Rxjava - How to get the current and the previous item?

user_7479173 0
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How do to use operators so that i always get the previous and the current value? If possible i want to avoid creating state outside the pipe.

- time ->
1      2      3      4
|      |      |      |
Operations
       |      |      |
       (1,2)  (2,3)  (3,4)

Note that every value besides the first and the last one have to appear twice, so a simple buffer won't do.

I thought about combining skip with merge and buffer but merge does not seem to guarantee ordering.

val s = PublishSubject.create<Int>()
s.mergeWith(s.skip(1)).buffer(2).subscribe{i -> print(i)}
s.onNext(1)
s.onNext(2)
s.onNext(3)
s.onNext(4)

output:
[1, 2][2, 3][3, 4]

val o = Observable.just(1,2,3,4)
o.mergeWith(o.skip(1)).buffer(2).subscribe{i -> print(i)}

output:
[1, 2][3, 4][2, 3][4]

(the sole 4 is fine, and expected)

rx-kotlin rx-java kotlin
1 Answers

Looks like you still can use buffer:

Observable.just(1, 2, 3, 4)
    .buffer(2, 1)
    .subscribe { println(it) }

// prints
// [1, 2]
// [2, 3]
// [3, 4]
// [4]
user_15546330
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