How can I filter Flux with state?
I'd like to apply filter on my Flux based on a state calculated from previous values. However, it is recommended to avoid using state in operators according to the javadoc
Note that using state in the java.util.function / lambdas used within Flux operators should be avoided, as these may be shared between several Subscribers.
For example, Flux#distinct filters items that appears earlier. How can we implement our own version of distinct?
project-reactor
asked Nov 12 at 6:38
hota911
12
add a comment |
I'd like to apply filter on my Flux based on a state calculated from previous values. However, it is recommended to avoid using state in operators according to the javadoc
Note that using state in the java.util.function / lambdas used within Flux operators should be avoided, as these may be shared between several Subscribers.
For example, Flux#distinct filters items that appears earlier. How can we implement our own version of distinct?
project-reactor
asked Nov 12 at 6:38
hota911
12
add a comment |
I'd like to apply filter on my Flux based on a state calculated from previous values. However, it is recommended to avoid using state in operators according to the javadoc
Note that using state in the java.util.function / lambdas used within Flux operators should be avoided, as these may be shared between several Subscribers.
For example, Flux#distinct filters items that appears earlier. How can we implement our own version of distinct?
project-reactor
asked Nov 12 at 6:38
hota911
12
I'd like to apply filter on my Flux based on a state calculated from previous values. However, it is recommended to avoid using state in operators according to the javadoc
Note that using state in the java.util.function / lambdas used within Flux operators should be avoided, as these may be shared between several Subscribers.
For example, Flux#distinct filters items that appears earlier. How can we implement our own version of distinct?
project-reactor
project-reactor
asked Nov 12 at 6:38
hota911
12
asked Nov 12 at 6:38
hota911
12
asked Nov 12 at 6:38
hota911
12
asked Nov 12 at 6:38
hota911
12
asked Nov 12 at 6:38
hota911
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I have found an answer to my question. Flux#distinct can take a Supplier which provides initial state and a BiPredicate which performs "distinct" check, so we can store arbitrary state in the store and decide whether to keep each element.
Following code shows how to keep the first 3 elements of each mod2 group without changing the order.
// Get first 3 elements per mod 2.
Flux<Integer> first3PerMod2 =
Flux.fromIterable(ImmutableList.of(9, 3, 7, 4, 5, 10, 6, 8, 2, 1))
.distinct(
// Group by mod2
num -> num % 2,
// Counter to store how many elements have been processed for each group.
() -> new HashMap<Integer, Integer>(),
// Increment or set 1 to the counter,
// and return whether 3 elements are published.
(map, num) -> map.merge(num, 1, Integer::sum) <= 3,
// Clean up the state.
map -> map.clear());
StepVerifier.create(first3PerMod2).expectNext(9, 3, 7, 4, 10, 6).verifyComplete();
answered Nov 12 at 14:12
hota911
12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I have found an answer to my question. Flux#distinct can take a Supplier which provides initial state and a BiPredicate which performs "distinct" check, so we can store arbitrary state in the store and decide whether to keep each element.
Following code shows how to keep the first 3 elements of each mod2 group without changing the order.
// Get first 3 elements per mod 2.
Flux<Integer> first3PerMod2 =
Flux.fromIterable(ImmutableList.of(9, 3, 7, 4, 5, 10, 6, 8, 2, 1))
.distinct(
// Group by mod2
num -> num % 2,
// Counter to store how many elements have been processed for each group.
() -> new HashMap<Integer, Integer>(),
// Increment or set 1 to the counter,
// and return whether 3 elements are published.
(map, num) -> map.merge(num, 1, Integer::sum) <= 3,
// Clean up the state.
map -> map.clear());
StepVerifier.create(first3PerMod2).expectNext(9, 3, 7, 4, 10, 6).verifyComplete();
answered Nov 12 at 14:12
hota911
12
add a comment |
I have found an answer to my question. Flux#distinct can take a Supplier which provides initial state and a BiPredicate which performs "distinct" check, so we can store arbitrary state in the store and decide whether to keep each element.
Following code shows how to keep the first 3 elements of each mod2 group without changing the order.
// Get first 3 elements per mod 2.
Flux<Integer> first3PerMod2 =
Flux.fromIterable(ImmutableList.of(9, 3, 7, 4, 5, 10, 6, 8, 2, 1))
.distinct(
// Group by mod2
num -> num % 2,
// Counter to store how many elements have been processed for each group.
() -> new HashMap<Integer, Integer>(),
// Increment or set 1 to the counter,
// and return whether 3 elements are published.
(map, num) -> map.merge(num, 1, Integer::sum) <= 3,
// Clean up the state.
map -> map.clear());
StepVerifier.create(first3PerMod2).expectNext(9, 3, 7, 4, 10, 6).verifyComplete();
answered Nov 12 at 14:12
hota911
12
add a comment |
I have found an answer to my question. Flux#distinct can take a Supplier which provides initial state and a BiPredicate which performs "distinct" check, so we can store arbitrary state in the store and decide whether to keep each element.
Following code shows how to keep the first 3 elements of each mod2 group without changing the order.
// Get first 3 elements per mod 2.
Flux<Integer> first3PerMod2 =
Flux.fromIterable(ImmutableList.of(9, 3, 7, 4, 5, 10, 6, 8, 2, 1))
.distinct(
// Group by mod2
num -> num % 2,
// Counter to store how many elements have been processed for each group.
() -> new HashMap<Integer, Integer>(),
// Increment or set 1 to the counter,
// and return whether 3 elements are published.
(map, num) -> map.merge(num, 1, Integer::sum) <= 3,
// Clean up the state.
map -> map.clear());
StepVerifier.create(first3PerMod2).expectNext(9, 3, 7, 4, 10, 6).verifyComplete();
answered Nov 12 at 14:12
hota911
12
I have found an answer to my question. Flux#distinct can take a Supplier which provides initial state and a BiPredicate which performs "distinct" check, so we can store arbitrary state in the store and decide whether to keep each element.
Following code shows how to keep the first 3 elements of each mod2 group without changing the order.
// Get first 3 elements per mod 2.
Flux<Integer> first3PerMod2 =
Flux.fromIterable(ImmutableList.of(9, 3, 7, 4, 5, 10, 6, 8, 2, 1))
.distinct(
// Group by mod2
num -> num % 2,
// Counter to store how many elements have been processed for each group.
() -> new HashMap<Integer, Integer>(),
// Increment or set 1 to the counter,
// and return whether 3 elements are published.
(map, num) -> map.merge(num, 1, Integer::sum) <= 3,
// Clean up the state.
map -> map.clear());
StepVerifier.create(first3PerMod2).expectNext(9, 3, 7, 4, 10, 6).verifyComplete();
answered Nov 12 at 14:12
hota911
12
answered Nov 12 at 14:12
hota911
12
answered Nov 12 at 14:12
hota911
12
answered Nov 12 at 14:12
hota911
12
12
add a comment |
add a comment |
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