Ramda：在另一个数组中查找一个数组的最低时机

Question

我有两个包含字符串的数组。第二个数组只能包含第一个中的字符串，但是可以多次。我正在尝试在第二个数组中找到出现次数最少的单词。这里有一些例子：

const original = ['foo', 'bar', 'baz'];

const arr1 = ['foo', 'bar']; // => ['baz'] because 'baz' is 0 times in arr1.
const arr2 = ['foo', 'foo', 'bar']; // => ['baz'] because 'baz' is 0 times in arr2.
const arr3 = ['foo', 'foo', 'bar', 'bar', 'baz']; // => ['baz'] because 'baz' in 1 time in arr3.
const arr4 = ['foo', 'bar', 'baz']; // => ['foo', 'bar', 'baz']; because they are all the lowest (1 time).
const arr5 = ['foo', 'foo', 'bar', 'bar', 'bar', 'baz', 'baz']; // => ['foo', 'baz'] because they are both only 2 times in arr5

您如何使用Ramda做到这一点？ （或在JS中通常）？我觉得可以用R.countBy来解决，但是我不知道如何获取所有具有最低值的键。

Answer 1

您可以使用R.countBy，但是您必须创建一个默认对象，其中originals中的每个项目的值为0，然后合并它们。然后，找到最小值，并过滤original数组：

const { chain, flip, zipObj, map, always, curry, pipe, countBy, identity, mergeRight, values } = R;

const getDefaults = chain(flip(zipObj), map(always(0)));

const countWithDefaults = curry((dflts, arr) => pipe(
  countBy(identity),
  mergeRight(dflts),
)(arr));

const fn = curry((orig, arr) => {
  const counts = countWithDefaults(getDefaults(orig), arr);
  
  const min = Math.min(...values(counts));

  return original.filter(k => counts[k] === min);
});

const original = ['foo', 'bar', 'baz'];

const fnO = fn(original);

console.log(fnO(['foo', 'bar'])); // => ['baz'] because 'baz' is 0 times in arr1.
console.log(fnO(['foo', 'foo', 'bar'])); // => ['baz'] because 'baz' is 0 times in arr2.
console.log(fnO(['foo', 'foo', 'bar', 'bar', 'baz'])); // => ['baz'] because 'baz' in 1 time in arr3.
console.log(fnO(['foo', 'bar', 'baz'])); // => ['foo', 'bar', 'baz']; because they are all the lowest (1 time).
console.log(fnO(['foo', 'foo', 'bar', 'bar', 'bar', 'baz', 'baz'])); // => ['foo', 'baz'] because they are both only 2 times in arr5

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