Is there a best way to sort the headers into a different order?

[TomzBench]

Hello,

Im making a kind of conversion application, where I have a csv file and I want to convert it into another csv file, allowing to represent data on different columns. I noticed the StringRecord type stores the data continuously in a single byte vector. I would like to split this vector and sort them based on some header information I have. But I think that this StringRecord does not expose the inner byte array this way.

There is probably a better way to do what im doing however?

[BurntSushi]

A StringRecord's representation isn't really amenable to this sort of thing. Internally it's roughly two allocations: a Vec<u8> and a Vec<usize>, where the latter represents the end offsets of every field in the record. This representation is optimized for reading since it avoids creating a bunch of tiny little allocs for every single field. This is especially helpful for CSV data with dozens or even hundreds of columns.

So even if the internals were exposed, there's no easy way to sort it. It could be sorted, but I think you'd need to write your own sort function that knows about the representation, but it'd be a little tricky I think.

Your best bet it to convert your StringRecord into a Vec<&str> via StringRecord::iter().collect(). Then do your sorting or whatever and convert back to a StringRecord.

I would also be open to implementing a StringRecord::swap routine, that lets you swap two fields. The problem I think is that this will need to potentially allocate, which is a bit of a bummer.

[TomzBench]

I'm doing something like:

let mut sorted = StringRecord::new();
for header in headers {
  match header.as_str() {
    "field_a" => sorted.push_field(&unsorted[3]),
    field => warn!(field, "unknown field")
  }
}

headers has an array of keys in strings, and "unsorted" is basically the order of the struct i define. and I map them via the match. I think this works how i want.