Python 3 - 元组超出范围错误。但我用的是字典？

Question

我的战舰程序中有一个元组超出了范围。我感到困惑的原因是我没有使用元组，除非我错了并且我愿意被纠正。元组是不可变的，并且它所指的字典在启动程序时已添加了项目。 我是初学者，所以如果我犯了一个愚蠢的错误，请不要评判!

from random import randint
#empty list to generate the board.
board = []
messages = {
  "win" : "Nooo you won!",
  "lose" : "Not my ship haha",
  "out" : "Oops, that's not even in the ocean.",
  "repeat" : "You guessed that one already"

}

ships = {
  'shiprows' : [0]
  'shipcols' : [0]
}

#generate board and append to board[] As of now it is a 10*10 grid.
for x in range(0, 10):
  board.append(["O"] * 10)

#prints the board every turn.
def print_board(board):
  for row in board:
    print(" ".join(row))

print_board(board)

#computer chooses where to put battleships' rows
def random_row1(board):
  return randint(0, len(board) - 1)

def random_col1(board):
 return randint(0, len(board) - 1)
#calling above two functions and storing their values for 5 ships.
#creating variables for 5 ships.
vars = 0
for vars in range(0, 5):
  print(vars)
  if len(ships.keys()) >= 4:
    while ships["shiprow{}".format(vars - 2)] == ships["shiprow{}".format(vars - 1)] and ships["shipcol{}".format(vars - 2)] == ships["shipcol{}".format(vars - 1)]:
      ships["shiprow{}".format(vars)] = random_row1(board)
      ships["shipcol{}".format(vars)] = random_col1(board)
    ships["shiprow{}".format(vars)] = random_row1(board)
    ships["shipcol{}".format(vars)] = random_col1(board)
  else:
    ships["shiprow{}".format(vars)] = random_row1(board)
    ships["shipcol{}".format(vars)] = random_col1(board)


#program itself
turn = 0
#enforces four turns before game over. Will possibly extend to unlimited with multiple ships.
print(ships)
for turn in range(20):
  turn = turn + 1
  print ("Turn {}".format(turn))
  print ("Ships Left: {}".format(int(len(ships.keys()) / 2))) 
  guess_row = int(input("Guess Row: "))
  guess_col = int(input("Guess Col: "))

#checking stuff.
  i = 0
  if guess_row == ships["shiprow{}".format(i = range(0, 10))] and guess_col == ships["shipcol{}".format(i)]:
    print (messages["win"])
    board[guess_col][guess_row] = u"#"
    print_board(board)

  elif board[guess_col][guess_row] == "X":
    print ("You guessed that one already.")
  elif guess_row not in range(len(board)) and guess_col not in range(len(board[0])):
    print(messages["out"])
  else:
    print(messages["lose"])
    board[guess_col][guess_row] = "X"
    print_board(board)
  if turn >= 20:
    print ("Game Over")
    board[ships["ship_col{}".format(range(0, 10))]][ships["ship_row{}".format(range(0, 10))]] = u"#"
    print_board(board)
    break

可疑行似乎是第 62 行 - 这看起来很粗略，但我实际上不知道该怎么做。请告知该怎么做: 顺便说一句，这是错误:

Traceback (most recent call last):
  File "battleship3.py", line 62, in <module>
    if guess_row == ships["shiprow{}".format(i = range(0, 10))] and guess_col == ships["shipcol{}".format(i)]:
IndexError: tuple index out of range

谢谢。

Answer 1

只要您使用具有位置格式规范的格式字符串(例如 {} 或 {1})，但仅传递关键字参数，您就会收到此错误消息。

类似地，当您使用仅包含关键字格式规范的格式字符串(如 {v})但仅传递位置参数时，您会收到 KeyError:

>>> '{}'.format(i=1)
IndexError: tuple index out of range
>>> '{i}'.format(1)
KeyError: 'i'

修复只是为了使您的规范符合您的论点。无论您喜欢哪种方式都可以，只需保持一致即可:

>>> '{i}'.format(i=1)
1
>>> '{}'.format(1)
1

<小时/>

话虽如此，我不确定这是什么意思:

"shiprow{}".format(i = range(0, 10))

您可以通过任何一种方式修复它，但这真的是您想要的字符串吗？

>>> "shiprow{i}".format(i = range(0, 10))
'shiprowrange(0, 10)'
>>> "shiprow{}".format(range(0, 10))
'shiprowrange(0, 10)'

<小时/>

如果您好奇为什么会出现此错误，稍微简化一下格式，它的工作原理如下:

def format(self, *args, **kwargs):
    result = ''
    index = 0
    bits = self.parse_format_stuff()
    for bit in bits:
        if bit is a regular string:
            result += bit
        elif bit is empty braces:
            result = args[index]
            index += 1
        elif bit is a number in braces:
            result += args[number]
        elif bit is a valid identifier string in braces:
            result += kwargs[identifier]
        else:
            raise a ValueError
    return result

因此，当它看到 {} 格式规范时，它会查找 args[0]。由于您没有传递任何位置参数，因此 args 是空元组 ()，因此 args[0] 是一个 IndexError .

可以说，如果 format 处理这些错误并将其转化为更好的东西，可能会更好，但有时能够以编程方式处理 KeyError 会很有用。 (IndexError 并不常见，但显然两者必须以相同的方式工作。)