结合具有RxJava可观测数据的并行下载图像地图
我正在尝试并行下载一个图像列表,将它们组合成一张地图 起初我试着做一个这样的观察:结合具有RxJava可观测数据的并行下载图像地图,java,android,rx-java,Java,Android,Rx Java,我正在尝试并行下载一个图像列表,将它们组合成一张地图 起初我试着做一个这样的观察: Observable<Map<Integer, Bitmap>> getImages(final List<Activity> activities) { return Observable.create(new Observable.OnSubscribe<Map<Integer, Bitmap>>() { @Override
Observable<Map<Integer, Bitmap>> getImages(final List<Activity> activities) {
return Observable.create(new Observable.OnSubscribe<Map<Integer, Bitmap>>() {
@Override
public void call(Subscriber<? super Map<Integer, Bitmap>> subscriber) {
try {
Map<Integer, Bitmap> result = new HashMap<Integer, Bitmap>();
for (Activity act : activities) {
result.put(act.getId(), downloadImage(act.getImage()));
}
subscriber.onNext(result);
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
可观察的getImages(最终列表活动){
返回Observable.create(newobservable.OnSubscribe(){
@凌驾
public void call(订户创建一个包含位图和活动的包装类。比如说ActivityBitmap
。将getImage
替换为getActivityBitmap
:
Observable<ActivityBitmap> getActivityBitmap(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<ActivityBitmap>() {
@Override
public void call(Subscriber<? super ActivityBitmap> subscriber) {
try {
subscriber.onNext(new ActivityBitmap(activity, downloadImage(activity.getImage())));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
创建一个包含位图和活动的包装类。比如说ActivityBitmap
。将getImage
替换为getActivityBitmap
:
Observable<ActivityBitmap> getActivityBitmap(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<ActivityBitmap>() {
@Override
public void call(Subscriber<? super ActivityBitmap> subscriber) {
try {
subscriber.onNext(new ActivityBitmap(activity, downloadImage(activity.getImage())));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
创建一个包含位图和活动的包装类。比如说ActivityBitmap
。将getImage
替换为getActivityBitmap
:
Observable<ActivityBitmap> getActivityBitmap(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<ActivityBitmap>() {
@Override
public void call(Subscriber<? super ActivityBitmap> subscriber) {
try {
subscriber.onNext(new ActivityBitmap(activity, downloadImage(activity.getImage())));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
创建一个包含位图和活动的包装类。比如说ActivityBitmap
。将getImage
替换为getActivityBitmap
:
Observable<ActivityBitmap> getActivityBitmap(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<ActivityBitmap>() {
@Override
public void call(Subscriber<? super ActivityBitmap> subscriber) {
try {
subscriber.onNext(new ActivityBitmap(activity, downloadImage(activity.getImage())));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
我有一个可能的解决方案。它使用reduce操作符转换为map。不过,我不确定在Observable中订阅Observable是否是一个好的实践
Observable<Bitmap> getImage(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<Bitmap>() {
@Override
public void call(Subscriber<? super Bitmap> subscriber) {
try {
subscriber.onNext(downloadImage(activity.getImage()));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
Observable<HashMap<Integer, Bitmap>> getImages(final List<Activity> activities) {
return Observable
.from(activities)
.reduce(new HashMap<Integer, Bitmap>(), new Func2<HashMap<Integer, Bitmap>, Activity, HashMap<Integer, Bitmap>>() {
@Override
public HashMap<Integer, Bitmap> call(final HashMap<Integer, Bitmap> bitmaps, final Activity activity) {
getImage(activity)
.observeOn(Schedulers.io())
.subscribeOn(Schedulers.io())
.subscribe(new Action1<Bitmap>() {
@Override
public void call(Bitmap bitmap) {
bitmaps.put(activity.getId(), bitmap);
}
});
return bitmaps;
}
});
}
可观察的getImage(最终活动){
返回Observable.create(newobservable.OnSubscribe(){
@凌驾
public void call(Subscriber我有一个可能的解决方案。它使用reduce操作符转换为map。不过,我不确定在Observable中订阅Observable是否是一个好的做法
Observable<Bitmap> getImage(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<Bitmap>() {
@Override
public void call(Subscriber<? super Bitmap> subscriber) {
try {
subscriber.onNext(downloadImage(activity.getImage()));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
Observable<HashMap<Integer, Bitmap>> getImages(final List<Activity> activities) {
return Observable
.from(activities)
.reduce(new HashMap<Integer, Bitmap>(), new Func2<HashMap<Integer, Bitmap>, Activity, HashMap<Integer, Bitmap>>() {
@Override
public HashMap<Integer, Bitmap> call(final HashMap<Integer, Bitmap> bitmaps, final Activity activity) {
getImage(activity)
.observeOn(Schedulers.io())
.subscribeOn(Schedulers.io())
.subscribe(new Action1<Bitmap>() {
@Override
public void call(Bitmap bitmap) {
bitmaps.put(activity.getId(), bitmap);
}
});
return bitmaps;
}
});
}
可观察的getImage(最终活动){
返回Observable.create(newobservable.OnSubscribe(){
@凌驾
public void call(Subscriber我有一个可能的解决方案。它使用reduce操作符转换为map。不过,我不确定在Observable中订阅Observable是否是一个好的做法
Observable<Bitmap> getImage(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<Bitmap>() {
@Override
public void call(Subscriber<? super Bitmap> subscriber) {
try {
subscriber.onNext(downloadImage(activity.getImage()));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
Observable<HashMap<Integer, Bitmap>> getImages(final List<Activity> activities) {
return Observable
.from(activities)
.reduce(new HashMap<Integer, Bitmap>(), new Func2<HashMap<Integer, Bitmap>, Activity, HashMap<Integer, Bitmap>>() {
@Override
public HashMap<Integer, Bitmap> call(final HashMap<Integer, Bitmap> bitmaps, final Activity activity) {
getImage(activity)
.observeOn(Schedulers.io())
.subscribeOn(Schedulers.io())
.subscribe(new Action1<Bitmap>() {
@Override
public void call(Bitmap bitmap) {
bitmaps.put(activity.getId(), bitmap);
}
});
return bitmaps;
}
});
}
可观察的getImage(最终活动){
返回Observable.create(newobservable.OnSubscribe(){
@凌驾
public void call(Subscriber我有一个可能的解决方案。它使用reduce操作符转换为map。不过,我不确定在Observable中订阅Observable是否是一个好的做法
Observable<Bitmap> getImage(final Activity activity) {
return Observable.create(new Observable.OnSubscribe<Bitmap>() {
@Override
public void call(Subscriber<? super Bitmap> subscriber) {
try {
subscriber.onNext(downloadImage(activity.getImage()));
subscriber.onCompleted();
} catch (Exception e) {
subscriber.onError(e);
}
}
});
}
Observable<HashMap<Integer, Bitmap>> getImages(final List<Activity> activities) {
return Observable
.from(activities)
.reduce(new HashMap<Integer, Bitmap>(), new Func2<HashMap<Integer, Bitmap>, Activity, HashMap<Integer, Bitmap>>() {
@Override
public HashMap<Integer, Bitmap> call(final HashMap<Integer, Bitmap> bitmaps, final Activity activity) {
getImage(activity)
.observeOn(Schedulers.io())
.subscribeOn(Schedulers.io())
.subscribe(new Action1<Bitmap>() {
@Override
public void call(Bitmap bitmap) {
bitmaps.put(activity.getId(), bitmap);
}
});
return bitmaps;
}
});
}
可观察的getImage(最终活动){
返回Observable.create(newobservable.OnSubscribe(){
@凌驾
公共无效呼叫(用户我会这样说:
Observable<Map<Integer, Bitmap>> getImages(List<Activity> activities) {
return Observable.from(activities)
.map(activity -> new Pair(activity.getId(), downloadImage(activity.getImage())))
.toMap(pair -> pair.first, pair -> pair.second);
}
(假设downloadImage
返回一个异步的可观察的
)我会这样做:
Observable<Map<Integer, Bitmap>> getImages(List<Activity> activities) {
return Observable.from(activities)
.map(activity -> new Pair(activity.getId(), downloadImage(activity.getImage())))
.toMap(pair -> pair.first, pair -> pair.second);
}
(假设downloadImage
返回一个异步的可观察的
)我会这样做:
Observable<Map<Integer, Bitmap>> getImages(List<Activity> activities) {
return Observable.from(activities)
.map(activity -> new Pair(activity.getId(), downloadImage(activity.getImage())))
.toMap(pair -> pair.first, pair -> pair.second);
}
(假设downloadImage
返回一个异步的可观察的
)我会这样做:
Observable<Map<Integer, Bitmap>> getImages(List<Activity> activities) {
return Observable.from(activities)
.map(activity -> new Pair(activity.getId(), downloadImage(activity.getImage())))
.toMap(pair -> pair.first, pair -> pair.second);
}
(前提是downloadImage
返回一个异步可观察的)优雅的解决方案,但这似乎仍然不能并行运行。有没有办法用这种结构来解决这个问题?问题是map
预期返回的序列与输入的顺序相同zip
在这里可能会有所帮助。可能是使用zip
减少?另外,请参阅优雅的解决方案,但不是his似乎仍然不能并行运行。有没有办法用这种结构来解决这个问题?问题是,map
预期会返回一个与输入相同顺序的序列,zip
在这里可能会有所帮助。可能用zip
来减少?另外,请参阅优雅的解决方案,但这似乎仍然不适用于run并行。有没有办法解决这个问题?问题是,map
预期会返回与输入相同顺序的序列可能zip
在这里会有所帮助。可能是reduce
和zip
?另外,请参阅优雅的解决方案,但这似乎仍然不能并行运行。是否有使用此结构解决此问题的方法?问题在于,map
预期返回的序列与输入的顺序相同,可能zip
在这里会有所帮助。可能使用zip
来减少。另外,请参阅