CODEWITHSUNDEEP
regexlualua-patterns
Wallboy
Wallboy

Reputation: 154

Format long number to shorter version in Lua

I'm trying to figure out how I would go about formatting a large number to the shorter version by appending 'k' or 'm' using Lua. Example:

17478    => 17.5k
2832     => 2.8k
1548034  => 1.55m

I would like to have the rounding in there as well as per the example. I'm not very good at Regex, so I'm not sure where I would begin. Any help would be appreciated. Thanks.

Upvotes: 2

Views: 4515

Answers (2)

Mijago
Mijago

Reputation: 1639

Here a longer form, which uses the hint from Tom Blodget. Maybe its not the perfect form, but its a little more specific. For Lua 5.0, replace #steps with table.getn(steps).

function shortnumberstring(number)
    local steps = {
        {1,""},
        {1e3,"k"},
        {1e6,"m"},
        {1e9,"g"},
        {1e12,"t"},
    }
    for _,b in ipairs(steps) do
        if b[1] <= number+1 then
            steps.use = _
        end
    end
    local result = string.format("%.1f", number / steps[steps.use][1])
    if tonumber(result) >= 1e3 and steps.use < #steps then
        steps.use = steps.use + 1
        result = string.format("%.1f", tonumber(result) / 1e3)
    end
    --result = string.sub(result,0,string.sub(result,-1) == "0" and -3 or -1) -- Remove .0 (just if it is zero!)
    return result .. steps[steps.use][2]
end



print(shortnumberstring(100))
print(shortnumberstring(200))
print(shortnumberstring(999))
print(shortnumberstring(1234567))
print(shortnumberstring(999999))
print(shortnumberstring(9999999))
print(shortnumberstring(1345123))

Result:

> dofile"test.lua"
100.0
200.0
1.0k
1.2m
1.0m
10.0m
1.3m
>

And if you want to get rid of the "XX.0", uncomment the line before the return. Then our result is: 

> dofile"test.lua"
100
200
1k
1.2m
1m
10m
1.3m
>

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122493

Pattern matching doesn't seem like the right direction for this problem.

Assuming 2 digits after decimal point are kept in the shorter version, try:

function foo(n)
    if n >= 10^6 then
        return string.format("%.2fm", n / 10^6)
    elseif n >= 10^3 then
        return string.format("%.2fk", n / 10^3)
    else
        return tostring(n)
    end
end

Test:

print(foo(17478))
print(foo(2832))
print(foo(1548034))

Output:

17.48k
2.83k
1.55m

Upvotes: 5

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