Comparing 2 nested data-structures,target+source,what are appropriate merge-strategies for missing target values compared to their source counterpart?

Question

What is a better way of doing this. I'am assigning either of two property values (from two different objects), depending on their existence, to a third data-structure.

In case the args object's value is nullish a non nullish value gets accessed from the default object and assigned to the final structure.

return {
  first: {
    visible: args.first?.visible ?? defaulttest.first?.visible,
    emoji: args.first?.emoji ?? defaulttest.first?.emoji,
    style: args.first?.style ?? defaulttest.first?.style,
  },
  back: {
    visible: args.back?.visible ?? defaulttest.back?.visible,
    emoji: args.back?.emoji ?? defaulttest.back?.emoji,
    style: args.back?.style ?? defaulttest.back?.style,
  },
  page: {
    visible: args.page?.visible ?? defaulttest.page?.visible,
    emoji: args.page?.emoji ?? defaulttest.page?.emoji,
    style: args.page?.style ?? defaulttest.page?.style,
  },
  forward: {
    visible: args.forward?.visible ?? defaulttest.forward?.visible,
    emoji: args.forward?.emoji ?? defaulttest.forward?.emoji,
    style: args.forward?.style ?? defaulttest.forward?.style,
  },

  last: {
    visible: args.last?.visible ?? defaulttest.last?.visible,
    emoji: args.last?.emoji ?? defaulttest.last?.emoji,
    style: args.last?.style ?? defaulttest.last?.style,
  },
  Mdelete: {
    visible: args.Mdelete?.visible ?? defaulttest.Mdelete?.visible,
    emoji: args.Mdelete?.emoji ?? defaulttest.Mdelete?.emoji,
    style: args.Mdelete?.style ?? defaulttest.Mdelete?.style,
  },
  removeBtn: {
    visible: args.removeBtn?.visible ?? defaulttest.removeBtn?.visible,
    emoji: args.removeBtn?.emoji ?? defaulttest.removeBtn?.emoji,
    style: args.removeBtn?.style ?? defaulttest.removeBtn?.style,
  },
};

Answer 1

If the structure of your object is the one you presented you can do:

function normalize(input, defaultValue) {
    // Loop on the outer keys
    Object.keys(input).forEach(mainKey => {
        // Loop on the inner keys
        Object.keys(input[mainKey]).forEach(key => {
            // set the value of the key as itself or default if null
            input[mainKey][key] = input[mainKey]?.[key] ?? defaultValue[mainKey]?.[key]
        })
    })
    return input;
}

Calling normalize(args, defaulttest) you will loop on each inner key, check if it exist and if it does not exist you substitute it with the default in the same path.

Example:

const x = {
  a: {a1: '1', a2: '2'},
  b: {b1: '1', b2: null}
}

const y = {b: {b2: '5'}}

console.log(normalize(x,y))

Output:

{
    "a": {
        "a1": "1",
        "a2": "2"
    },
    "b": {
        "b1": "1",
        "b2": "5"
    }
}

With this approach you must have the key in the args input. If the key is missing, it will not be substituted with the default. To make it work even with not-present keys you need to use a third structure with all the possible path for example.