如何使用下划线在 JavaScript 数组中获取重复项

How to get duplicates in a JavaScript Array using Underscore

提问by Abhishek Dutta

I have an array for which I need to the items that are duplicates and print the items based on a specific property. I know how to get the unique items using underscore.js but I need to find the duplicates instead of the unique values

我有一个数组，我需要它的重复项并根据特定属性打印这些项。我知道如何使用 underscore.js 获取唯一项，但我需要找到重复项而不是唯一值

var somevalue=[{name:"john",country:"spain"},{name:"jane",country:"spain"},{name:"john",country:"italy"},{name:"marry",country:"spain"}]


var uniqueList = _.uniq(somevalue, function (item) {
        return item.name;
    })

This returns:

这将返回：

[{name:"jane",country:"spain"},{name:"marry",country:"spain"}] 

but I actually need the opposite

但我实际上需要相反的

[{name:"john",country:"spain"},{name:"john",country:"italy"}]

回答by chmac

A purely underscore based approach is:

纯粹基于下划线的方法是：

_.chain(somevalue).groupBy('name').filter(function(v){return v.length > 1}).flatten().value()

This would produce an array of all duplicate, so each duplicate will be in the output array as many times as it is duplicated. If you only want 1 copy of each duplicate, you can simply add a .uniq()to the chain like so:

这将产生一个包含所有重复项的数组，因此每个重复项将在输出数组中出现的次数与它被复制的次数相同。如果您只想要每个副本的 1 个副本，您可以简单地将一个添加.uniq()到链中，如下所示：

_.chain(somevalue).groupBy('name').filter(function(v){return v.length > 1}).uniq().value()

No idea how this performs, but I do love my one liners... :-)

不知道它的表现如何，但我确实喜欢我的一个衬垫...... :-)

回答by Igor Semin

Use .filter() and .where() for source array by values from uniq array and getting duplicate items.

根据 uniq 数组中的值使用 .filter() 和 .where() 作为源数组并获取重复项。

var uniqArr = _.uniq(somevalue, function (item) {
    return item.name;
});

var dupArr = [];
somevalue.filter(function(item) {
    var isDupValue = uniqArr.indexOf(item) == -1;

    if (isDupValue)
    {
        dupArr = _.where(somevalue, { name: item.name });
    }
});

console.log(dupArr);

Fiddle

小提琴

UpdatedSecond way if you have more than one duplicate item, and more clean code.

如果您有多个重复项和更干净的代码，则更新第二种方式。

var dupArr = [];
var groupedByCount = _.countBy(somevalue, function (item) {
    return item.name;
});

for (var name in groupedByCount) {
    if (groupedByCount[name] > 1) {
        _.where(somevalue, {
            name: name
        }).map(function (item) {
            dupArr.push(item);
        });
    }
};

Look fiddle

看小提琴

回答by Leo

Here how I have done the same thing:

在这里，我是如何做同样的事情的：

_.keys(_.pick(_.countBy(somevalue, b=> b.name), (value, key, object) => value > 1))

_.keys(_.pick(_.countBy(somevalue, b=> b.name), (value, key, object) => value > 1))

回答by user10522292

var somevalue=[{name:"john",country:"spain"},{name:"jane",country:"spain"},{name:"john",country:"italy"},{name:"marry",country:"spain"}];
var uniqueList = _.uniq(somevalue, function (item) {return item.country;})

//from this you will get the required output

//由此您将获得所需的输出

回答by Ravi Chaudhary

Correction to chmac'sanswer.

更正chmac的答案。

_.chain(somevalue).groupBy('name').filter(function(v){return v.length > 1}).flatten().uniq().value()

The values need to be flattened before using unique function

在使用唯一函数之前需要将值展平