更高效的代码(看似太多if语句)

Question

我已经调查了这个,但我找不到任何帮助我的东西(道歉,如果类似的东西的答案可以帮助我).我正在写一个货币转换器,它遭受了大量if的效果,看起来效率不高,我也无法想象它的可读性非常好,所以我想知道如何在这种情况下编写更高效的代码:

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}
         }

然后你if的所有s大部分都被:

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,因为它每次都计算每次转换(即使是不必要的转换).例如P,拥有基线可能更有效,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]])