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
主程序：
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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 sys
from random import choice
import pygame
from pygame.locals import *
from block import O, I, S, Z, L, J, T
COLS = 16
ROWS = 20
CELLS = COLS * ROWS
CELLPX = 32 # 每个cell的像素宽度
POS_FIRST_APPEAR = COLS / 2
SCREEN_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 True
def merge(grid1, grid2):
# 合并两个grid
grid = grid1[:]
for i, c in enumerate(grid2):
if c:
grid[i] = c
return grid
def 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] * CELLS
speed = 500
screen.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
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调用的模块：
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#coding:utf-8
#! /usr/bin/env python
COLS = 16
ROWS = 20
class 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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以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/zz958712568/article/details/10225975