拉姆达滤镜零道具

Question

我有事件对象数组，看起来像：

{
    date: "2015-06-03T19:29:01.000Z",
    description: "Test",
    talks: [{
        author: "Nick",
        tags: ["tag1", "tag2", "tag3"]
    }]
} 

我只想从该对象获取标签，所以我使用Ramda像这样：

let eventTags = pipe(prop('talks'), map(prop('tags')), flatten, uniq)
... 
eventTags(event); //and call function on event object

但是在某些情况下，事件对象如下所示：

{
    date: "2015-06-03T19:29:01.000Z",
    description: "Test",
    talks: [{
        author: "Nick",
        tags: null
    }]
} 

所以我在eventTags数组中得到了[null] ，但我想改为得到一个空数组。 那么如何过滤null？

Answer 1

我会提倡一种解决方案，该解决方案可以使用镜头访问tags并使用Ramda和Sanctuary将undefined视为Maybe Nothing

const x = [{
  date: "2015-06-03T19:29:01.000Z",
  description: "Test",
  talks: [{
    author: "Nick",
    tags: ["tag1", "tag2", "tag3"]
  }]
}, {
  date: "2015-06-03T19:29:01.000Z",
  description: "Test",
  talks: [{
    author: "Nick",
    tags: null
  }]
}]

const viewTalks = S.compose ( S.toMaybe ) ( 
  R.view ( R.lensProp( 'talks' ) ) 
)

const viewTags = S.compose ( S.toMaybe ) ( 
  R.view ( R.lensProp ( 'tags' ) ) 
)

const allTalkTags = S.map ( S.pipe ( [
  S.map ( viewTags ),
  S.justs,
  R.unnest
] ) )

const allTalksTags = S.pipe( [
  S.map ( S.pipe( [
    viewTalks,
    allTalkTags
  ] ) ),
  S.justs,
  R.unnest,
  R.uniq
] )

// outputs: ['tag1', 'tag2', 'tag3']
allTalksTags ( x )

单击以运行工作示例

Answer 2

您可以在此处利用R.defaultTo([])创建一个函数，如果接收到null或未定义的值，则该函数将返回一个空数组，否则将值传递给未修改的对象。

const eventTags = pipe(
  prop('talks'),
  map(pipe(prop('tags'), defaultTo([]))),
  flatten,
  uniq
)

Answer 3

在MatíasFidemraizer的帮助下，我将eventTags函数更改为：

const viewTags = talk => !!talk.tags ? talk.tags : [];

export let eventTags = pipe(prop('talks'), map(viewTags), flatten, uniq)