从数组中删除相似的对象
Remove similar objects from array
我正在尝试从具有 4 个相同值和 1 个唯一值的数组中删除 objectst。
快速 google 搜索提供了很多选项,用于从数组中删除 相同的 对象,但不包括 相似[=25 的对象=].
数组示例:
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
]
我试过使用 lodash _uniq
函数,但只需要一个 属性 就可以匹配:
var non_duplidated_data = _.uniq(data, 'Name');
除日期外,所有值都相同。如何根据 4 个属性删除相同的对象?
使用 .uniqBy https://lodash.com/docs#uniqBy,它允许自定义标准来统一您的数组。
您可以通过两个嵌套循环来完成此操作,这两个循环将针对自身迭代数组。
var newArray = []; //Result array
//iterate
data.forEach(function(item, index){
var found = false;
//Go through the rest of elements and see if any matches
for (var i = index; i < data.length, i++) {
if (item.name == data[i].name && item.title == data[i].title) // add more fields here
found = true;
}
//Add to the result array if no duplicates found
if (!found)
newArray.push(item);
})
这可以帮助你,在这段代码中,我首先映射所有应该比较的属性,然后我减少数组以便只获得唯一的记录。
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
];
var uniqueRecord = data.map(function(item){
return {Name:item.Name,Title:item.Title,Department: item.Department,Company:item.Company};
}).reduce(function(acc,curr){
if(!Array.isArray(acc)){
var copyOfAcc = JSON.parse(JSON.stringify(acc));
acc = [];
acc.push(copyOfAcc);
}
return (acc.indexOf(curr) > -1) ? acc.push(curr) : acc;
});
结果将是:
[ { Name: 'Tito',
Title: 'Developer',
Department: 'IT',
Company: 'XXX' } ]
然后您需要定义要保留的日期(第一个、最后一个、哪个?)并再次查找记录,为此,一个简单的 'forEach' 就可以了。
您可以使用普通 Javascript 解决此问题。
只需检查每个对象的每个 属性:
Array.prototype.allValuesSame = function() {
for(var i = 1; i < this.length; i++) {
if(this[i] !== this[0])
return false;
}
return true;
}
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
]
var tmpPropValues = {};
for (var property in data[0]) {
tmpPropValues[property] = [];
for(var i = 0; i < data.length; i++) {
var obj = data[i];
// store temporary in array
if (obj.hasOwnProperty(property)) {
var value = obj[property];
tmpPropValues[property].push(value);
}
}
if(tmpPropValues[property].allValuesSame()) {
delete tmpPropValues[property];
}
}
for(var prop in tmpPropValues) {
var tmp = document.getElementById("result").innerHTML;
document.getElementById("result").innerHTML = tmp+" "+prop;
}
我给你做了一个fiddle
我正在尝试从具有 4 个相同值和 1 个唯一值的数组中删除 objectst。
快速 google 搜索提供了很多选项,用于从数组中删除 相同的 对象,但不包括 相似[=25 的对象=].
数组示例:
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
]
我试过使用 lodash _uniq
函数,但只需要一个 属性 就可以匹配:
var non_duplidated_data = _.uniq(data, 'Name');
除日期外,所有值都相同。如何根据 4 个属性删除相同的对象?
使用 .uniqBy https://lodash.com/docs#uniqBy,它允许自定义标准来统一您的数组。
您可以通过两个嵌套循环来完成此操作,这两个循环将针对自身迭代数组。
var newArray = []; //Result array
//iterate
data.forEach(function(item, index){
var found = false;
//Go through the rest of elements and see if any matches
for (var i = index; i < data.length, i++) {
if (item.name == data[i].name && item.title == data[i].title) // add more fields here
found = true;
}
//Add to the result array if no duplicates found
if (!found)
newArray.push(item);
})
这可以帮助你,在这段代码中,我首先映射所有应该比较的属性,然后我减少数组以便只获得唯一的记录。
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
];
var uniqueRecord = data.map(function(item){
return {Name:item.Name,Title:item.Title,Department: item.Department,Company:item.Company};
}).reduce(function(acc,curr){
if(!Array.isArray(acc)){
var copyOfAcc = JSON.parse(JSON.stringify(acc));
acc = [];
acc.push(copyOfAcc);
}
return (acc.indexOf(curr) > -1) ? acc.push(curr) : acc;
});
结果将是:
[ { Name: 'Tito',
Title: 'Developer',
Department: 'IT',
Company: 'XXX' } ]
然后您需要定义要保留的日期(第一个、最后一个、哪个?)并再次查找记录,为此,一个简单的 'forEach' 就可以了。
您可以使用普通 Javascript 解决此问题。 只需检查每个对象的每个 属性:
Array.prototype.allValuesSame = function() {
for(var i = 1; i < this.length; i++) {
if(this[i] !== this[0])
return false;
}
return true;
}
var data = [
{Date: "2016-04-27T09:03:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T08:07:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"},
{Date: "2016-04-27T10:23:45Z", Name: "Tito", Title: "Developer", Department:"IT", Company: "XXX"}
]
var tmpPropValues = {};
for (var property in data[0]) {
tmpPropValues[property] = [];
for(var i = 0; i < data.length; i++) {
var obj = data[i];
// store temporary in array
if (obj.hasOwnProperty(property)) {
var value = obj[property];
tmpPropValues[property].push(value);
}
}
if(tmpPropValues[property].allValuesSame()) {
delete tmpPropValues[property];
}
}
for(var prop in tmpPropValues) {
var tmp = document.getElementById("result").innerHTML;
document.getElementById("result").innerHTML = tmp+" "+prop;
}
我给你做了一个fiddle