Ignoring fast changes on a Boolean variable with RxSwift
I have a boolean Variable which is switching true/false. I need to ignore any changes which settles less than 1 second. In other words, I need just values which stays more than 1 second. How to get such output from RxSwift?
Sample1: If value is true, and changes to false and then after 0.5 seconds become true again, I need my output sequence to show nothing.
Sample2: Above scenario with 1.2 seconds lag between false and true, makes a false event in my output sequence.
boolean rx-swift ripple
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
add a comment |
I have a boolean Variable which is switching true/false. I need to ignore any changes which settles less than 1 second. In other words, I need just values which stays more than 1 second. How to get such output from RxSwift?
Sample1: If value is true, and changes to false and then after 0.5 seconds become true again, I need my output sequence to show nothing.
Sample2: Above scenario with 1.2 seconds lag between false and true, makes a false event in my output sequence.
boolean rx-swift ripple
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
add a comment |
I have a boolean Variable which is switching true/false. I need to ignore any changes which settles less than 1 second. In other words, I need just values which stays more than 1 second. How to get such output from RxSwift?
Sample1: If value is true, and changes to false and then after 0.5 seconds become true again, I need my output sequence to show nothing.
Sample2: Above scenario with 1.2 seconds lag between false and true, makes a false event in my output sequence.
boolean rx-swift ripple
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
I have a boolean Variable which is switching true/false. I need to ignore any changes which settles less than 1 second. In other words, I need just values which stays more than 1 second. How to get such output from RxSwift?
Sample1: If value is true, and changes to false and then after 0.5 seconds become true again, I need my output sequence to show nothing.
Sample2: Above scenario with 1.2 seconds lag between false and true, makes a false event in my output sequence.
boolean rx-swift ripple
boolean rx-swift ripple
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
edited Nov 19 '18 at 14:55
Abbas Sabeti
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
asked Nov 19 '18 at 14:38
Abbas SabetiAbbas Sabeti
1516
1516
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can use a combination of skip, debounce and distinctUntilChanged to achieve that:
aSwitch.rx.isOn
.skip(1) // 1
.debounce(1, scheduler: MainScheduler.instance) // 2
.distinctUntilChanged() // 3
.subscribe(onNext: { isOn in
print("IS ON: (isOn)")
})
.disposed(by: disposeBag)
A little explanation:
// 1: This prevents the Switch from emitting its initial value. You can remove this if you need the initial value of the UISwitch
// 2: This waits until the UISwitch has not changed its value for at least 1 second before emitting the new value
// 3: This filters out values that have not changed. So no value is emitted if after one second of inactivity the value of the switch is the same as before the user started changing the value.
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
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
You can use a combination of skip, debounce and distinctUntilChanged to achieve that:
aSwitch.rx.isOn
.skip(1) // 1
.debounce(1, scheduler: MainScheduler.instance) // 2
.distinctUntilChanged() // 3
.subscribe(onNext: { isOn in
print("IS ON: (isOn)")
})
.disposed(by: disposeBag)
A little explanation:
// 1: This prevents the Switch from emitting its initial value. You can remove this if you need the initial value of the UISwitch
// 2: This waits until the UISwitch has not changed its value for at least 1 second before emitting the new value
// 3: This filters out values that have not changed. So no value is emitted if after one second of inactivity the value of the switch is the same as before the user started changing the value.
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
add a comment |
You can use a combination of skip, debounce and distinctUntilChanged to achieve that:
aSwitch.rx.isOn
.skip(1) // 1
.debounce(1, scheduler: MainScheduler.instance) // 2
.distinctUntilChanged() // 3
.subscribe(onNext: { isOn in
print("IS ON: (isOn)")
})
.disposed(by: disposeBag)
A little explanation:
// 1: This prevents the Switch from emitting its initial value. You can remove this if you need the initial value of the UISwitch
// 2: This waits until the UISwitch has not changed its value for at least 1 second before emitting the new value
// 3: This filters out values that have not changed. So no value is emitted if after one second of inactivity the value of the switch is the same as before the user started changing the value.
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
add a comment |
You can use a combination of skip, debounce and distinctUntilChanged to achieve that:
aSwitch.rx.isOn
.skip(1) // 1
.debounce(1, scheduler: MainScheduler.instance) // 2
.distinctUntilChanged() // 3
.subscribe(onNext: { isOn in
print("IS ON: (isOn)")
})
.disposed(by: disposeBag)
A little explanation:
// 1: This prevents the Switch from emitting its initial value. You can remove this if you need the initial value of the UISwitch
// 2: This waits until the UISwitch has not changed its value for at least 1 second before emitting the new value
// 3: This filters out values that have not changed. So no value is emitted if after one second of inactivity the value of the switch is the same as before the user started changing the value.
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
You can use a combination of skip, debounce and distinctUntilChanged to achieve that:
aSwitch.rx.isOn
.skip(1) // 1
.debounce(1, scheduler: MainScheduler.instance) // 2
.distinctUntilChanged() // 3
.subscribe(onNext: { isOn in
print("IS ON: (isOn)")
})
.disposed(by: disposeBag)
A little explanation:
// 1: This prevents the Switch from emitting its initial value. You can remove this if you need the initial value of the UISwitch
// 2: This waits until the UISwitch has not changed its value for at least 1 second before emitting the new value
// 3: This filters out values that have not changed. So no value is emitted if after one second of inactivity the value of the switch is the same as before the user started changing the value.
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
edited Nov 19 '18 at 15:38
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
answered Nov 19 '18 at 15:33
joernjoern
20.7k66382
20.7k66382
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
add a comment |
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
Great solution. Thanks!
– Abbas Sabeti
Nov 19 '18 at 20:04
add a comment |
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