这段脚本有没有对游戏速度影响最小的写法

typhon_524 · 发表于 2009-12-2 15:04:10

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本帖最后由 typhon_524 于 2009-12-2 15:12 编辑

下面这段脚本是加在 Scene_Map 里的，在地图上会用事件随时调用 $scene.defence ，但是写得很长，对游戏速度有一些影响，求教有没有提高游戏速度的脚本写法

def defence
$game_variables[41] = 0 # 抗火属性归零
$game_variables[42] = 0 # 抗冰属性归零
$game_variables[43] = 0 # 抗雷属性归零
$game_variables[44] = 0 # 抗水属性归零
$game_variables[45] = 0 # 抗土属性归零
$game_variables[46] = 0 # 抗风属性归零
$game_variables[47] = 0 # 抗光属性归零
$game_variables[48] = 0 # 抗暗属性归零
for i in [32,36,40]
$game_variables[41] += 4 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[41] += 4 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[41] += 4 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
for i in [33,37,41]
$game_variables[41] += 6 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[41] += 6 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[41] += 6 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
for i in [34,38,42]
$game_variables[41] += 8 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[41] += 8 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[41] += 8 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
$game_variables[42] += 10 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(2)
$game_variables[42] += 10 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(2)
$game_variables[42] += 10 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(2)
$game_variables[42] += 10 if $game_party.actors[0].armor4_id != 0 and $data_armors[$game_party.actors[0].armor4_id].guard_element_set.include?(2)
$game_variables[42] += 10 if $game_party.actors[0].armor5_id != 0 and $data_armors[$game_party.actors[0].armor5_id].guard_element_set.include?(2)
$game_variables[42] += 10 if $game_party.actors[0].armor6_id != 0 and $data_armors[$game_party.actors[0].armor6_id].guard_element_set.include?(2)
for i in [43,47,51]
$game_variables[42] += 2 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[42] += 2 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[42] += 2 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
for i in [44,48,52]
$game_variables[42] += 4 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[42] += 4 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[42] += 4 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
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紫苏 · 发表于 2009-12-2 18:49:19

本帖最后由紫苏于 2009-12-2 19:12 编辑

要 Ruby 最快的写法，可以从减少函数的调用上着手，命令式编程语言中，自然是函数调用越少效率越高～在你这段代码中，有很多“常量”，直到在这个函数结束前都不会改变，而每次你都调用一个函数去获取该“常量”，这个函数调用的步骤多次重复，效率就从中丢失了

比如：$game_party.actors[0]——每调用一次 $game_party.actors[0]，实际上就是调用了一次 Game_Party#actors 方法获取了队员数组，并再通过这个数组调用了 Array#[] 方法去获取索引为 0 的数组元素。然而从这段代码中不难看出，变量的运算仅仅是针对队伍中的第一个队员，即 $game_party.actors[0]，那完全可以用一个变量保存这个队员对象，这样之后仅仅是引用这个对象，而不会再次重复 $game_party.actors[0] 的调用了

总地来说，就是重复了很多次的相同代码，尽量把它返回的结果保存在另一个局部变量中，这样其实是用空间换取了时间～

以前两个循环为例，我们分别把优化后的代码和未经优化的代码放到两个函数中：

def defence1
$game_variables[41] = 0 # 抗火属性归零
$game_variables[42] = 0 # 抗冰属性归零
$game_variables[43] = 0 # 抗雷属性归零
$game_variables[44] = 0 # 抗水属性归零
$game_variables[45] = 0 # 抗土属性归零
$game_variables[46] = 0 # 抗风属性归零
$game_variables[47] = 0 # 抗光属性归零
$game_variables[48] = 0 # 抗暗属性归零
actor = $game_party.actors[0]
armor1_id = actor.armor1_id
armor2_id = actor.armor2_id
armor3_id = actor.armor3_id
armed1 = armor1_id != 0
armed2 = armor2_id != 0
armed3 = armor3_id != 0
armor1 = $data_armors[armor1_id]
armor2 = $data_armors[armor2_id]
armor3 = $data_armors[armor3_id]
ges1 = armor1.guard_element_set
ges2 = armor2.guard_element_set
ges3 = armor3.guard_element_set
for i in [32,36,40]
$game_variables[41] += 4 if armed1 and ges1.include?(i)
$game_variables[41] += 4 if armed2 and ges2.include?(i)
$game_variables[41] += 4 if armed3 and ges3.include?(i)
end
for i in [33,37,41]
$game_variables[41] += 6 if armed1 and ges1.include?(i)
$game_variables[41] += 6 if armed2 and ges2.include?(i)
$game_variables[41] += 6 if armed3 and ges3.include?(i)
end
# ...
end
def defence2
$game_variables[41] = 0 # 抗火属性归零
$game_variables[42] = 0 # 抗冰属性归零
$game_variables[43] = 0 # 抗雷属性归零
$game_variables[44] = 0 # 抗水属性归零
$game_variables[45] = 0 # 抗土属性归零
$game_variables[46] = 0 # 抗风属性归零
$game_variables[47] = 0 # 抗光属性归零
$game_variables[48] = 0 # 抗暗属性归零
for i in [32,36,40]
$game_variables[41] += 4 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[41] += 4 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[41] += 4 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
for i in [33,37,41]
$game_variables[41] += 6 if $game_party.actors[0].armor1_id != 0 and $data_armors[$game_party.actors[0].armor1_id].guard_element_set.include?(i)
$game_variables[41] += 6 if $game_party.actors[0].armor2_id != 0 and $data_armors[$game_party.actors[0].armor2_id].guard_element_set.include?(i)
$game_variables[41] += 6 if $game_party.actors[0].armor3_id != 0 and $data_armors[$game_party.actors[0].armor3_id].guard_element_set.include?(i)
end
end

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然后再在事件中测试运行时间：

t = Time.now
for i in 0..100000
defence1
end
p Time.now - t
t = Time.now
for i in 0..100000
defence2
end
p Time.now - t

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未经优化的 defence2 执行的时间约略是优化后的 defence1 的两倍有余，所以在 defence 经常反复调用时，速度的差距也就明显了

另外一个着手点是 include? 方法——一般的 Array 是乱序线性数组，include? 自然也就只能是线性搜索，而这厮在数组很长时是一个骨灰级的效率杀手——在最坏的场合，include? 会一直从数组开头搜索到数组结尾。你的函数中，由于 guard_element_set 本身默认就是顺序数组（除非你用了什么脚本改变了从数据库中读取的防具属性防御）的这个特殊性，include? 完全可以被优化为一个一次性的遍历（即从头到尾检查整个 guard_element_set 数组一遍），每次枚举其中的一个元素，按照自然顺序检查元素（即属性的 ID）是否为某值，若是此值如何如何，如是彼值怎样怎样……也就是说，本来的这数个 for 循环，都会变成三个 for 循环（不管你有多少个属性需要判断），每个循环分别检查 armor1、armor2 和 armor3 的三个不同的 guard_element_set。这样下来，一共只会有三次数组的遍历，最坏的场合也不过是循环中的 if 或 case 语句判断到最后一个而已

另外，在循环外，你有很长一段代码重复调用 guard_element_set.include?(2)，由于参数总是 2，且 guard_element_set 也并未改变，这些 include? 都会返回相同的结果。因此，将 include? 的结果保存在变量中引用也会提高效率～

typhon_524 · 发表于 2009-12-3 09:07:38

本帖最后由 typhon_524 于 2009-12-3 09:08 编辑

非常感谢，如果我还有一个$game_party.actors[1]要添加，最好的办法是不是就直接加在后面了？这游戏最多就两个角色

def defence1
$game_variables[41] = 0 # 抗火属性归零
$game_variables[42] = 0 # 抗冰属性归零
$game_variables[43] = 0 # 抗雷属性归零
$game_variables[44] = 0 # 抗水属性归零
$game_variables[45] = 0 # 抗土属性归零
$game_variables[46] = 0 # 抗风属性归零
$game_variables[47] = 0 # 抗光属性归零
$game_variables[48] = 0 # 抗暗属性归零
actor = $game_party.actors[0]
armor1_id = actor.armor1_id
armor2_id = actor.armor2_id
armor3_id = actor.armor3_id
armed1 = armor1_id != 0
armed2 = armor2_id != 0
armed3 = armor3_id != 0
armor1 = $data_armors[armor1_id]
armor2 = $data_armors[armor2_id]
armor3 = $data_armors[armor3_id]
ges1 = armor1.guard_element_set
ges2 = armor2.guard_element_set
ges3 = armor3.guard_element_set
for i in [32,36,40]
$game_variables[41] += 4 if armed1 and ges1.include?(i)
$game_variables[41] += 4 if armed2 and ges2.include?(i)
$game_variables[41] += 4 if armed3 and ges3.include?(i)
end
for i in [33,37,41]
$game_variables[41] += 6 if armed1 and ges1.include?(i)
$game_variables[41] += 6 if armed2 and ges2.include?(i)
$game_variables[41] += 6 if armed3 and ges3.include?(i)
end
# ...
end
def petdefence
$game_variables[41] = 0 # 抗火属性归零
$game_variables[42] = 0 # 抗冰属性归零
$game_variables[43] = 0 # 抗雷属性归零
$game_variables[44] = 0 # 抗水属性归零
$game_variables[45] = 0 # 抗土属性归零
$game_variables[46] = 0 # 抗风属性归零
$game_variables[47] = 0 # 抗光属性归零
$game_variables[48] = 0 # 抗暗属性归零
actor = $game_party.actors[1]
armor1_id = actor.armor1_id
armor2_id = actor.armor2_id
armor3_id = actor.armor3_id
armed1 = armor1_id != 0
armed2 = armor2_id != 0
armed3 = armor3_id != 0
armor1 = $data_armors[armor1_id]
armor2 = $data_armors[armor2_id]
armor3 = $data_armors[armor3_id]
ges1 = armor1.guard_element_set
ges2 = armor2.guard_element_set
ges3 = armor3.guard_element_set
for i in [32,36,40]
$game_variables[41] += 4 if armed1 and ges1.include?(i)
$game_variables[41] += 4 if armed2 and ges2.include?(i)
$game_variables[41] += 4 if armed3 and ges3.include?(i)
end
for i in [33,37,41]
$game_variables[41] += 6 if armed1 and ges1.include?(i)
$game_variables[41] += 6 if armed2 and ges2.include?(i)
$game_variables[41] += 6 if armed3 and ges3.include?(i)
end
# ...
end

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[已经解决] 这段脚本有没有对游戏速度影响最小的写法

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