如何使用 Ramda 从对象中过滤出特定键？

Question

http://ramdajs.com/0.21.0/docs/#prop

Ramda Repl

var myObject = {a: 1, b: 2, c: 3, d: 4};

var newObject = R.filter(R.props('a'), myObject);
//var newObject = R.filter(R.equals(R.props('a')), myObject);

console.log('newObject', newObject);

Right now the code above is returning the entire object:

newObject {"a":1,"b":2,"c":3,"d":4}

What I would like to do is just return a new object with just the 'a' key. Or a new object with the a and b keys.

Answer 1

Use pick :

let newObj = R.pick(['a'], oldObj);

If your filtering criteria is more complex than just existence, you can use pickBy to select via arbitrary predicate functions.

Answer 2

The answer from Jared Smith is great. I just wanted to add a note on why your code did not work. You tried

R.filter(R.props('a'), {a: 1, b: 2, c: 3, d: 4});

First of all, you pointed to the documentation for prop , but used props . These are different, but related, functions. prop looks like

// prop :: k -> {k: v} -> v
prop('c', {a: 1, b: 2, c: 3, d: 4}); //=> 3

(there is some added complexity regarding undefined .)

props on the other hand takes multiple values

// props :: [k] -> {k: v} -> [v]
props(['a', 'd'], {a: 1, b: 2, c: 3, d: 4}); //=> [1, 4]

But neither of these is going to be useful input to filter , which for these purposes we can think of as

// filter :: (k -> Bool) -> {k: v} -> {k: v}

The first parameter to filter is a function from a (string) key to a boolean; it works with Javascript's idea that everything is truth-y except for a few specific values. It will be called with each key in turn. So for example, when deciding whether to include {c: 3} , it calls props('a')('c') , which for another slightly odd reason ^* , actually works, returning [3] , which is treated as truth-y, and the filter function will include {c: 3} in its output. So too every key will be included.

---

^* The reason props('a', obj) works when it really should be props(['a'], obj) is that in JS, strings are close enough to lists, having a length property and indexed values. 'a'.length; ==> 1 'a'.length; ==> 1 , 'a'[0]; //=> 'a' 'a'[0]; //=> 'a' . Hence props can treat single-character strings as though they were one-element lists of character strings. But it can be a bit bizarre, too: R.props('cabby', {a: 1, b: 2, c: 3, d: 4}); //=> [3, 1, 2, 2, undefined] R.props('cabby', {a: 1, b: 2, c: 3, d: 4}); //=> [3, 1, 2, 2, undefined] .

Answer 3

pickBy

 const getOnlyGoods = R.pickBy((_, key) => key.includes('good')); const onlyGoods = getOnlyGoods({ "very good": 90, "good": 80, "good enough": 60, "average": 50, "bad": 30, "😡": 10, }); console.log(onlyGoods);

 <script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>