网上搜到一个Pygame写的俄罗斯方块（tetris），大部分看懂的前提下增加了注释，Fedora19下运行OK的
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#coding:utf8#! /usr/bin/env python# 注释说明：shape表示一个俄罗斯方块形状 cell表示一个小方块import sysfrom random import choiceimport pygamefrom pygame.locals import *from block import O, I, S, Z, L, J, TCOLS = 16ROWS = 20CELLS = COLS * ROWSCELLPX = 32 # 每个cell的像素宽度POS_FIRST_APPEAR = COLS / 2SCREEN_SIZE = (COLS * CELLPX, ROWS * CELLPX)COLOR_BG = (0, 0, 0)def draw(grid, pos=None): # grid是一个list，要么值为None，要么值为'Block' # 非空值在eval（）的作用下，用于配置颜色 if pos: # 6x5 s = pos - 3 - 2 * COLS # upper left position for p in range(0, COLS): q = s + p * COLS for i in range(q, q + 6): if 0 <= i < CELLS: # 0 <=i < CELLS:表示i这个cell在board内部。 c = eval(grid[i] + ".color") if grid[i] else COLOR_BG # 执行着色。shape的cell涂对应的class设定好的颜色，否则涂黑（背景色） a = i % COLS * CELLPX b = i / COLS * CELLPX screen.fill(c, (a, b, CELLPX, CELLPX)) else: # all screen.fill(COLOR_BG) for i, occupied in enumerate(grid): if occupied: c = eval(grid[i] + ".color") # 获取方块对应的颜色 a = i % COLS * CELLPX # 横向长度 b = i / COLS * CELLPX # 纵向长度 screen.fill(c, (a, b, CELLPX, CELLPX)) # fill:为cell上色, 第二个参数表示rect pygame.display.flip() # 刷新屏幕def phi(grid1, grid2, pos): # 4x4# 两个grid之4*4区域内是否会相撞（冲突） s = pos - 2 - 1 * COLS # upper left position for p in range(0, 4): q = s + p * COLS for i in range(q, q + 4): try: if grid1[i] and grid2[i]: return False except: pass return Truedef merge(grid1, grid2): # 合并两个grid grid = grid1[:] for i, c in enumerate(grid2): if c: grid[i] = c return griddef complete(grid): # 减去满行 n = 0 for i in range(0, CELLS, COLS): # 步长为一行。 if not None in grid[i:i + COLS]: #这一句很容易理解错误。 #实际含义是：如果grid[i:i + COLS]都不是None,那么执行下面的语句 grid = [None] * COLS + grid[:i] + grid[i + COLS:] n += 1 return grid, n#n表示减去的行数，用作统计分数pygame.init()pygame.event.set_blocked(None)pygame.event.set_allowed((KEYDOWN, QUIT))pygame.key.set_repeat(75, 0)pygame.display.set_caption('Tetris')screen = pygame.display.set_mode(SCREEN_SIZE)pygame.display.update()grid = [None] * CELLSspeed = 500screen.fill(COLOR_BG)while True: # spawn a block block = choice([O, I, S, Z, L, J, T])() pos = POS_FIRST_APPEAR if not phi(grid, block.grid(pos), pos): break # you lose pygame.time.set_timer(KEYDOWN, speed) # repeatedly create an event on the event queue # speed是时间间隔。。。speed越小，方块下落的速度越快。。。speed应该换为其他名字 while True: # move the block draw(merge(grid, block.grid(pos)), pos) event = pygame.event.wait() if event.type == QUIT: sys.exit() try: aim = { K_UNKNOWN: pos+COLS, K_UP: pos, K_DOWN: pos+COLS, K_LEFT: pos-1, K_RIGHT: pos+1, }[event.key] except KeyError: continue if event.key == K_UP: # 变形 block.rotate() elif event.key in (K_LEFT, K_RIGHT) and pos / COLS != aim / COLS: # pos/COLS表示当前位置所在行 # aim/COLS表示目标位置所在行 # 此判断表示，当shape在左边界时，不允许再向左移动（越界。。）,在最右边时向右也禁止 continue grid_aim = block.grid(aim) if grid_aim and phi(grid, grid_aim, aim): pos = aim else: if event.key == K_UP: block.rotate(times=3) elif not event.key in (K_LEFT, K_RIGHT): break grid = merge(grid, block.grid(pos)) grid, n = complete(grid) if n: draw(grid) speed -= 5 * n if speed < 75: speed = 75 |
调用的模块：
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#coding:utf-8#! /usr/bin/env pythonCOLS = 16ROWS = 20class Block(): color = (255,255,255) def __init__(self): self._state = 0 def __str__(self): return self.__class__.__name__ def _orientations(self): raise NotImplementedError() def rotate(self, times=1): for i in range(times): if len(self._orientations())-1 == self._state: self._state = 0 #只要_state比_orientations长度-1还要小，就让_state加1 else: self._state += 1 def blades(self): # 返回对应形状的一种旋转形状。(返回一个list,list中每个元素是一个(x,y)) return self._orientations()[self._state] def grid(self, pos, cols=COLS, rows=ROWS): # grid()函数：对于一个形状，从它的cell中的pos位置，按照orientations的位置提示，把所有cell涂色 # pos表示的是shape中的一个cell，也就是(0,0) if cols*rows <= pos: return None # 这种情况应该不可能出现吧。如果出现<=的情况 # 那么，pos都跑到界外了。。 grid = [None] * cols * rows grid[pos] = str(self) for b in self.blades(): x, y = b # pos/cols表示pos处于board的第几行 if pos/cols != (pos+x)/cols: return None i = pos + x + y * cols if i < 0: continue elif cols*rows <= i: return None grid[i] = str(self) # 给相应的其他位置都“涂色”,比如对于方块，是O型的，那么pos肯定是有值的，pos位于有上角。。 return grid# 以下每个形状class，_orientations（）都返回形状的列表。(0,0)一定被包含在其中，为了省略空间所以都没有写出.class O(Block): color = (207,247,0) def _orientations(self): return ( [(-1,0), (-1,1), (0,1)], )class I(Block): color = (135,240,60) def _orientations(self): return ( [(-2,0), (-1,0), (1,0)], [(0,-1), (0,1), (0,2)], )class S(Block): color = (171,252,113) def _orientations(self): return ( [(1,0), (-1,1), (0,1)], [(0,-1), (1,0), (1,1)], )class Z(Block): color = (243,61,110) def _orientations(self): return ( [(-1,0), (0,1), (1,1)], [(1,-1), (1,0), (0,1)], )class L(Block): color = (253,205,217) def _orientations(self): return ( [(-1,1), (-1,0), (1,0)], [(0,-1), (0,1), (1,1)], [(-1,0), (1,0), (1,-1)], [(-1,-1), (0,-1), (0,1)], )class J(Block): color = (140,180,225) def _orientations(self): return ( [(-1,0), (1,0), (1,1)], [(0,1), (0,-1), (1,-1)], [(-1,-1), (-1,0), (1,0)], [(-1,1), (0,1), (0,-1)], )class T(Block): color = (229,251,113) def _orientations(self): return ( [(-1,0), (0,1), (1,0)], [(0,-1), (0,1), (1,0)], [(-1,0), (0,-1), (1,0)], [(-1,0), (0,-1), (0,1)], ) |
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原文链接：https://blog.csdn.net/zz958712568/article/details/10225975
