删除嵌套数组中具有键值的对象
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【中文标题】删除嵌套数组中具有键值的对象【英文标题】：delete an object in a nested array which has key value 【发布时间】：2019-03-29 15:01:15 【问题描述】：我想删除每个部分下唯一的低级对象（例如在下面的代码中，在个人数据下有两个对象...我想删除一个对象，其中操作：旧）在每个部分下，其中“操作”： “旧”
我在我的项目中使用 lodash
[
"clientDetails":
"personalData": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientAddress":
"primaryAddress": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"secondaryAddress": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientDemise":
"deathDetails": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientMarital":
"divorceInformation": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"marraigeInformation": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
]
抱歉发错了，这是我第一次发帖
【问题讨论】：
这个的o/p应该是什么？ 【参考方案1】：如果您的数据结构相当一致（即类似于您在问题中包含的内容），您可以执行以下操作：
const mapObj = (f, obj) =>
return Object.keys(obj).reduce((acc, key) =>
acc[key] = f(obj[key], key)
return acc
, )
const filterData = data =>
// the data itself is an array, so iterate over each item in the array
return data.map(x1 =>
// x1 is an object, so need to iterate over each item in the object
return mapObj(x2 =>
// x2 is an object, so need to iterate over each item in the object
return mapObj(x3 =>
// x3 is an array of objects. each item in the array has an action key which could equal "NEW" or "OLD". get rido of the items with action === "OLD"
return x3.filter(x4 => x4.action !== "OLD")
, x2)
, x1)
)
const data = [
"clientDetails":
"personalData": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientAddress":
"primaryAddress": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"secondaryAddress": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientDemise":
"deathDetails": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientMarital":
"divorceInformation": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"marraigeInformation": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
]
const result = filterData(data)
console.log(result)如果您想要一个更通用的解决方案，该解决方案可以获取任何结构的数据，并且只需使用等于“OLD”的操作删除所有对象：
const reduceObj = (f, initial, obj) =>
return Object.keys(obj).reduce((acc, key) =>
return f(acc, obj[key], key)
, initial)
const isObject = x => x !== null && typeof x === 'object'
const removeAllOld = data =>
if(Array.isArray(data))
return data.reduce((acc, value) =>
// don't include the item if it has a key named 'action' that is equal to 'OLD'
if(value.action && value.action === 'OLD') return acc
acc.push(removeAllOld(value))
return acc
, [])
else if(isObject(data))
return reduceObj((acc, value, key) =>
// don't include the item if it has a key named 'action' that is equal to 'OLD'
if(value.action && value.action === 'OLD') return acc
acc[key] = removeAllOld(value)
return acc
, , data)
else
return data
const data = [
"clientDetails":
"personalData": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientAddress":
"primaryAddress": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"secondaryAddress": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientDemise":
"deathDetails": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientMarital":
"divorceInformation": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"marraigeInformation": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
]
console.log(removeAllOld(data))【讨论】：
【参考方案2】：您可以使用 javascript 过滤器。不使用 lodash 来减少你的 bundle 大小。
// it's upto you, you can use new Array() as well and insert if(ktm.action==='NEW')
clients = clients.filter(function(itm)
Object.keys(itm).forEach(function(Okey, Ovalue)
Object.keys(itm[Okey]).forEach(function(inkey, invalue)
itm[Okey][inkey].filter(function(ktm)
if (ktm.action === 'OLD')
// perform your logic, either you can insert into new Array() or
// delete that object and return clients
);
);
);
);
【讨论】：
【参考方案3】：考虑到这一点，只需几行就可以实现这一点
input = your input
这种和平的代码将完成工作
for (var i of input)
for (var j in i)
var ob = i[j];
for (var k in ob)
var index = _.findIndex(ob[k], 'action': 'OLD');
if (index > -1)
ob[k].splice(index, 1);
【讨论】：
这是一个不错的简短解决方案。您可以删除toRemove 变量，顺便说一句。此外，您可以将其转换为代码 sn-p，并将 lodash 作为外部库包含在内。
非常感谢您的建议，我删除了不必要的变量，如果需要，将创建代码 sn-p。
但这不能用于删除不需要的数据的多个条目（如果需要）
你的代码可以工作...谢谢.. 但是如果没有主地址@ShyamTayal，它会抛出错误 ob[k].splice is not a function 例如在上面的代码中
我认为最后一级字段是一个数组，如果不是这样，请在第三个循环中添加类型检查以避免任何错误。希望对你有帮助【参考方案4】：
你可以像这样做一个深拷贝：
const array = [
"clientDetails":
"personalData": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientAddress":
"primaryAddress": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"secondaryAddress": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientDemise":
"deathDetails": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
,
"clientMarital":
"divorceInformation": [
"action": "OLD",
"id": "12345"
,
"action": "NEW",
"id": "12445"
],
"marraigeInformation": [
"action": "NEW",
"id": "12345"
,
"action": "OLD",
"id": "12445"
]
]
function removeOldAction(a)
if (a instanceof Array)
let copiee = [];
for (let item in a)
const propValue = removeOldAction(a[item]);
if(propValue)
copiee.push(propValue);
return copiee;
if (a instanceof Object)
if (a['action'] === 'OLD')
return;
let copiee = ;
for (let key in a)
copiee[key] = removeOldAction(a[key]);
return copiee;
return a;
console.log(removeOldAction(array));
【讨论】：
【参考方案5】：你可以通过类似这样的方式来实现这一点，而无需 lodash：
var data = [ "clientDetails": "personalData": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientAddress": "primaryAddress": [ "action": "OLD", "id": "12345" , "action": "NEW", "id": "12445" ], "secondaryAddress": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientDemise": "deathDetails": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientMarital": "divorceInformation": [ "action": "OLD", "id": "12345" , "action": "NEW", "id": "12445" ], "marraigeInformation": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] ]
const removeOld = (data) => data.map(x =>
Object.entries(x).reduce((r, [k,v]) =>
r[k] = Object.entries(v).map(([o,p]) =>
([o]: p.filter(n => n.action != 'OLD')))
return r
,))
console.log(removeOld(data))<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>使用map、Object.entries、reduce 和filter。
另一种方法是在 ES6 中使用类似于 @Vanojx1 方法的递归：
var data = [ "clientDetails": "personalData": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientAddress": "primaryAddress": [ "action": "OLD", "id": "12345" , "action": "NEW", "id": "12445" ], "secondaryAddress": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientDemise": "deathDetails": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] , "clientMarital": "divorceInformation": [ "action": "OLD", "id": "12345" , "action": "NEW", "id": "12445" ], "marraigeInformation": [ "action": "NEW", "id": "12345" , "action": "OLD", "id": "12445" ] ]
const removeOld = (data) =>
Array.isArray(data) ? data.filter(x => x.action != 'OLD').map(x => removeOld(x)) :
typeof(data) == 'object' ? Object.entries(data).reduce((r, [k,v]) => (r[k] = removeOld(v), r), ) :
data
console.log(removeOld(data))【讨论】：
【参考方案6】：结构独立的解决方案检查每个对象节点中的操作
var data=[clientDetails:personalData:[action:"NEW",id:"12345",action:"OLD",id:"12445"],clientAddress:primaryAddress:[action:"OLD",id:"12345",action:"NEW",id:"12445"],secondaryAddress:[action:"NEW",id:"12345",action:"OLD",id:"12445"],clientDemise:deathDetails:[action:"NEW",id:"12345",action:"OLD",id:"12445"],clientMarital:divorceInformation:[action:"OLD",id:"12345",action:"NEW",id:"12445"],marraigeInformation:[action:"NEW",id:"12345",action:"OLD",id:"12445"]];
const reducer = (curr) =>
if(_.isArray(curr))
return _(curr)
.filter(el => !('action' in el && el.action == 'OLD'))
.map(el => reducer(el))
.value()
else if(_.isObject(curr))
return _(curr)
.mapValues(el => reducer(el))
.value()
else
return curr;
;
console.log(reducer(data));<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>【讨论】：
【参考方案7】：让我们不要改变原始输入数据，使用定制器克隆它，并拒绝定制器中不需要的东西（如果它们存在），以便按预期获得更干净的克隆输出。你可以使用lodash#cloneDeepWith
_.cloneDeepWith(input, v => _.find(v, action: "OLD") ? _.reject(v, action: "OLD") : undefined);
这只是对您想要拒绝的内容进行（硬编码）的示例。但是您可以将其包装在回调中并将拒绝标准作为参数以使其动态化。
所以我们开始：
let input = ["clientDetails":"personalData":["action":"NEW","id":"12345","action":"OLD","id":"12445"],"clientAddress":"primaryAddress":["action":"OLD","id":"12345","action":"NEW","id":"12445"],"secondaryAddress":["action":"NEW","id":"12345","action":"OLD","id":"12445"],"clientDemise":"deathDetails":["action":"NEW","id":"12345","action":"OLD","id":"12445"],"clientMarital":"divorceInformation":["action":"OLD","id":"12345","action":"NEW","id":"12445"],"marraigeInformation":["action":"NEW","id":"12345","action":"OLD","id":"12445"]],
clear = (input, rej) => (
_.cloneDeepWith(input, v => _.find(v, rej) ? _.reject(v, rej) : undefined)
),
res;
res = clear(input, action: "OLD"); //you can filter out action: OLD
console.log(res);
res = clear(input, action: "NEW"); //you can filter out action: NEW
console.log(res);
res = clear(input, d => d.action==="OLD"); //you can filter with custom callback with complex logic
console.log(res);<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>【讨论】：
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