在python中重复一些东西

Question

我目前正面临着试图让自己重复一遍的问题。 下面的代码用于游戏，用户必须在网格中定位宝箱（如战舰）。 代码在网格中放置1个随机宝箱，用户必须找到它。 （这些都来自变量名称treasure_row = random_row(board).......treasure_col = random_col(board) ）但是我想要这样，不仅仅有1个宝箱被放置在网格中（例如10或20）

我该怎么办？

from random import randint

board = []

for x in range(8):
    board.append(["•"] * 8)

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

print ("You have chosen the option to play the game")
print_board(board)

def random_row(board):
    return randint(1, len(board) - 1)

def random_col(board):
    return randint(1, len(board[0]) - 1)

treasure_row = random_row(board)
treasure_col = random_col(board)

turn = 0
for turn in range(64):
    turn = turn + 1
    guess_row = int(input("Guess Row:"))
    guess_col = int(input("Guess Col:"))

    if guess_row == treasure_row and guess_col == treasure_col:
        print ("Congratulations! you have found a treasure chest!")
        break
    else:
        if (guess_row>8) or (guess_col>8):
                print ("Sorry, that isn't on the grid.")
        elif(board[guess_row][guess_col] == "X"):
                print ("You guessed that one already.")
        else:
            if turn == 64:
                print ("Game Over")
            else:
                print ("Sorry there is no treasure here!")
                board[guess_row][guess_col] = "X"
                print  ("Turn", turn + 1)
    # Print (turn + 1) here!
        print_board(board)

Answer 1

最简单的方法是通过元组列表存储宝箱：

treasure_chests = [(1, 2), (4, 5), (1, 4)]

然后使用x in y的x in y比较猜测：

if (guess_row, guess_col) in treasure_chests:

如果你想生成宝藏，我会使用这样的函数：

def generate_treasure(num, max_x, max_y):
    treasure_chests = []
    while len(treasure_chests) < num:
        new_chest = (random.randint(1, max_x), random.randint(1, max_y))
        if new_chest not in treasure_chests:
            treasure_chests.append(new_chest)
    return treasure_chests

哪个输出：

>>> generate_treasure(5, 10, 10)
[(10, 5), (2, 3), (2, 9), (9, 4), (2, 2)]

Answer 2

在你的情况下，只需创建两个不同的treasure_row和treasure_col 。 像这样：

treasure_row_1 = random_row(board)
treasure_col_1 = random_col(board)

treasure_row_2 = random_row(board)
treasure_col_2 = random_col(board)

turn = 0
for turn in range(64):
    turn = turn + 1
    guess_row = int(input("Guess Row:"))
    guess_col = int(input("Guess Col:"))

    if guess_row == treasure_row_1 and guess_col == treasure_col_1 or guess_row == treasure_row_2 and guess_col == treasure_col_2:

或者通过元组数组来完成。 如你所愿

Answer 3

你总是可以有一个list的宝藏地点treasure_locations您核对：

#...
def random_col(board):
    return randint(1, len(board[0]) - 1)

treasure_locations = []
   while len(treasure_locations < 10): #Creates 10 locations
      treasure = random_row(board), random_col(board)
      if treasure in treasure_locations: #duplicate, try again
         continue
      treasure_locations.append(treasure)

turn = 0
for turn in range(64):
    if( not treasure_locations ):
       print "You found all the treasure!"
       break
    turn = turn + 1
    guess_row = int(input("Guess Row:"))
    guess_col = int(input("Guess Col:"))
    guess_loc = guess_row, guess_col
    if guess_loc in treasure_locations:
        print ("Congratulations! you have found a treasure chest!")
        treasure_locations.remove(guess_loc ) #remove once found
        break
    else:
        #...

Answer 4

使用列表推导来快速和在线表达。

• 创建董事会
• 使用1行和listcomp创建可能的coords列表，使用tuple列表而不是2个单独的列表
• 使用random.sample确保选择正确数量的项目（重复random.randint调用，你可以点击相同的位置：将有比预期更少的宝藏）
• 打印宝藏清单（这样你就可以测试）
• 在宝藏元组列表中使用in来立即检查x＆y

简单的概念证明创造最多5件宝物：

import random

board = [["*"]*8 for _ in range(8)]

possible_coords = [(x+1,y+1) for x in range(len(board[0])) for y in range(len(board))]

treasure_coords = random.sample(possible_coords,5)

print(treasure_coords)

x=int(input("x coord"))
y=int(input("y coord"))

if (x,y) in treasure_coords:
    print("you have found the treasure")

宝藏地点的例子：

[(2, 3), (8, 8), (4, 1), (8, 7), (6, 4)]
[(1, 1), (5, 8), (3, 7), (4, 4), (8, 7)]

对您的代码的评论：

• randint(1, len(board[0]) - 1)缺少最后一个元素。
• 你也turn在for循环增加turn内增加turn 。