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如何根據對象中的屬性合并對象數組中的對象

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如何根據對象中的屬性合并對象數組中的對象

JavaScript

慕哥9229398 2021-08-26 15:18:18

我有一系列的活動，比如this.activities = [{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: disregard,timestamp: 14356787},{id: 54,event_type: 'NA',user_id: 56,user_name: 'Anand',purpose: 'privacy',timestamp: ''},{id: 54,event_type: 'FA',user_id: 56,user_name: 'Anand',purpose: 'Call',timestamp: ''},{id: 54,event_type: 'FA',user_id: 56,user_name: 'Anand',purpose: 'Listen',timestamp: ''},{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: 'Not allowed',timestamp: 14356784},{id: 54,event_type: 'NA',user_id: 56,user_name: 'Anand',purpose: 'data',timestamp: 14356786},{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: 'voicemail',timestamp: 14356785},{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: 'phone',timestamp: 14356775},{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: 'email',timestamp: 14356776},{id: 54,event_type: 'CA',user_id: 56,user_name: 'Anand',purpose: 'letter',timestamp: 14356777}]我想根據每個活動的5ms時間戳差異，僅通過 event_type 'CA' 對該數組進行分組/過濾。因此，如果 event_type 為“CA”的活動的時間戳彼此相差在 5ms 之內，那么將這些活動分組并形成一個新活動，除了稱為 CA_combined 的俱樂部活動的新 event_type 之外，所有內容都相同。

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2 回答

慕尼黑5688855

TA貢獻1848條經驗獲得超2個贊

解決了這個 O(n) 解決方案

sortedCAarray.forEach((act, index) => {

const currActivityTime = moment(act.timestamp).valueOf()

console.log(currActivityTime)

// First element is already added, avoid duplicates

if (index < 1) return

// Grab the last key from the timestampMap

const timestampKeys = Object.keys(timestampMap)

const key = timestampKeys[timestampKeys.length - 1]

// Add the activity and purpose to the respective arrays and set for that key if the key is within 5 units of the current activity timestamp

if (Math.abs(key - currActivityTime) < 5) {

let activities = timestampMap[key].activities

let purposes = timestampMap[key].purposes

activities.push(act)

purposes.add(act.purpose)

let timestamp = moment(activities[activities.length - 1].timestamp).valueOf()

// Edge case to to include all activities in the same bucket if they were within 5 ms of any element within the bucket

if (timestamp > key) {

Object.defineProperty(timestampMap, timestamp,

Object.getOwnPropertyDescriptor(timestampMap, key))

delete timestampMap[key]

}

} else {

// If not just create a new key with the current activity timestamp as the key

timestampMap[currActivityTime] = { activities: [act], purposes: new Set([act.purpose]) }

}

})

希望對此有任何改進！謝謝

反對回復 2021-08-26

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如何根據對象中的屬性合并對象數組中的對象

如何根據對象中的屬性合并對象數組中的對象

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