从 table 获取价值的最优化方式
Most optimized way to get value from table
我需要检查第一个值是否 >= 'from' 并且第二个值是否为 <= 'to',如果为真,那么我的函数将返回数字。它正在工作,但我不知道这是否是获得价值的最佳和最优化的方式(来自 table 的数字)。
local table = {
{from = -1, to = 12483, number = 0},
{from = 12484, to = 31211, number = 1},
{from = 31212, to = 53057, number = 2},
{from = 53058, to = 90200, number = 3},
{from = 90201, to = 153341, number = 4},
{from = 153342, to = 443162, number = 5},
{from = 443163, to = 753380, number = 6},
{from = 753381, to = 1280747, number = 7},
{from = 1280748, to = 2689570, number = 8},
{from = 2689571, to = 6723927, number = 9},
{from = 6723928, to = 6723928, number = 10}
}
local exampleFromValue = 31244
local exampleToValue = 42057
local function getNumber()
local number = 0
for k, v in pairs(table) do
if (v.from and exampleFromValue >= v.from) and (v.to and exampleToValue <= v.to) then
number = v.number
break
end
end
return number
end
print(getNumber())
对于这么小的数据量,这样的功能似乎不是性能问题。不过,您可以稍微压缩一下数据:
local t = {
12484, 31212, 53058, 90201, 153342, 443163, 753381, 1280748, 2689571, 6723928
}
local exampleFromValue = 31244
local exampleToValue = 42057
local function getNumber()
local last = -1
for i, v in ipairs(t) do
if exampleFromValue >= last and exampleToValue < v then
return i - 1
end
last = v
end
return 0
end
我需要检查第一个值是否 >= 'from' 并且第二个值是否为 <= 'to',如果为真,那么我的函数将返回数字。它正在工作,但我不知道这是否是获得价值的最佳和最优化的方式(来自 table 的数字)。
local table = {
{from = -1, to = 12483, number = 0},
{from = 12484, to = 31211, number = 1},
{from = 31212, to = 53057, number = 2},
{from = 53058, to = 90200, number = 3},
{from = 90201, to = 153341, number = 4},
{from = 153342, to = 443162, number = 5},
{from = 443163, to = 753380, number = 6},
{from = 753381, to = 1280747, number = 7},
{from = 1280748, to = 2689570, number = 8},
{from = 2689571, to = 6723927, number = 9},
{from = 6723928, to = 6723928, number = 10}
}
local exampleFromValue = 31244
local exampleToValue = 42057
local function getNumber()
local number = 0
for k, v in pairs(table) do
if (v.from and exampleFromValue >= v.from) and (v.to and exampleToValue <= v.to) then
number = v.number
break
end
end
return number
end
print(getNumber())
对于这么小的数据量,这样的功能似乎不是性能问题。不过,您可以稍微压缩一下数据:
local t = {
12484, 31212, 53058, 90201, 153342, 443163, 753381, 1280748, 2689571, 6723928
}
local exampleFromValue = 31244
local exampleToValue = 42057
local function getNumber()
local last = -1
for i, v in ipairs(t) do
if exampleFromValue >= last and exampleToValue < v then
return i - 1
end
last = v
end
return 0
end