我想在我的电晕应用程序中评估一个数学字符串。现在我专注于三角函数,所以让我们让这个例子成为我们可能面临的最困难的例子:
local expr = "2sin(4pi+2)+7"
我的目标是以某种方式(或者)用一个pi --> math.pi
开关来评估它,或者甚至将它分解。然而,分手会更加困难,因为它可能像上面一样复杂,但也可能只是sin(1)
。
所以我宁愿尽可能地接近pythoneval(expr)
函数,但如果这不能发生,我很灵活。
这应该可以解决问题,它能够使用 lua 数学函数而无需放置“math.function”,因此 sqrt(100) 可以正常工作。我把它放在一起是因为我已经看到这个问题被问了太多次了。希望这会有所帮助:)
如果您有任何问题,请随时通过 rayaman99@gmail.com 与我联系
function evaluate(cmd,v) -- this uses recursion to solve math equations
--[[ We break it into pieces and solve tiny pieces at a time then put them back together
Example of whats going on
Lets say we have "5+5+5+5+5"
First we get this:
5+5+5+5 + 5
5+5+5 + 5
5+5 + 5
5 + 5
Take all the single 5's and do their opperation which is addition in this case and get 25 as our answer
if you want to visually see this with a custom expression, uncomment the code below that says '--print(l,o,r)'
]]
v=v or 0
local count=0
local function helper(o,v,r)-- a local helper function to speed things up and keep the code smaller
if type(v)=="string" then
if v:find("%D") then
v=tonumber(math[v]) or tonumber(_G[v]) -- This section allows global variables and variables from math to be used feel free to add your own enviroments
end
end
if type(r)=="string" then
if r:find("%D") then
r=tonumber(math[r]) or tonumber(_G[r]) -- A mirror from above but this affects the other side of the equation
-- Think about it as '5+a' and 'a+5' This mirror allows me to tackle both sides of the expression
end
end
local r=tonumber(r) or 0
if o=="+" then -- where we handle different math opperators
return r+v
elseif o=="-" then
return r-v
elseif o=="/" then
return r/v
elseif o=="*" then
return r*v
elseif o=="^" then
return r^v
end
end
for i,v in pairs(math) do
cmd=cmd:gsub(i.."(%b())",function(a)
a=a:sub(2,-2)
if a:sub(1,1)=="-" then
a="0"..a
end
return v(evaluate(a))
end)
end
cmd=cmd:gsub("%b()",function(a)
return evaluate(a:sub(2,-2))
end)
for l,o,r in cmd:gmatch("(.*)([%+%^%-%*/])(.*)") do -- iteration this breaks the expression into managable parts, when adding pieces into
--print(":",l,o,r) -- uncomment this to see how it does its thing
count=count+1 -- keep track for certain conditions
if l:find("[%+%^%-%*/]") then -- if I find that the lefthand side of the expression contains lets keep breaking it apart
v=helper(o,r,evaluate(l,v))-- evaluate again and do the helper function
else
if count==1 then
v=helper(o,r,l) -- Case where an expression contains one mathematical opperator
end
end
end
if count==0 then return (tonumber(cmd) or tonumber(math[cmd]) or tonumber(_G[cmd])) end
-- you can add your own enviroments as well... I use math and _G
return v
end
a=5
print(evaluate("2+2+2*2")) -- This still has work when it comes to pemdas; however, the use parentheses can order things!
print(evaluate("2+2+(2*2)"))-- <-- As seen here
print(evaluate("sqrt(100)"))
print(evaluate("sqrt(100)+abs(-100)"))
print(evaluate("sqrt(100+44)"))
print(evaluate("sqrt(100+44)/2"))
print(evaluate("5^2"))
print(evaluate("a")) -- that we stored above
print(evaluate("pi")) -- math.pi
print(evaluate("pi*2")) -- math.pi