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Parsing out multiple lines with LPeg in Lua

I have some text file with multiple lines block like

2011/01/01 13:13:13,<AB>, Some Certain Text,=,
[    
certain text
         [
                  0: 0 0 0 0 0 0 0 0 
                  8: 0 0 0 0 0 0 0 0 
                 16: 0 0 0 9 343 3938 9433 8756 
                 24: 6270 4472 3182 2503 1768 1140 836 496 
                 32: 326 273 349 269 144 121 94 82 
                 40: 64 80 66 59 56 47 50 46 
                 48: 64 35 42 53 42 40 41 34 
                 56: 35 41 39 39 47 30 30 39 
                 Total count: 12345
        ]
    certain text
]
some text
2011/01/01 14:14:14,<AB>, Some Certain Text,=,
[
 certain text
   [
              0: 0 0 0 0 0 0 0 0 
              8: 0 0 0 0 0 0 0 0 
             16: 0 0 0 4 212 3079 8890 8941 
             24: 6177 4359 3625 2420 1639 974 594 438 
             32: 323 286 318 296 206 132 96 85 
             40: 65 73 62 53 47 55 49 52 
             48: 29 44 44 41 43 36 50 36 
             56: 40 30 29 40 35 30 25 31 
             64: 47 31 25 29 24 30 35 31 
             72: 28 31 17 37 35 30 20 33 
             80: 28 20 37 25 21 23 25 36 
             88: 27 35 22 23 15 24 34 28
             Total count: 123456 
    ]
    certain text
some text
]

Those variant-length blocks exist between text. I want to read out all numbers after : and keep them in individual arrays. In this case, there will be two arrays:

array1 = { 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 343 3938 9433 8756 6270 4472 3182 2503 1768 1140 836 496 326 273 349 269 144 121 94 82 64 80 66 59 56 47 50 46 64 35 42 53 42 40 41 34 35 41 39 39 47 30 30 39 12345 }

array2 = { 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 212 3079 8890 8941 6177 4359 3625 2420 1639 974 594 438 323 286 318 296 206 132 96 85 65 73 62 53 47 55 49 52 29 44 44 41 43 36 50 36 40 30 29 40 35 30 25 31 47 31 25 29 24 30 35 31 28 31 17 37 35 30 20 33 28 20 37 25 21 23 25 36 27 35 22 23 15 24 34 28 123456 }

I found lpeg may be a light-weighted way to achieve it. But I'm totally new to PEGs and LPeg. Please help!

like image 446
Decula Avatar asked Dec 08 '22 12:12

Decula


2 Answers

LPEG version:

local lpeg            = require "lpeg"
local lpegmatch       = lpeg.match
local C, Ct, P, R, S  = lpeg.C, lpeg.Ct, lpeg.P, lpeg.R, lpeg.S
local Cg              = lpeg.Cg

local data_to_arrays

do
  local colon    = P":"
  local lbrak    = P"["
  local rbrak    = P"]"
  local digits   = R"09"^1
  local eol      = P"\n\r" + P"\r\n" + P"\n" + P"\r"
  local ws       = S" \t\v"
  local optws    = ws^0
  local getnum   = C(digits) / tonumber * optws
  local start    = lbrak * optws * eol
  local stop     = optws * rbrak
  local line     = optws * digits * colon * optws
                 * getnum * getnum * getnum * getnum
                 * getnum * getnum * getnum * getnum
                 * eol
  local count    = optws * P"Total count:" * optws * getnum * eol
  local inner    = Ct(line^1 * count^-1)
--local inner    = Ct(line^1 * Cg(count, "count")^-1)
  local array    = start * inner * stop
  local extract  = Ct((array + 1)^0)

  data_to_arrays = function (data)
    return lpegmatch (extract, data)
  end
end

This actually works only if there are exactly eight integers on each line of the data block. Depending on how well formed your input is this may be a curse or a blessing ;-)

And a test file:

data = [[
some text
[    
some text
         [
                  0: 0 0 0 0 0 0 0 0 
                  8: 0 0 0 0 0 0 0 0 
                 16: 0 0 0 9 343 3938 9433 8756 
                 24: 6270 4472 3182 2503 1768 1140 836 496 
                 32: 326 273 349 269 144 121 94 82 
                 40: 64 80 66 59 56 47 50 46 
                 48: 64 35 42 53 42 40 41 34 
                 56: 35 41 39 39 47 30 30 39 
                 Total count: 12345
        ]
    some text
]
some text
[
 some text
   [
              0: 0 0 0 0 0 0 0 0 
              8: 0 0 0 0 0 0 0 0 
             16: 0 0 0 4 212 3079 8890 8941 
             24: 6177 4359 3625 2420 1639 974 594 438 
             32: 323 286 318 296 206 132 96 85 
             40: 65 73 62 53 47 55 49 52 
             48: 29 44 44 41 43 36 50 36 
             56: 40 30 29 40 35 30 25 31 
             64: 47 31 25 29 24 30 35 31 
             72: 28 31 17 37 35 30 20 33 
             80: 28 20 37 25 21 23 25 36 
             88: 27 35 22 23 15 24 34 28 
    ]
    some text
some text
]
]]

local arrays = data_to_arrays (data)

for n = 1, #arrays do
  local ar   = arrays[n]
  local size = #ar
  io.write (string.format ("[%d] = { --[[size: %d items]]\n  ", n, size))
  for i = 1, size do
    io.write (string.format ("%d,%s", ar[i], (i % 5 == 0) and "\n  " or " "))
  end
  if ar.count ~= nil then
    io.write (string.format ("\n  [\"count\"] = %d,", ar.count))
  end
  io.write (string.format ("\n}\n"))
end
like image 111
Philipp Gesang Avatar answered Jan 18 '23 19:01

Philipp Gesang


Try this code, which does no use LPEG:

-- assume T contains the text
local a={}
local i=0
for b in T:gmatch("%b[]") do
        b=b:gsub("%d+:","")
        i=i+1
        local t={}
        local j=0
        for n in b:gmatch("%d+") do
                j=j+1; t[j]=tonumber(n)
        end
        a[i]=t
end
like image 40
lhf Avatar answered Jan 18 '23 20:01

lhf