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How would I math.floor with 2 decimals remaining?

Asked by

Aepeus 59

8 years ago

I have a math.floor script which rounds the number and adds SI prefixes at the end, but im trying to figure out to leave 2 decimals remaining. For example 1240.1245435, would be 1.24k+, instead of 1k+. Please help!

01local plr = game.Players.LocalPlayer
02local lstats = plr:WaitForChild("leaderstats")
03local doritos = lstats:WaitForChild("Doritos")
04local coinslabel = script.Parent
05 
06while true do -- always add bigger numbers first
07    wait()
08 
09    if doritos.Value >= 1000000000000000000000000000 then
10    coinslabel.Text = math.floor(doritos.Value / 1000000000000000000000000000) .. "o+ Doritos!" -- 1,000,000,000,000,000,000,000,000,000 / 1o+      
11 
12    elseif doritos.Value >= 1000000000000000000000000 then
13    coinslabel.Text = math.floor(doritos.Value / 1000000000000000000000000) .. "sp+ Doritos!" -- 1,000,000,000,000,000,000,000,000 / 1sp+   
14 
15    elseif doritos.Value >= 1000000000000000000000 then
16    coinslabel.Text = math.floor(doritos.Value / 1000000000000000000000) .. "sx+ Doritos!" -- 1,000,000,000,000,000,000,000 / 1sx+  
17 
18    elseif doritos.Value >= 1000000000000000000 then
19    coinslabel.Text = math.floor(doritos.Value / 1000000000000000000) .. "qn+ Doritos!" -- 1,000,000,000,000,000,000 / 1qn+ 
20 
21    elseif doritos.Value >= 1000000000000000 then
22    coinslabel.Text = math.floor(doritos.Value / 1000000000000000) .. "qd+ Doritos!" -- 1,000,000,000,000,000 / 1qd+    
23 
24    elseif doritos.Value >= 1000000000000 then
25    coinslabel.Text = math.floor(doritos.Value / 1000000000000) .. "t+ Doritos!" -- 1,000,000,000,000 / 1t+
26 
27    elseif doritos.Value >= 1000000000 then
28    coinslabel.Text = math.floor(doritos.Value / 1000000000) .. "b+ Doritos!" -- 1,000,000,000 / 1b+
29 
30    elseif doritos.Value >= 1000000 then
31    coinslabel.Text = math.floor(doritos.Value / 1000000) .. "m+ Doritos!" -- 1,000,000 / 1m+
32 
33    elseif doritos.Value >= 1000 then
34    coinslabel.Text = math.floor(doritos.Value / 1000 ) .. "k+ Doritos!" -- 1,000 / 1k+
35 
36    elseif doritos.Value <= 1000 then
37    coinslabel.Text = doritos.Value .. " Doritos!" -- <1,000 / <1k+
38 
39    end
40end
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1 answer

Answered by

Validark 1580

8 years ago

I think you're looking for a function like this:

1local function floor2Decimals(x)
2    return math.floor(x / .01) * .01
3end

math.floor(x / .01) * .00001 would be for 1k? Aepeus 59 — 8y

Close, but the goal is to shift the decimal place over to the right to the length of precision you want. This is done in his division operation, where he divides by the number of decimal places you desire. In this case, the tenths place (2 decimals). You simply copy whatever you multiply by, over to whatever you divide by. It cancels out. ScriptGuider 5640 — 8y

For 1k: math.floor(x / .00001) * .00001 Validark 1580 — 8y

How would I math.floor with 2 decimals remaining?

1 answer

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