簡單的Python戰艦遊戲

Question

我最近開始學習python，並決定嘗試做我的第一個項目。我正在嘗試製作一個戰艦遊戲，隨機在棋盤上放置兩塊3塊長的船。但它並不正確。我爲船2做了一段時間循環，應該檢查它旁邊的兩個空格是否空閒，然後在那裏建立它自己。但有時它只是在＃1號船的頂端放上了自己的力量。有人可以幫我嗎？

下面的代碼的第一部分：

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)] 

這裏的地方船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

我建議您允許您的代碼在沒有if語句的情況下運行，它會更乾淨。然後，最後，您可以檢查是否有任何部分重疊，以及是否重置。

根據你最終決定存儲單個船隻的位置，可能是一個元組列表。你可以這樣做

的地方船方法可以返回的元組（點）名單

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 

把你放置代碼的單一功能，使得它可以被簡單地稱爲這種方式。你的代碼開始變得混亂，我建議採用這樣的方法來避免遇到意大利式麪條代碼問題。

你也可以給placeShip()一個你想添加的船的尺寸的參數，然後這個方法可以是你的一體船配置器。只要讓你的功能看起來像這樣placeShip(size)，然後在你的網格內隨機生成許多點

Answer 2

分配到row_2和col_2在make_it_different不分配通過這些名字到全局變量。 Python用於確定函數局部變量的規則是函數在沒有聲明global的情況下分配的任何東西都是本地的;分配給row_2和col_2將創建新的局部變量，而不是更改全局變量。你可以通過聲明row_和col_2global來解決這個問題，但它可能會更好地將新值傳遞給調用者並讓調用者分配它們。

（爲什麼make_it_different採取row_2和col_2在所有的初始值？爲什麼不只是有它產生的座標，如果發現直到一些工作？）

Answer 3

您在編寫大量的代碼對於這一點，也許另一種方法會更好。嘗試編寫一個功能，如

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 

現在您可以爲每艘船隻生成點並直接檢查相同的點。代碼少得多，方式更可重用。