比较underscorejs uniq iteratee中的多个属性
Compare more than one properties in underscorejs uniq iteratee
我正在使用 underscorejs uniq 方法。
_.uniq(
[{ name: 'a', family: 't' }, { name: 'b', family: 'n' }],
false,
function (item, key, a) { return item.name; }
);
Iteratee 函数只是 return 一个 属性。但我想比较姓名和家庭以创建唯一列表。怎么做到的?
试试这个代码:
_.uniq(
[
{name: 'a', family: 't'},
{name: 'b', family: 'n'},
{name: 'a', family: 'd'},
{name: 'a', family: 't'}
],
false,
function (item, key, a) { return 'n=' + item.name + ';f=' + item.family; }
);
它将生成一个只有三个对象的数组:
[{name: 'a', family: 't'}, {name: 'b', family: 'n'}, {name: 'a', family: 'd'}]
我正在使用 underscorejs uniq 方法。
_.uniq(
[{ name: 'a', family: 't' }, { name: 'b', family: 'n' }],
false,
function (item, key, a) { return item.name; }
);
Iteratee 函数只是 return 一个 属性。但我想比较姓名和家庭以创建唯一列表。怎么做到的?
试试这个代码:
_.uniq(
[
{name: 'a', family: 't'},
{name: 'b', family: 'n'},
{name: 'a', family: 'd'},
{name: 'a', family: 't'}
],
false,
function (item, key, a) { return 'n=' + item.name + ';f=' + item.family; }
);
它将生成一个只有三个对象的数组:
[{name: 'a', family: 't'}, {name: 'b', family: 'n'}, {name: 'a', family: 'd'}]