Question

我被赋予此算法以转换为伪代码：

Name[1] = 'Ben'
Name[2] = 'Thor'
Name[3] = 'Zoe'
Name[4] = 'Kate'
Max <- 4
Current <- 1
Found <- False
OUTPUT 'What player are you looking for?'
INPUT PlayerName
WHILE (Found = False) AND (Current <= Max)
    IF Names[Current] = PlayerName
        THEN Found <- True
        ELSE Current <- Current + 1
    ENDIF
ENDWHILE
IF Found = True
    THEN OUTPUT 'Yes, they have a top score'
    ELSE OUTPUT 'No, they do not have a top score'
ENDIF

所以我把它转换成Python就像这样：

def Names():
    list = ['Ben', 'Thor', 'Zoe', 'Kate']
    Max = 4
    Current = 1
    Found = False
    PlayerName = str(input("What Player are you looking for? "))
    while (Found == False) and (Current <= Max):
        if Names(Current) == PlayerName:
            Found = True
        else:
            Current = Current + 1
        if Found == True:
            print("Yes, they have a top score.")
        else:
            print("No, they do not have a top score.")

但是出现了一个错误：

'TypeError: Names() takes 0 positional arguments but 1 was given'

我哪里出错了？

Answer 1

让我们分析你得到的错误：

'TypeError: Names() takes 0 positional arguments but 1 was given'

它说，你所做的函数，名为Names()，取零参数，但你给它一个参数。

在您的代码中，当您在函数Names(Current)中使用Names()时，可以看到这一点。实际上，这是再次调用函数，因此您无意中编写了一个您不想要的递归函数。

您要做的是检查名称列表，以查看是否有任何名称与用户输入的名称相匹配。

要执行此操作，请使用for循环而不是while循环。这样，您就不必专门说明列表的起点和终点，如果您愿意，可以添加其他名称。

这就是转化的样子：

def Names():
    names_list= ['Ben', 'Thor', 'Zoe', 'Kate']
    Found = False
    PlayerName = str(raw_input("What Player are you looking for? "))
    for name in names_list:
        if name == PlayerName:
            Found = True

    if Found == True:   #place this if-else outside your for-loop
        print("Yes, they have a top score.")
    else:
        print("No, they do not have a top score.")


Names()

这将输出：

>>>
What Player are you looking for? Thor
Yes, they have a top score.

>>>
What Player are you looking for? Bob
No, they do not have a top score.

Answer 2

您收到type错误，因为您试图将参数传递给Names函数。 Names不接受任何参数，因为那里没有列出任何参数。您需要在不使用参数的情况下调用函数，或在声明Names函数时向括号中添加参数。此外，除非返回函数，否则无法从外部访问函数本地变量。所以不要这样做：

Names["Something"] == Another_Thing

但是，相反传递你需要你的函数吐出作为参数。如果您需要全局访问其中的所有内容，请不要在函数内部定义所有内容，而是在其外部定义，并将您尝试执行的所有内容作为参数传递给函数。

def Names(A_Parameter, Another_Parameter, A_Default_Parameter = Some_Default):
    # do something

Answer 3

@logic有一个解决方案，但它可以简化。无需str，for循环或Found变量。

def Names():
    names_list = ['Ben', 'Thor', 'Zoe', 'Kate']
    PlayerName = raw_input("What Player are you looking for? ")
    if PlayerName in names_list:
        print "Yes, they have a top score."
    else:
        print "No, they do not have a top score."

Names()

Python 2.x中的

print也不应该有括号。在上述情况下，这并不重要，但会有以下几点：

PlayerName = "Ben"
print("Name:",PlayerName)
print "Name:",PlayerName

输出：

('Name:', 'Ben')
Name: Ben

将伪代码转换为Python时出错

3 个答案: