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

我尝试了Adam Boduch的代码来输出一个很深的差异-这完全没有经过测试，但碎片在那里:

function diff (obj1, obj2, path) {
    obj1 = obj1 || {};
    obj2 = obj2 || {};

    return _.reduce(obj1, function(result, value, key) {
        var p = path ? path + '.' + key : key;
        if (_.isObject(value)) {
            var d = diff(value, obj2[key], p);
            return d.length ? result.concat(d) : result;
        }
        return _.isEqual(value, obj2[key]) ? result : result.concat(p);
    }, []);
}

diff({ foo: 'lol', bar: { baz: true }}, {}) // returns ["foo", "bar.baz"]

2017-03-06 01:56:40

其他回答

这是一个简单的带有Lodash深度差异检查器的Typescript，它将生成一个新对象，只包含旧对象和新对象之间的差异。

例如，如果我们有:

const oldData = {a: 1, b: 2};
const newData = {a: 1, b: 3};

结果对象将是:

const result: {b: 3};

它还兼容多层深层对象，对于数组，它可能需要一些调整。

import * as _ from "lodash";

export const objectDeepDiff = (data: object | any, oldData: object | any) => {
  const record: any = {};
  Object.keys(data).forEach((key: string) => {
    // Checks that isn't an object and isn't equal
    if (!(typeof data[key] === "object" && _.isEqual(data[key], oldData[key]))) {
      record[key] = data[key];
    }
    // If is an object, and the object isn't equal
    if ((typeof data[key] === "object" && !_.isEqual(data[key], oldData[key]))) {
      record[key] = objectDeepDiff(data[key], oldData[key]);
    }
  });
  return record;
};

2019-03-27 21:10:01

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

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

我需要知道它们的某个嵌套属性是否不同

其他答案为这个问题提供了可能令人满意的解决方案，但它非常困难和常见，似乎有一个非常受欢迎的包来帮助解决这个问题。

要使用这个包，你需要npm i deep-object-diff，然后:

const { diff } = require('deep-object-diff');
var a = {};
var b = {};

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

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

if (!_.isEqual(a, b)) {
  const abDiff = diff(a, b);
  console.log(abDiff);
  /*
  {
    prop2: {
      prop3: 3
    }
  }
  */
}

// or alternatively
const abDiff = diff(a, b);
if(!_.isEmpty(abDiff)) {
  // if a diff exists then they aren't deeply equal
  // perform needed actions with diff...
}

下面是一个更详细的案例，直接从他们的文档中删除属性:

const lhs = {
  foo: {
    bar: {
      a: ['a', 'b'],
      b: 2,
      c: ['x', 'y'],
      e: 100 // deleted
    }
  },
  buzz: 'world'
};

const rhs = {
  foo: {
    bar: {
      a: ['a'], // index 1 ('b')  deleted
      b: 2, // unchanged
      c: ['x', 'y', 'z'], // 'z' added
      d: 'Hello, world!' // added
    }
  },
  buzz: 'fizz' // updated
};

console.log(diff(lhs, rhs)); // =>
/*
{
  foo: {
    bar: {
      a: {
        '1': undefined
      },
      c: {
        '2': 'z'
      },
      d: 'Hello, world!',
      e: undefined
    }
  },
  buzz: 'fizz'
}
*/

有关实现细节和其他使用信息，请参阅该repo。

2020-08-04 17:52:52

要递归地显示一个对象与其他对象的不同之处，可以使用_。Reduce与_结合。isEqual和_.isPlainObject。在这种情况下，你可以比较a与b的不同，或者b与a的不同:

2018-01-10 06:00:27

如何做一个深层比较2个对象与lodash?

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