如何输出多列

Question

我想处理这样的数据文件：

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

并输出一些列。

到目前为止我做到了：

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()

这适用于一列：lua column.lua test.dat new.dat 2

我想要的是 lua column.lua test.dat new.dat 1,2,4 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

现在您可以使用 ofile:write(cols(data[i], arg[3]) .. "\n") 而不是 ofile:write(data[i][column] .. "\n")。由于它会在每一行上进行解析，因此对于大量行来说效率可能很低，因此如果您遇到这种情况，您可能需要在脚本开头解析一次。

如何输出多列

How to output more than one column

csv

lua