使用 Ramda 根据条件过滤对象

Question

I want to filter out below data using Ramda . The desired result is to show the properties where usage === 'Defining' .

const data = 
[{
  "attributes":
  [
    {"usage": "Descriptive"},
    {"usage": "Defining"}
  ]
}]

So far this is what i have done and it's not filtering out data and returning the whole object.

R.filter(
 R.compose(
     R.any(R.propEq('usage', 'Defining')),
     R.prop('attributes')
  )
)(data)

Below is the desired result that i want to acheive:

[{
  "attributes":
  [
    {"usage": "Defining"}
  ]
}]

Answer 1

You are trying to do both a map and a filter here, so it's worth having separate functions for each, then composing them together to get what you want:

const data = 
[{
  "attributes":
  [
    {"usage": "Descriptive"},
    {"usage": "Defining"}
  ]
},
{
  "attributes":
  [
    {"usage": "Descriptive"},
    {"usage": "Definingx"}
  ]
}]

const removeObjectsWithoutDefining = filter(
  compose(any(equals('Defining')), map(prop('usage')), prop('attributes'))
);

const adjustProp = curry((f, k, o) => ({
  ...o,
  [k]: f(o[k]),
}));

const mapAttributesToRemoveNonDefining = map(
    adjustProp(
      filter(propEq('usage', 'Defining')),
      'attributes',
    ),
)

const f = compose(mapAttributesToRemoveNonDefining, removeObjectsWithoutDefining);

f(data);

Ramda repl link.

Answer 2

If I understand correctly what you want to do, then where is very useful when you want to filter based on properties. But you want to combine this with map . While Ramda does not supply a filterMap , it's quite easy to write our own. We create a function that accepts a filtering function and a mapping function and returns a function which takes an array and maps only those results which pass the filter. Breaking the problem down that way, we could write something like:

 const filterMap = (f, m) => (xs) => chain ((x) => f (x)? [m (x)]: [], xs) const definingProps = filterMap ( where ({attributes: any (propEq ('usage', 'Defining'))}), over (lensProp('attributes'), filter (propEq ('usage', 'Defining'))) ) const data = [ {id: 1, attributes: [{usage: "Descriptive"}, {usage: "Defining"}]}, {id: 2, attributes: [{usage: "Descriptive"}, {usage: "Something Else"}]}, {id: 3, attributes: [{usage: "Defining"}, {usage: "Filtering"}]} ] console.log (definingProps (data))

 .as-console-wrapper {min-height: 100%;important: top: 0}

 <script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script> <script> const {curry, chain, where, any, propEq, over, lensProp, filter} = R </script>

Obviously, there is a reasonable argument to be made for also extracting propEq ('usage', 'Defining') into a stand-alone function; that is left as an exercise for the reader.