Question

我有一个提供数组数组的源代码。所有数组都具有相同数量的元素。如何使用rxjs zip运算符转换源数组，如下所述，以便我可以轻松地进行映射？提前谢谢！

var source = Rx.Observable.from([[1, 2, 3], ["a", "b", "c"], ["do", "re", "mi"]]);
// I would like to transform to:
// [[1, "a", "do"], [2, "b", "re"], [3, "c", "mi"]]

Answer 1

您可以将每个数组转换为Observable，然后压缩：

var myArrays = [
  [1, 2, 3],
  ['a', 'b', 'c'],
  ['do', 're', 'mi']
];

var myObservables = myArrays.map(xs => Rx.Observable.from(xs));
var zipped$ = Rx.Observable.zip(myObservables);

zipped$.subscribe(x => console.log(x));

或者，您可以放弃从Array到Observable的转换，并使用lodash的zip来组合数组：

var zipped = _.zip(...myArrays)

如何为数组数组实现rxjs zip运算符？

1 个答案: