': ccint*100, 'E': ccint*20, 'G': ccint*10, 'P': ccint}
         }

然后，所有if大部分都被以下内容覆盖：

print(lookup[ccalpha[1]][ccalpha[3]])

但是可能包含其他字符吗？然后你需要引入一个后备：

try:
    print(lookup[ccalpha[1]][ccalpha[3]])
except KeyError:
    # Failed to find an entry for the characters:
    print(ccalpha[1], ccalpha[3], "combination wasn't found")

如上所述，它不是最有效的方式，因为它每次都会计算每次转换（即使是不必要的转换）。拥有基线可能更有效，例如P并保存因子：

lookup = {'C': 1000,
          'S': 100,
          'E': 50,
          'G': 10,
          'P': 1,
          }

# I hope I have them the right way around... :-)
print(ccint * lookup[ccalpha[3]] / lookup[ccalpha[1]])

Answer 2

您应该分两步完成，而不是直接从源单元转换到目标单元：

factors = { 'CP': 1, 'SP': 10, and so on }

def convert_currency(amount, from_unit, to_unit):
    copper = amount * factors[from_unit]
    return copper / factors[to_unit]

这段代码就是您所需要的。你可以这样称呼它：

print(convert_currency(12345, 'SP', 'EP'))

Answer 3

另一种方法，使用矩阵（实际上只是一个列表列表）：

conversion_rate = [[1, 1 / 10, 1 / 50, 1 / 100, 1 / 1000],
                   [1 * 10, 1, 1 / 10, 1 / 10, 1 / 100],
                   [1 * 50, 1 * 5, 1, 1 / 2, 1 / 20],
                   [1 * 100, 1 * 10, 1 * 2, 1, 1 / 10],
                   [1 * 1000, 1 * 100, 1 * 20, 1 * 10, 1]]
currency_value =  {'C':0,'S': 1, 'E': 2, 'G': 3, 'P': 4}

from_ = currency_value[ccalpha[1]]
to = currency_value[ccalpha[3]]

print(ccint*conversion_rate[from_][to])

首先创建货币转换矩阵。然后你将货币与一个数字匹配（非常像其他语言中的enum，如C或Java）。为此你使用一个字典：它就像一个列表，除了你定义索引（它不是从0到长度 - 1）。然后，您将获得适当的转换率，将其与您的数字相乘并打印出来。

这与MSeifert的答案非常相似，只是你使用较少的字典，所以如果你对这些字典感到不舒服，可能会更容易理解。

更高效的代码（看似过多的if语句）

3 个答案: