Ramda:删除具有动态键的空对象

Question

我有这个对象，其中addon_sizes键是动态的，即"1","2", "3", "4":

const sizes = {
"addon_sizes": {
"1": ["a", "b"],
"2": ["c"],
"3": null,
"4": []
}
}

我需要删除该对象中值为null/undefined/空数组的所有键/值对。

键"3"one_answers"4";应该从列表中删除

到目前为止，我所尝试的是:

const newObj = R.reject(R.anyPass([R.isEmpty, R.isNil]))(sizes.addon_sizes);

但这不会删除null或空值。

Answer 1

通过选择所有非空或nil的属性创建一个新对象:

const { pickBy, complement, anyPass, isEmpty, isNil } = R
const fn = pickBy(complement(anyPass([isEmpty, isNil])))
const sizes = {"addon_sizes":{"1":["a","b"],"2":["c"],"3":null,"4":[]}}
const result = fn(sizes.addon_sizes)
console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.28.0/ramda.min.js" integrity="sha512-t0vPcE8ynwIFovsylwUuLPIbdhDj6fav2prN9fEu/VYBupsmrmk9x43Hvnt+Mgn2h5YPSJOk7PMo9zIeGedD1A==" crossorigin="anonymous" referrerpolicy="no-referrer"></script>

Answer 2

既然你已经标记了javascript:

您可以遍历对象条目，并使用delete关键字删除负或负长度属性。

const sizes = {
"addon_sizes": {
"1": ["a", "b"],
"2": ["c"],
"3": null,
"4": []
}
};
Object.entries(sizes.addon_sizes).forEach(([ key, value ]) => {
if(!value || !value.length)
delete sizes.addon_sizes[key];
});
console.log(sizes);

Answer 3

这取决于您对remove的定义。如果你想改变你原来的目标，拉姆达是不会帮你的。如果您想要创建一个没有这些属性的副本，那么您的代码似乎已经可以工作了，尽管它仍然需要应用于您的输入对象。这里我们用over和lensProp

const cleanAddons = over (
lensProp ('addon_sizes'), 
reject (anyPass ([isNil, isEmpty]))
)
const sizes = { addon_sizes: {1: ["a", "b"], 2: ["c"], 3: null, 4: []}}
console .log (cleanAddons (sizes))

.as-console-wrapper {max-height: 100% !important; top: 0}

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.28.0/ramda.min.js"></script>
<script> const {over, lensProp, reject, anyPass, isNil, isEmpty} = R     </script>

lensProp ('addon_sizes')创建一个lens，聚焦于addon_sizes属性，over使用该镜头和基于reject的函数创建一个函数，该函数将创建原始对象的副本，并从addon_sizes中删除空值或nil值。

当只使用两个函数时，我发现either (f, g)比anyPass ([f, g])更具语义性，所以我更喜欢这个版本，但它做同样的事情:

const cleanAddons = over (
lensProp ('addon_sizes'), 
reject (either (isNil, isEmpty))
)