Load only N lines of a CSV-CodePudding

I've a function for load OHCLV data from CSV, at the moment the function grabs all values, and return them, and that's works

What i would like is return only the last N values.

For example of 1000 values i need only the last 100.

CSV appears like that:

date,open,high,low,close,volume
2013-03-25 16:00:00,76.31,77.5,76.31,77.5,117.65500211
2013-03-25 16:15:00,77.61,77.89,77.1,77.1,36.04786927
[...]
2022-12-22 17:15:00,16646.0,16649.0,16611.0,16623.0,37.57992574
2022-12-22 17:30:00,16626.0,16626.0,16585.0,16608.0,14.26339254

Here's the function

const getOHCLVData = async (path) => {
  const res  = await fetch(path);
  const resp = await res.text();
  const cdata = resp.split('\n').slice(1).map((row) => {   // slice(1) skip the header
    const [date, open, high, low, close, volume] = row.split(',');
    return {
      time: new Date(`${date}`).getTime() / 1000,
      open: open,
      high: high,
      low: low,
      close: close,
      volume: volume,
    };
  });
  return cdata;
};

// Call function
getOHCLVData(myfile.csv)

i should try add a new argument to function and do something like that: cdata.lenght - NumberValue

Any suggestion?

CodePudding user response：

This is an amendment to Randy Casburn's comment (new user = low rep = cannot comment).

"change .slice(1) to .slice(-100) - negative values pulls from the end of the array. – Randy Casburn "

You will likely want to add a default variable to your function.

. . . = async (path, nValues=-1)

and alter your slice to be

. . . ('\n').slice(-nValues).map((row)

Note that the default of -1 is the same as .slice(1) because we use -nValues

This should add the functionality you want without forcing the user to know how many lines they want. You do not want to needlessly limit your function.