简单Python战舰

Question

我最近开始学习python，并决定尝试制作我的第一个项目。我正在尝试做一个战舰游戏，随机放置两个3块长的船在一块板上。但效果不太好。我为2号飞船做了一个临时的循环，它应该检查一下旁边的两个空间是否空闲，然后在那里建立自己的空间。但有时它只是把自己放在船1已经在的位置上。有人能帮我吗？

下面是代码的第一部分：

from random import randint

###board:

board = []

for x in range(7):
    board.append(["O"] * 7)

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

###ships' positions:
#ship 1
def random_row(board):
    return randint(0, len(board) - 1)
def random_col(board):
    return randint(0, len(board[0]) - 1)
row_1 = random_row(board)
col_1 = random_col(board)

#ship 2
row_2 = random_row(board)
col_2 = random_col(board)
def make_it_different(r,c):
    while r == row_1 and c == col_1:
        r = random_row(board)
        c = random_col(board)
        row_2 = r
        col_2 = c
make_it_different(row_2,col_2)


### Makes the next two blocks of the ships:
def random_dir():
    n = randint(1,4)
    if n == 1:
        return "up"
    elif n == 2:
        return "right"
    elif n == 3:
        return "down"
    elif n == 4:
        return "left"
#ship one:
while True:
    d = random_dir() #reset direction
    if d == "up":
        if row_1 >= 2:
            #building...
            row_1_2 = row_1 - 1
            col_1_2 = col_1
            row_1_3 = row_1 - 2
            col_1_3 = col_1
            break
    if d == "right":
        if col_1 <= len(board[0])-3:
            #building...
            row_1_2 = row_1
            col_1_2 = col_1 + 1
            row_1_3 = row_1
            col_1_3 = col_1 + 2
            break
    if d == "down":
        if row_1 <= len(board)-3:
            #building...
            row_1_2 = row_1 + 1
            col_1_2 = col_1
            row_1_3 = row_1 + 2
            col_1_3 = col_1
            break
    if d == "left":
        if col_1 >= 2:
            #building...
            row_1_2 = row_1
            col_1_2 = col_1 - 1
            row_1_3 = row_1
            col_1_3 = col_1 - 2
            break
ship_1 = [(row_1,col_1),(row_1_2,col_1_2),(row_1_3,col_1_3)]

这里是ship 2部分：

#ship two:
while True:
    d = random_dir() #reset direction
    if d == "up":
        if row_2 >= 2:
            if (row_2 - 1,col_2) not in ship_1 and (row_2 - 2,col_2) not in ship_1:
                #building...
                row_2_2 = row_2 - 1
                col_2_2 = col_2
                row_2_3 = row_2 - 2
                col_2_3 = col_2
                break
    if d == "right":
        if col_2 <= len(board[0])-3:
             if (row_2 ,col_2 + 1) not in ship_1 and (row_2,col_2 + 2) not in ship_1:
                #building...
                row_2_2 = row_2
                col_2_2 = col_2 + 1
                row_2_3 = row_2
                col_2_3 = col_2 + 2
                break
    if d == "down":
        if row_2 <= len(board)-3:
            if (row_2 + 1 ,col_2) not in ship_1 and (row_2 + 2,col_2) not in ship_1:
                #building...
                row_2_2 = row_2 + 1
                col_2_2 = col_2
                row_2_3 = row_2 + 2
                col_2_3 = col_2
                break
    if d == "left":
        if col_2 >= 2:
            if (row_2 ,col_2 - 1) not in ship_1 and (row_2,col_2 - 2) not in ship_1:
                #building...
                row_2_2 = row_2
                col_2_2 = col_2 - 1
                row_2_3 = row_2
                col_2_3 = col_2 - 2
                break

###test
board[row_1][col_1] = "X"
board[row_1_2][col_1_2] = "X"
board[row_1_3][col_1_3] = "X"
board[row_2][col_2] = "Y"
board[row_2_2][col_2_2] = "Y"
board[row_2_3][col_2_3] = "Y"
#Ship1 = X's and Ship2 = Y's
print_board(board)

Answer 1

你为此写了很多代码，也许另一种方法更好。试着写一个函数

def ship_points(rowcol = (3,3), shiplength = 4, direction = (1,0), boardsize=(7,7)):
    points = []
    for i in xrange(shiplength):
        points.append((rowcol[0]+i*direction[0], rowcol[1]+i*direction[1]))
        if points[i][0] >= boardsize[0] or points[i][1] >= boardsize[1]:
            return None
    return points

现在，您只需为每艘船生成点并直接检查相同的点。代码更少，可重用性更强。

Answer 2

赋值给make_it_different中的row_2和col_2并不是按这些名称赋值给全局变量。Python确定函数局部变量的规则是，函数在没有声明的情况下赋值给任何对象都是局部变量；赋值给row_2和col_2将创建新的局部变量，而不是更改全局变量。您可以通过声明row_和col_2global来解决这个问题，但是最好将新值传递给调用者并让调用者分配它们。

（为什么make_it_different要取row_2和col_2的初始值？为什么不让它生成坐标，直到找到一些有用的东西？）

Answer 3

我建议允许您的放置代码简单地运行，而不使用if语句，这样会更干净。然后，在最后，您可以检查是否有任何部分重叠，以及它们是否重置。

取决于您最终决定如何存储各个船只所在的点，可能是一个元组列表。你可以这么做

place ship方法可以返回元组（点）的列表

def placeShip():
    points = []

    # put random point generation here

    for point in points:
        if point in otherShipPoints:
            return placeShip()         # overlap detected, redo ship placement

    return points

将放置代码放在一个函数中，这样就可以简单地调用它。您的代码开始变得混乱，我建议您使用这样的方法来避免遇到意面代码问题。

你也可以给placeShip()一个你想要添加的船只大小的参数，然后这个方法可以是你的一体式船只放置器。只要让函数看起来像这样，然后在网格中随机生成这么多点