如何转换List <Optional<Type> &gt;进入清单 <Type>

Question

I have extracted values from a Map into a List but got a List<Optional<TXN_PMTxnHistory_rb>> , and I want to convert it into List<TXN_PMTxnHistory_rb> .

My code:

List<Optional<TXN_PMTxnHistory_rb>> listHistory_rb6 = 
    listHistory_rb5.values()
                   .stream()
                   .collect(Collectors.toList());

I'd like to obtain a List<TXN_PMTxnHistory_rb> .

Answer 1

Filter out all the empty values and use map to obtain the non-empty values:

List<TXN_PMTxnHistory_rb> listHistory_rb6 = 
    listHistory_rb5.values()
                   .stream()
                   .filter(Optional::isPresent)
                   .map(Optional::get)
                   .collect(Collectors.toList());

Answer 2

It's possible to do this using a method called flatMap on the stream of Optional s which will remove any 'empty' Optional s.

List<TXN_PMTxnHistory_rb> listHistory_rb6 = 
    listHistory_rb5.values()
                   .stream()
                   .flatMap(Optional::stream)
                   .collect(Collectors.toList());

Flatmap is essentially performing two things - "mapping" and "flattening". In the mapping phase it calls whatever method you've passed in and expects back a new stream - in this case each Optional in your original List will become a Stream containing either 1 or 0 values.

The flatten phase will then create a new Stream containing the results of all the mapped Stream s. Thus, if you had 2 Optional items in your List , one empty and one full, the resulting Stream would contain 0 elements from the first mapped Stream , and 1 value from the second.

Answer 3

Another option is to get all values and then filter out nulls :

List<TXN_PMTxnHistory_rb> listHistory_rb6 =
        listHistory_rb5.values().stream()
                       .map(opt -> opt.orElse(null))
                       .filter(Objects::nonNull)
                       .collect(Collectors.toList());