如何做一个深层比较2个对象与lodash?

我有2个不同的嵌套对象，我需要知道它们是否在其中一个嵌套属性中有不同。

var a = {};
var b = {};

a.prop1 = 2;
a.prop2 = { prop3: 2 };

b.prop1 = 2;
b.prop2 = { prop3: 3 };

对象可以更复杂，有更多嵌套的属性。但这是一个很好的例子。我可以选择使用递归函数或lodash的东西…

当前回答

As it was asked, here's a recursive object comparison function. And a bit more. Assuming that primary use of such function is object inspection, I have something to say. Complete deep comparison is a bad idea when some differences are irrelevant. For example, blind deep comparison in TDD assertions makes tests unnecessary brittle. For that reason, I'd like to introduce a much more valuable partial diff. It is a recursive analogue of a previous contribution to this thread. It ignores keys not present in a

var bdiff = (a, b) =>
    _.reduce(a, (res, val, key) =>
        res.concat((_.isPlainObject(val) || _.isArray(val)) && b
            ? bdiff(val, b[key]).map(x => key + '.' + x) 
            : (!b || val != b[key] ? [key] : [])),
        []);

BDiff允许检查期望值，同时容忍其他属性，这正是您想要的自动检查。这允许构建各种高级断言。例如:

var diff = bdiff(expected, actual);
// all expected properties match
console.assert(diff.length == 0, "Objects differ", diff, expected, actual);
// controlled inequality
console.assert(diff.length < 3, "Too many differences", diff, expected, actual);

回到完整的解决方案。使用bdiff构建一个完整的传统diff是很简单的:

function diff(a, b) {
    var u = bdiff(a, b), v = bdiff(b, a);
    return u.filter(x=>!v.includes(x)).map(x=>' < ' + x)
    .concat(u.filter(x=>v.includes(x)).map(x=>' | ' + x))
    .concat(v.filter(x=>!u.includes(x)).map(x=>' > ' + x));
};

在两个复杂对象上运行上述函数将输出类似于下面的内容:

 [
  " < components.0.components.1.components.1.isNew",
  " < components.0.cryptoKey",
  " | components.0.components.2.components.2.components.2.FFT.min",
  " | components.0.components.2.components.2.components.2.FFT.max",
  " > components.0.components.1.components.1.merkleTree",
  " > components.0.components.2.components.2.components.2.merkleTree",
  " > components.0.components.3.FFTResult"
 ]

最后，为了了解值之间的差异，我们可能需要直接eval() diff输出。为此，我们需要一个更丑的bdiff版本，输出语法正确的路径:

// provides syntactically correct output
var bdiff = (a, b) =>
    _.reduce(a, (res, val, key) =>
        res.concat((_.isPlainObject(val) || _.isArray(val)) && b
            ? bdiff(val, b[key]).map(x => 
                key + (key.trim ? '':']') + (x.search(/^\d/)? '.':'[') + x)
            : (!b || val != b[key] ? [key + (key.trim ? '':']')] : [])),
        []);

// now we can eval output of the diff fuction that we left unchanged
diff(a, b).filter(x=>x[1] == '|').map(x=>[x].concat([a, b].map(y=>((z) =>eval('z.' + x.substr(3))).call(this, y)))));

这将输出类似于下面的内容:

[" | components[0].components[2].components[2].components[2].FFT.min", 0, 3]
[" | components[0].components[2].components[2].components[2].FFT.max", 100, 50]

“一族”许可

2018-05-10 17:11:58

其他回答

作为对亚当·博杜赫的回答的补充，这个问题考虑到了性质的差异

const differenceOfKeys = (...objects) =>
  _.difference(...objects.map(obj => Object.keys(obj)));
const differenceObj = (a, b) => 
  _.reduce(a, (result, value, key) => (
    _.isEqual(value, b[key]) ? result : [...result, key]
  ), differenceOfKeys(b, a));

2016-10-01 23:12:09

这是基于@JLavoie，使用lodash

let differences = function (newObj, oldObj) {
      return _.reduce(newObj, function (result, value, key) {
        if (!_.isEqual(value, oldObj[key])) {
          if (_.isArray(value)) {
            result[key] = []
            _.forEach(value, function (innerObjFrom1, index) {
              if (_.isNil(oldObj[key][index])) {
                result[key].push(innerObjFrom1)
              } else {
                let changes = differences(innerObjFrom1, oldObj[key][index])
                if (!_.isEmpty(changes)) {
                  result[key].push(changes)
                }
              }
            })
          } else if (_.isObject(value)) {
            result[key] = differences(value, oldObj[key])
          } else {
            result[key] = value
          }
        }
        return result
      }, {})
    }

https://jsfiddle.net/EmilianoBarboza/0g0sn3b9/8/

2018-04-20 18:11:10

我知道这并不能直接回答OP的问题，但我是通过搜索如何删除lodash被引导到这里的。希望这能帮助到和我处境相似的人。

这要归功于@JohanPersson。我在这个答案的基础上实现了对深度嵌套值的比较，并获得对差异的键引用

getObjectDiff = (obj1, obj2) => { const obj1Props = Object.keys(obj1); const obj2Props = Object.keys(obj2); const keysWithDiffValue = obj1Props.reduce((keysWithDiffValueAccumulator, key) => { const propExistsOnObj2 = obj2.hasOwnProperty(key); const hasNestedValue = obj1[key] instanceof Object && obj2[key] instanceof Object; const keyValuePairBetweenBothObjectsIsEqual = obj1[key] === obj2[key]; if (!propExistsOnObj2) { keysWithDiffValueAccumulator.push(key); } else if (hasNestedValue) { const keyIndex = keysWithDiffValueAccumulator.indexOf(key); if (keyIndex >= 0) { keysWithDiffValueAccumulator.splice(keyIndex, 1); } const nestedDiffs = getObjectDiff(obj1[key], obj2[key]); for (let diff of nestedDiffs) { keysWithDiffValueAccumulator.push(`${key}.${diff}`); } } else if (keyValuePairBetweenBothObjectsIsEqual) { const equalValueKeyIndex = keysWithDiffValueAccumulator.indexOf(key); keysWithDiffValueAccumulator.splice(equalValueKeyIndex, 1); } return keysWithDiffValueAccumulator; }, obj2Props); return keysWithDiffValue; } const obj1 = {a0: {a1: {a2: {a3: 'Im here'}}}}; const obj2 = {a0: {a1: {a2: {a3: 'Not here', b3: 'some'}}}}; console.log('final', getObjectDiff(obj1, obj2));

2021-07-06 22:54:28

基于Adam Boduch的回答，我写了这个函数，它在最深层的意义上比较两个对象，返回具有不同值的路径以及从一个或另一个对象中缺失的路径。

代码的编写并没有考虑到效率，在这方面的改进是非常受欢迎的，但这里是基本形式:

var compare = function (a, b) {

  var result = {
    different: [],
    missing_from_first: [],
    missing_from_second: []
  };

  _.reduce(a, function (result, value, key) {
    if (b.hasOwnProperty(key)) {
      if (_.isEqual(value, b[key])) {
        return result;
      } else {
        if (typeof (a[key]) != typeof ({}) || typeof (b[key]) != typeof ({})) {
          //dead end.
          result.different.push(key);
          return result;
        } else {
          var deeper = compare(a[key], b[key]);
          result.different = result.different.concat(_.map(deeper.different, (sub_path) => {
            return key + "." + sub_path;
          }));

          result.missing_from_second = result.missing_from_second.concat(_.map(deeper.missing_from_second, (sub_path) => {
            return key + "." + sub_path;
          }));

          result.missing_from_first = result.missing_from_first.concat(_.map(deeper.missing_from_first, (sub_path) => {
            return key + "." + sub_path;
          }));
          return result;
        }
      }
    } else {
      result.missing_from_second.push(key);
      return result;
    }
  }, result);

  _.reduce(b, function (result, value, key) {
    if (a.hasOwnProperty(key)) {
      return result;
    } else {
      result.missing_from_first.push(key);
      return result;
    }
  }, result);

  return result;
}

您可以使用以下代码段(建议以全页模式运行)尝试代码:

2017-01-02 18:28:26

var bdiff = (a, b) =>
    _.reduce(a, (res, val, key) =>
        res.concat((_.isPlainObject(val) || _.isArray(val)) && b
            ? bdiff(val, b[key]).map(x => key + '.' + x) 
            : (!b || val != b[key] ? [key] : [])),
        []);