[英]How to covert an array of Object into Object with key and value as array
我正在寻找使用 reddduced 函数将对象数组转换为具有数组值的对象的优化解决方案
const companies=[ {name:'Company One',email:'trishabh885@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'X Name',email:'grfgr@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'A one',email:'grfgt@yahoo.com',phone:'943548213', category:'Lecture',start:1981,end:2003,date:'21-06-2009'}, {name:' Name',email:'gtrgt@gmail.com',phone:'9656013653', category:'Finance',start:1981,end:2003}, {name:'Company Name',email:'grgtrgg@gmail.com', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'sdksdjd',email:'bfbrg@gbmail.com',phone:'7485788475898548', category:'Sciennce',start:1981,end:2003,date:'21-06-2009'}, {name:'kdjfkf',email:'gfggfb@gmail.com',phone:'0959898860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjf',email:'g5rgf@gmail.com',phone:'050498860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjvk',email:'trishabh885@gmail.com',phone:'958958860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdkv',email:'trishabh885@gmail.com',phone:'5434886013821543543', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'zeeshan',email:'trishabh885@gmail.com',phone:'8864535213', category:'Finance',start:1981,end:2003,date:'21-06-2009'} ] // Looking for better and optimised solution to convert array of object into object with. values as an array using map methods let reduced_object={} let reduced_array=companies.reduce((arr,current)=>{ let entries=Object.entries(current) entries.forEach(element=>{ if(reduced_object.hasOwnProperty(element[0])){ reduced_object[element[0]].push(element[1]) }else{ reduced_object[element[0]]=[element[1]]; } }) return reduced_object }) console.log(reduced_array)
此解决方案不使用减少:
let obj = {
name: companies.map(item => item.name).filter(value => value !== undefined),
email: companies.map(item => item.email).filter(value => value !== undefined),
phone: companies.map(item => item.phone).filter(value => value !== undefined),
category: companies.map(item => item.category).filter(value => value !== undefined),
start: companies.map(item => item.start).filter(value => value !== undefined),
end: companies.map(item => item.end).filter(value => value !== undefined),
date: companies.map(item => item.date).filter(value => value !== undefined)
}
编辑:添加过滤器以在源中缺少属性时删除“未定义”条目。
我会改用地图:
// ...
const reduced_object = {};
reduced_object.name = companies.map(item => item.name);
reduced_object.email = companies.map(item => item.email);
reduced_object.phone = companies.map(item => item.phone);
// ...
当然,一旦您知道(或确定)了每个项目的键列表,您就可以将其包装在 for-of 循环中:
// ...
const reduced_object = {};
for (let key of item_keys) {
reduced_object[key] = companies.map(item => item[key]);
}
另一种解决方案:
const companies=[ {name:'Company One',email:'trishabh885@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'X Name',email:'grfgr@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'A one',email:'grfgt@yahoo.com',phone:'943548213', category:'Lecture',start:1981,end:2003,date:'21-06-2009'}, {name:' Name',email:'gtrgt@gmail.com',phone:'9656013653', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'Company Name',email:'grgtrgg@gmail.com',phone:'954543656138256', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'sdksdjd',email:'bfbrg@gbmail.com',phone:'7485788475898548', category:'Sciennce',start:1981,end:2003,date:'21-06-2009'}, {name:'kdjfkf',email:'gfggfb@gmail.com',phone:'0959898860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjf',email:'g5rgf@gmail.com',phone:'050498860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjvk',email:'trishabh885@gmail.com',phone:'958958860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdkv',email:'trishabh885@gmail.com',phone:'5434886013821543543', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'zeeshan',email:'trishabh885@gmail.com',phone:'8864535213', category:'Finance',start:1981,end:2003,date:'21-06-2009'} ] const arr = companies.reduce((acc, company) => ({ ...acc, ...Object.entries(company).reduce((objAcc, [key, val]) => ({ ...objAcc, [key]: [...acc?.[key] ?? [], ...val ? [val] : []] }), {}) }), {}) console.log(arr)
如果你不想有重复的值,你也可以这样做:
const companies=[ {name:'Company One',email:'trishabh885@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'X Name',email:'grfgr@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'A one',email:'grfgt@yahoo.com',phone:'943548213', category:'Lecture',start:1981,end:2003,date:'21-06-2009'}, {name:' Name',email:'gtrgt@gmail.com',phone:'9656013653', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'Company Name',email:'grgtrgg@gmail.com',phone:'954543656138256', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'sdksdjd',email:'bfbrg@gbmail.com',phone:'7485788475898548', category:'Sciennce',start:1981,end:2003,date:'21-06-2009'}, {name:'kdjfkf',email:'gfggfb@gmail.com',phone:'0959898860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjf',email:'g5rgf@gmail.com',phone:'050498860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjvk',email:'trishabh885@gmail.com',phone:'958958860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdkv',email:'trishabh885@gmail.com',phone:'5434886013821543543', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'zeeshan',email:'trishabh885@gmail.com',phone:'8864535213', category:'Finance',start:1981,end:2003,date:'21-06-2009'} ] const arr = companies.reduce((acc, company) => ({ ...acc, ...Object.entries(company).reduce((objAcc, [key, val]) => ({ ...objAcc, [key]: [...new Set([...acc?.[key] ?? [], ...val ? [val] : []])] }), {}) }), {}) console.log(arr)
有趣的功能路线:
[...new Set(companies.flatMap(Object.keys))]
.reduce((result, key) => {result[key] = companies.map(c => c[key]); return result;}, {})
第一行找到所有的键。 第二行创建所需的结果。 但老实说,在现实生活中,我会跳过那个 reduce 并编写一个简单的 for 循环。 我总是会选择清晰度而不是“功能纯度”。
使用简单的forEach
循环,迭代并检查每个项目。
const process = (arr, output = {}) => { companies.forEach((obj) => Object.entries(obj).forEach(([key, value]) => (output[key] ||= []).push(value) ) ); return output; }; const companies=[ {name:'Company One',email:'trishabh885@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'X Name',email:'grfgr@gmail.com',phone:'8860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'A one',email:'grfgt@yahoo.com',phone:'943548213', category:'Lecture',start:1981,end:2003,date:'21-06-2009'}, {name:' Name',email:'gtrgt@gmail.com',phone:'9656013653', category:'Finance',start:1981,end:2003}, {name:'Company Name',email:'grgtrgg@gmail.com', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'sdksdjd',email:'bfbrg@gbmail.com',phone:'7485788475898548', category:'Sciennce',start:1981,end:2003,date:'21-06-2009'}, {name:'kdjfkf',email:'gfggfb@gmail.com',phone:'0959898860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjf',email:'g5rgf@gmail.com',phone:'050498860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdjvk',email:'trishabh885@gmail.com',phone:'958958860138213', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'vdkv',email:'trishabh885@gmail.com',phone:'5434886013821543543', category:'Finance',start:1981,end:2003,date:'21-06-2009'}, {name:'zeeshan',email:'trishabh885@gmail.com',phone:'8864535213', category:'Finance',start:1981,end:2003,date:'21-06-2009'} ] console.log(process(companies))
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