我需要比较两个对象数组(源/目标)的内容。

如果目标中不存在源中的项目,则应仅添加要包括的项目。

如果目标中确实存在源中的项目,则应仅添加要排除的项目。

当前使用ramda R.differenceWith(),但是当目标为空时出现问题。

我想知道differenceWith是否适合这里的海豚,或者我可以使用其他功能。请给我一个例子,谢谢!

注意:即使不使用ramda,也可以得到答案。

Demo

// source
  const pathHash1 = {
    hash: "c4ca4238a0b923820dcc509a6f75849b",
    path: "./source/file1.txt"
  };
  const pathHash2 = {
    hash: "c81e728d9d4c2f636f067f89cc14862c",
    path: "./source/file2.txt"
  };
  const pathHash3 = {
    hash: "eccbc87e4b5ce2fe28308fd9f2a7baf3",
    path: "./souce/file3.txt"
  };
  // target
  const pathHash4 = {
    hash: "eccbc87e4b5ce2fe28308fd9f2a7baf3",
    path: "./target/file3.txt"
  };

// works
// const source = [pathHash1, pathHash2, pathHash3]
// const target = [pathHash4]

// does no work
const source = [pathHash1, pathHash2, pathHash3]
const target = []

// result pathHash1, pathHash2
const resultInclude = R.differenceWith((x,y)=> x.hash === y.hash, source, target)
const resultExclude= R.differenceWith((x,y)=> x.hash !== y.hash, source, target)

console.log('include',resultInclude.map(x=>x.hash))
console.log('exclude',resultExclude.map(x=>x.hash))

最佳答案

对于排除情况,请使用R.innerJoin



const getInclude = R.differenceWith(R.eqProps('hash'))
const getExclude = R.innerJoin(R.eqProps('hash'))

const pathHash1 = {hash: "c4ca4238a0b923820dcc509a6f75849b",path: "./source/file1.txt"},pathHash2 = {hash: "c81e728d9d4c2f636f067f89cc14862c",path: "./source/file2.txt"},pathHash3 = {hash: "eccbc87e4b5ce2fe28308fd9f2a7baf3",path: "./souce/file3.txt"},pathHash4 = {hash: "eccbc87e4b5ce2fe28308fd9f2a7baf3",path: "./target/file3.txt"}

const source1 = [pathHash1, pathHash2, pathHash3]
const target1 = [pathHash4]

const source2 = [pathHash1, pathHash2, pathHash3]
const target2 = []

const getHash = R.map(R.prop('hash'))

console.log('include1', getHash(getInclude(source1, target1)))
console.log('exclude1', getHash(getExclude(source1, target1)))

console.log('include2', getHash(getInclude(source2, target2)))
console.log('exclude2', getHash(getExclude(source2, target2)))

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08-07 10:53