对于嵌套数组中不只有一个对象的情况：

const arrayOfArrays = [
  [{val1:1, val2:2, val3:3}, {val1:1, val2:2, val3:3, val4:4}, {val1:1, val2:2, val3:3, val4:4}],
  [{val1:1, val2:2, val3:3}, {val1:1, val2:2, val3:3, val4:4}],
  [{val4:4, val3:3, val2:2, val1:1}],
  [{val2:2, val1:1, val3:3}, {val3:3, val2:2, val1:1, val4:4}]
]

const arrayOfObjects = [].concat.apply([], arrayOfArrays);

console.log('Flattened:', arrayOfObjects);

const normalizeObject = obj => _(obj).toPairs().sortBy(0).fromPairs().value();

const removeDuplicates = function(array) {
  return [
    ...new Set(
      array.map(item => JSON.stringify(normalizeObject(item)))
    )
  ].map(item => JSON.parse(item));
};

const arrayOfUniqueObjects = removeDuplicates(arrayOfObjects);

console.log('Unique:', arrayOfUniqueObjects);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.10/lodash.min.js"></script>

奖励：支票 removeDuplicates 我的答案中的方法（；
额外：让我们用两种方法扩展数组的原型

const normalizeObject = obj => _(obj).toPairs().sortBy(0).fromPairs().value();

Array.prototype.flattenNested = function() {
  return [].concat.apply([], this);
}

Array.prototype.removeDuplicateObjects = function() {
  return [
    ...new Set(
      this.map(item => JSON.stringify(normalizeObject(item)))
    )
  ].map(item => JSON.parse(item));
}

const arrayOfArrays = [
  [{val1:1, val2:2, val3:3}, {val1:1, val2:2, val3:3, val4:4}, {val1:1, val2:2, val3:3, val4:4}],
  [{val1:1, val2:2, val3:3}, {val1:1, val2:2, val3:3, val4:4}],
  [{val1:1, val2:2, val3:3, val4:4}, {val3:3, val1:1, val2:2}, {val3:3, val2:2, val1:1, val4:4}]
]

const arrayOfObjects = arrayOfArrays.flattenNested();

console.log('Flattened:', arrayOfObjects);

const arrayOfUniqueObjects = arrayOfObjects.removeDuplicateObjects();

console.log('Unique:', arrayOfUniqueObjects);

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使用 [].concat.apply([], array) ```
var arr = [[{val1:1,val2:2,val3:3}],[{val1:1,val2:2,val3:3}],[{val1:1,val2:2,val3:3}]]

console.log([].concat.apply([], arr))