How do I shorten this numbers?

Help and Feedback Scripting Support
8

Not as elegant as it could be, but here’s a quick solution:

local abbreviations = {
    "K", -- 4 digits
    "M", -- 7 digits
    "B", -- 10 digits
    "T", -- 13 digits
    "QD", -- 16 digits
    "QT", -- 19 digits
    "SXT", -- 22 digits
    "SEPT", -- 25 digits
    "OCT", -- 28 digits
    "NON", -- 31 digits
    "DEC", -- 34 digits
    "UDEC", -- 37 digits
    "DDEC", -- 40 digits
}

local function Abbreviate(x, decimals)
    if decimals == nil then decimals = 0 end
    local visible = nil
    local suffix = nil
    if x < 1000 then
      visible = x * math.pow(10, decimals)
      suffix = ""
    else
      local digits = math.floor(math.log10(x)) + 1
      local index = math.min(#abbreviations, math.floor((digits - 1) / 3))
      visible = x / math.pow(10, index * 3 - decimals)
      suffix = abbreviations[index] .. "+"
    end
    local front = visible / math.pow(10, decimals)
    local back = visible % math.pow(10, decimals)

    if decimals > 0 then
      return string.format("%i.%0." .. tostring(decimals) .. "i%s", front, back, suffix)
    else
      return string.format("%i%s", front, suffix)
    end
end
Results 
-- tests
function test(x, decimals, expected)
    local res = Abbreviate(x, decimals)
    print(tostring(x), "to " .. tostring(decimals) .. " places ->", res)
    assert(res == expected)
end

test(0, nil, "0")
test(0, 4, "0.0000")
test(1001, 2, "1.00K+")
test(1001, 3, "1.001K+")
test(100234, 2, "100.23K+")
test(1002345678, 2, "1.00B+")
test(1.23456789e35, nil, "123DEC+")
test(1.23456789e35, 2, "123.45DEC+")
test(1.23456789e44, 2, "123456.78DDEC+")

--[[ results:
0   to nil places ->    0
0   to 4 places ->  0.0000
1001    to 2 places ->  1.00K+
1001    to 3 places ->  1.001K+
100234  to 2 places ->  100.23K+
1002345678  to 2 places ->  1.00B+
1.23456789e+35  to nil places ->    123DEC+
1.23456789e+35  to 2 places ->  123.45DEC+
1.23456789e+44  to 2 places ->  123456.78DDEC+
]]
7 Likes