Question

我想处理这样的数据文件：

DateTime

并输出部分列。

到目前为止，我设法：

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

这适用于一列：local infile = arg[1] local outfile = arg[2] local column = tonumber(arg[3]) local data = {} local row = 0 local ofile for line in io.lines(infile) do row = row + 1 data[row] = {} local i = 0 for value in string.gmatch(line, "%S+") do i = i + 1 data[row][i] = value end end ofile = assert(io.open(outfile, "w")) for i = 1,row do ofile:write(data[i][column] .. "\n") end ofile:close()

lua column.lua test.dat new.dat 2

我希望190 165 160 151 141 142在新文件中包含第1,2和4列。这可能吗？

Answer 1

您可以使用以下函数提取列列表：

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

现在可以使用ofile:write(data[i][column] .. "\n")代替ofile:write(cols(data[i], arg[3]) .. "\n")而不是<img>。由于它在每一行上都进行了解析，因此对于大量行来说效率可能不高，所以如果你的情况如此，那么你可能需要在脚本的开头解析一次。 / p>

如何输出多个列

1 个答案: