csv - How to output more than one column -

i want process data file this:

2015-02-23  190   170   131   14  8   9   130 85  102.0   12  68 2015-02-24  165   128   97    14  7   6   110 75  101.7   12  64 2015-02-25  160   123   129   11  5   7   130 85  101.3   12  68 2015-02-26  151   115   128   11  nan 7   120 80  100.9   12  64 2015-02-27  141   119   130   11  4   nan 130 85  101.6   12  68 2015-02-28  142   137   143   nan nan nan 120 80  101.2   12  64 

and output some columns.

so far managed:

local infile = arg[1] local outfile = arg[2] local column = tonumber(arg[3]) local data = {} local row = 0 local ofile  line in io.lines(infile)   row = row + 1   data[row] = {}   local = 0   value in string.gmatch(line, "%s+")     = + 1     data[row][i] = value   end end  ofile = assert(io.open(outfile, "w")) = 1,row   ofile:write(data[i][column] .. "\n") end ofile:close() 

this works fine 1 column: lua column.lua test.dat new.dat 2

190 165 160 151 141 142 

what have lua column.lua test.dat new.dat 1,2,4 have columns 1, 2 , 4 in new file. possible?

you can use following function extract list of columns:

function cols(t, colnums, prefix)   -- number of first column , rest of numbers   -- splits 1,2,3 1 , 2,3   local col, rest = colnums:match("^(%d+)[,%s]*(.*)")   -- if nothing provided return current `prefix` value (may `nil`)   if not col return prefix end   -- convert string column number number   -- needed because t[1] , t['1'] references different values   local val = t[tonumber(col)]   -- call same function recursively, using rest of columns   -- concatenates prefix (if any) current value   return cols(t, rest, prefix , prefix.."\t"..val or val) end 

now instead of ofile:write(data[i][column] .. "\n"), can use ofile:write(cols(data[i], arg[3]) .. "\n"). since parsing on every row, may inefficient large number of rows, if that's case you, you'll need parse once @ beginning of script.