How to convert CSVRecord to List?

Question

I need to convert a CSVRecord to a List to get the position of an element.

So I used this :

List<CSVRecord> records = csvParser.getRecords();
int i = records.get(0).indexOf(element);

However, records.get(0) returns a List<CSVRecord> and not a List<String>...

Moreover I cannot use csvParser.getHeaderNames(); because after that I need to copy the file with the header.

What kind of `List` do expect to get by calling `records.get(0)`? — T A, Oct 30 '19 at 13:45
@TA why not an `Array List` or even an `Array` converted in `List` with `Arrays.asList` — Royce, Oct 30 '19 at 13:46
so you want an `ArrayList`? I still dont quite get your problem. — T A, Oct 30 '19 at 13:48
I tried to answer this but realized I didn't understand your question properly, so what you want is not the value but only the index for a column? Or do you want to find a value? — Joakim Danielson, Oct 30 '19 at 13:54
If I have the value I can have the index. My main problem is conversion between CSVRecord and List. — Royce, Oct 30 '19 at 13:55
You want to convert the CSVRecord to a List of what? `List`? — T A, Oct 30 '19 at 13:56
@Royce dw, you should propably add that to your question though — T A, Oct 30 '19 at 14:05

score 2 · Accepted Answer · answered Oct 30 '19 at 13:48

2

Given the below function that I found on geeksforgeeks.com

public static <T> List<T> getListFromIterator(Iterator<T> iterator) { 
    Iterable<T> iterable = () -> iterator; 

    return StreamSupport 
              .stream(iterable.spliterator(), false) 
              .collect(Collectors.toList()); 
}

and the fact that you can get an iterator from CVSRecords you can do

List<String> list = getListFromIterator(records.get(0).iterator());

answered Oct 30 '19 at 13:48

Joakim Danielson

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This solutions seems to solve my problem. Thanks a lot. I have to check if this solution is not too much slow with a lot of data. – Royce Oct 30 '19 at 14:04

How to convert CSVRecord to List?

1 Answers1