[英]Based on one array object keys and other array object values create one more new array?
I have two arrays like this 我有两个这样的数组
arr1 = [
{
'name':'Victoria Cantrell',
'position':'Integer Corporation',
'office':'Croatia',
'ext':'0839',
'startDate':'2015-08-19',
'salary':208.178
},
{
'name':'Pearleeee',
'position':'In PC',
'office':'Cambodia',
'ext':'8262',
'startDate':'2014-10-08',
'salary':114.367
},
{
'name':'Pearl Crosby',
'position':'Integer',
'office':'Cambodia',
'ext':'8162',
'startDate':'2014-10-08',
'salary':114.367
}
]
arr2 =
[{
'name': 'name',
'checkfilter': false
},
{
'name': 'position',
'checkfilter': true
},
{
'name': 'office',
'checkfilter': true
},
{
'name': 'startDate',
'checkfilter': false
},
{
'name': 'ext',
'checkfilter': false
},
{
'name': 'salary',
'checkfilter': false
}]
based on checkfilter== true i want produce third array like this 基于checkfilter == true我想产生这样的第三个数组
arr3 = ` arr3 =`
[{
name: 'position',
values: [{
checkName: 'Integer Corporation'
},
{
checkName: 'In PC'
},
{
checkName: 'Integer'
}]
},
{
name:'office',
values: [{
checkName: 'Croatia'
},
{
checkName: 'Cambodia'
}]
}
]
` I tried to solve this scenario like this, but its not working perfect `我试图解决这种情况,但效果不理想
arr3=[]
this.arr2.forEach((column: any) => {
if (column.ischeckFilter === true) {
this.arr3.push({
name: column.name,
values: []
});
}
});
this.arr1.forEach((d: any) => {
this.arr2.forEach((column: any) => {
if (column.ischeckFilter === true) {
this.arr3.forEach((c: any) => {
// console.log(d[column.name], c.name, 'JJJJ');
// console.log(Object.keys(d), 'BBBBBBBBB');
let keys = Object.keys(d);
keys.forEach((k: any) => {
if (k === c.name) {
if (find( c.values, { 'checkName': d[column.name]}) === undefined) {
c.values.push({
checkName: d[column.name] ,
ischeck: false
});
}
}
});
});
}
});
});
console.log( this.arr3)
}
the output array values should not contain any duplicate, I used for each loops, is there any best practices to solve this scenario like decresing the loops, because above arrays length is somuch, if i use more loops it's incresing the loading time, so please let me know how to solve this issue smartly 输出数组值不应包含任何重复项,我在每个循环中都使用过,是否有解决此情况的最佳实践,例如减少循环数,因为上述数组长度很大,如果我使用更多循环,则会增加加载时间,所以请让我知道如何聪明地解决这个问题
Thanks in advance 提前致谢
Fairly easy with use of standard map
, filter
and reduce
: 使用标准
map
相当容易, filter
和reduce
:
var arr1 = [{ 'name':'Victoria Cantrell', 'position':'Integer Corporation', 'office':'Croatia', 'ext':'0839', 'startDate':'2015-08-19', 'salary':208.178 }, { 'name':'Pearleeee', 'position':'In PC', 'office':'Cambodia', 'ext':'8262', 'startDate':'2014-10-08', 'salary':114.367 }, { 'name':'Pearl Crosby', 'position':'Integer', 'office':'Cambodia', 'ext':'8162', 'startDate':'2014-10-08', 'salary':114.367 }]; const arr2 = [{ 'name': 'name', 'checkfilter': false },{ 'name': 'position', 'checkfilter': true },{ 'name': 'office', 'checkfilter': true },{ 'name': 'startDate', 'checkfilter': false },{ 'name': 'ext', 'checkfilter': false },{ 'name': 'salary', 'checkfilter': false }]; const isDuplicate = (arr, name) => !arr.find(({ checkname }) => checkname === name); const arr3 = arr2 .filter(({ checkfilter }) => checkfilter) .map(({ name }) => ({ name: name, values: arr1.reduce((names, item) => isDuplicate(names, item[name]) ? [...names, { checkname: item[name] } ] : names, []) })); console.log(arr3);
First lets filter our arr2. 首先让我们过滤arr2。
var x = arr2.filter((r) => r.checkfilter).
Our Name gets mapped.. 我们的名字被映射..
map((r) => ({name:r.name,
Get our matching values., and also remove duplicates. 获取我们的匹配值。并删除重复项。
values:([...new Set(arr1.map((x) => (x[r.name])))]).
Finally remap, so we get {checkName: ?}
最终重新映射,所以我们得到
{checkName: ?}
map((r) => ({checkName:r}))
Lets not forget to close the brackets & stuff.. 让我们不要忘记关闭括号和内容。
}));
const arr1 = [ { 'name':'Victoria Cantrell', 'position':'Integer Corporation', 'office':'Croatia', 'ext':'0839', 'startDate':'2015-08-19', 'salary':208.178 }, { 'name':'Pearleeee', 'position':'In PC', 'office':'Cambodia', 'ext':'8262', 'startDate':'2014-10-08', 'salary':114.367 }, { 'name':'Pearl Crosby', 'position':'Integer', 'office':'Cambodia', 'ext':'8162', 'startDate':'2014-10-08', 'salary':114.367 } ]; const arr2 = [{ 'name': 'name', 'checkfilter': false }, { 'name': 'position', 'checkfilter': true }, { 'name': 'office', 'checkfilter': true }, { 'name': 'startDate', 'checkfilter': false }, { 'name': 'ext', 'checkfilter': false }, { 'name': 'salary', 'checkfilter': false }]; var x = arr2.filter((r) => r.checkfilter). map((r) => ({name:r.name, values:([...new Set(arr1.map((x) => (x[r.name])))]). map((r) => ({checkName:r})) })); console.log(x);
function createArr3(target, source) {
for (var prop in source) {
if (source[prop]["checkfilter"]) {
var newArr3Element = {};
newArr3Element.name = source[prop].name;
newArr3Element.values = [];
target.forEach(function(element) {
if (newArr3Element.values.indexOf(element[source[prop].name]) < 0){
newArr3Element.values.push(element[source[prop].name]);
}
});
arr3.push(newArr3Element);
}
}
console.log(arr3);
}
https://jsfiddle.net/x6140nct/ https://jsfiddle.net/x6140nct/
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