Question

所以，我只是从Python开始，但是我试图在字典中实现函数（有点像C中的函数指针），但是我一直在坚持如何从函数中接收返回的值而没有一个错误。这是我的代码片段：

def main():
    numberOfDice = 5
    dice = rollDice(numberOfDice) #rolls dice thrice, returns a list of 5 integers
    scoreBoard(dice)   #displays Yahtzee scoreboard
    value = choice(dice)  #value = the category choice[1-13]
    whichCategory(value) 

def whichCategory(category):
    board = {
        6: numberOfSixes, # I have 1-13 filled, but just for example
        12: yahtzee       # this is just to save space.
        }
    board[category]()

def numberOfSixes(theDice):
    count = theDice.count(6)
    points = count * 6

    return points

def yahtzee(theDice):
    yahtzee = 0
    ones, twos, threes, fours, fives, sixes = countDice(theDice)
    # countDice() determines the amount that each number is present in the 
    # list and then returns those six variables.
    if any(x==5 for x in(ones, twos, threes, fours, fives, sixes)):
           yahtzee = 50

    return yahtzee

滚动3次后，我确定骰子（整数列表），然后从与whichCategory（）一起选择的值中选择我的值，然后执行正确的计算，从这些函数中吐出正确的返回值，但是然后我会得到这样的错误（用于输入两个六点（12点））：

Traceback (most recent call last):
File "C:\python\Lib\idlelib\Yahtzee.py", line 407, in <module>
main()   # My call to main()
File "C:\python\Lib\idlelib\Yahtzee.py", line 18, in main
whichCategory(value)
File "C:\python\Lib\idlelib\Yahtzee.py", line 90, in whichCategory
board[category]()
TypeError: yahtzee() missing 1 required positional argument: 'theDice

这很有趣，因为它说错误是yahtzee（）引起的，而不是6位，我相信这是因为它以某种方式得到12作为答案，并且在字典中调用了12th关键字。否则我会收到如下错误：

Traceback (most recent call last):
File "C:\python\Lib\idlelib\Yahtzee.py", line 407, in <module>
main()   # My call to main()
File "C:\python\Lib\idlelib\Yahtzee.py", line 18, in main
whichCategory(value)
File "C:\python\Lib\idlelib\Yahtzee.py", line 90, in whichCategory
board[category]()
KeyError: 50

请，我一直在寻找答案，但尚未找到答案。如果可能的话，我想将字典中函数的值返回给main（）。任何帮助将不胜感激。

Answer 1

这未经测试，但是我认为问题在于在行board[category]()上，您调用了一个函数，但没有调用它需要的值。尝试进行这些更改

def main():
    numberOfDice = 5
    dice = rollDice(numberOfDice) #rolls dice thrice, returns a list of 5 integers
    scoreBoard(dice)   #displays Yahtzee scoreboard
    value = choice(dice)  #value = the category choice[1-13]
    whichCategory(value, dice) # pass dice

def whichCategory(category, dice):  # receive dice
    board = {
        6: numberOfSixes, # I have 1-13 filled, but just for example
        12: yahtzee       # this is just to save space.
        }
    board[category](dice)   # call with dice

获取字典中函数值的一些问题

1 个答案: