在Python中解包的值太多

Question

我正试图制作一个战舰游戏，但我被困在船的一代。 当我试着运行这个程序时：

初始化

print "Welcome to Battleships!!!"
board = []
import os
from random import randint

for x in range(10):
    board.append(["O"] * 10)

def print_board(board):
    for row in board:
        print " | ".join(row)

ships = {"battleship" : [5], "cruiser1" : [4], "cruiser2" : [4], "frigate1" : [3], "frigate2" : [3], "frigate3" : [3], "frigate4" : [3], "minesweeper1" : [2], "minesweeper2" : [2], "minesweeper3" : [2], "minesweeper4" : [2]}
numbers1 = []
numbers2 = []
numbers = []

一些设置功能

^{pr2}$

主程序

setup_nums1()
setup_nums2()
setup_ships()
print numbers1
print numbers2
print ships

我得到这个错误：

Welcome to Battleships!!!
Traceback (most recent call last):
  File "/home/matthew/Documents/battleship2.py", line 77, in <module>
    setup_ships()
  File "/home/matthew/Documents/battleship2.py", line 41, in setup_ships
    for ship, info in ships:
ValueError: too many values to unpack

Answer 1

而不是for ship, info in ships尝试for ship, info in ships.iteritems()。在

使用in ships语法，您只能返回一个值（它们的键）。在

例如，您可以：

ships = { 'a': 'ship_a', 'b': 'ship_b', 'c': 'ship_c' }
for key in ships:
    print(key) # Outputs a b c

相反，如果你这样做：

^{pr2}$

有关python2.x和python3.x兼容性的更多信息，请查看this answer。在