CODEWITHSUNDEEP
stringluacountsymbols
GentleSir
GentleSir

Reputation: 85

LUA count repeating characters in string

I have a string "A001BBD0" and i want to know this info:

and that's it.

I found this pattern on web: "([a-zA-Z]).*(\1)" but it always returns nil for some reason

I guess i should split this string and check each symbol in several loops. I don't think this is a good idea (low performance)

i also found this topic but it doesn't give me any information

Upvotes: 4

Views: 4875

Answers (4)

cyclically
cyclically

Reputation: 11

Here's an even shorter answer than the previous ones that I just came up with.

This is the simplest, and most efficient way to count repeating characters in a string on Lua.

local str = "A001BBD0"

local count_0 = #string.gsub(str, "[^0]", "") -- count the 0's only
local count_B_and_0 = #string.gsub(str, "[^B0]", "") -- count the B's and 0's together

print(count_0, count_B_and_0)

Upvotes: 1

EzR1d3r
EzR1d3r

Reputation: 194

Shorter version of the previous answer about use of the ternary operator

local records = {}
s = "A001BBD0"
for c in string.gmatch(s, "%w") do
    records[c] = records[c] and records[c] + 1 or 1
end

for k,v in pairs(records) do
    if(v > 1) then -- print repeated chars
        print(k,v)
    end
end

Upvotes: 1

wsha
wsha

Reputation: 964

Creating a record for each alphanumeric char will give a more generic solution

local records = {} -- {['char'] = #number of occurances}
s = "A001BBD0"
for c in string.gmatch(s, "%w") do
    if records[c] then
        records[c] = records[c] + 1
    else
        records[c] = 1
    end
end

for k,v in pairs(records) do
    if(v > 1) then -- print repeated chars
        print(k,v)
    end
end
-- Output:
-- 0    3
-- B    2

Upvotes: 2

lhf
lhf

Reputation: 72312

gsub returns the number of substitutions. So, try this code:

function repeats(s,c)
    local _,n = s:gsub(c,"")
    return n
end

print(repeats("A001BBD0","0"))
print(repeats("A001BBD0","B"))

Upvotes: 9

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