我有一系列的人,我想过滤自己(以测试数组中的所有其他项目):
const people = [{
name: "James Cromwell",
region: "Australia"
}, {
name: "Janet April",
region: "Australia"
}, {
name: "David Smith",
region: "USA"
}, {
name: "Tracey Partridge",
region: "USA"
}]
在这种情况下我想做的是留下任何人:
name
starts用同一封信region
的价值是一样的在这种情况下,结果将是:
[{
name: "James Cromwell",
region: "Australia"
}, {
name: "Janet April",
region: "Australia"
}]
我看过做filter
和any
的组合,但没有喜悦。我在这里使用ramda的决定是我在现有的ramda compose
函数中使用它来转换数据。
按照region
生成的密钥和name
的第一个字母对元素进行分组。拒绝长度为1的任何组,然后使用R.value转换回数组,并展平。
注意:此解决方案将返回多组“相同”的人。如果你只想要一个组,你可以选择第一组,或者最大的一组......而不是获取值和展平。
const { compose, groupBy, reject, propEq, values, flatten } = R
const fn = compose(
flatten, // flatten the array - or R.head to get just the 1st group
values, // convert to an array of arrays
reject(propEq('length', 1)), // remove groups with 1 items
groupBy(({ name: [l], region }) => `${region}-${l.toLowerCase()}`) // collect into groups by the requested key
)
const people = [{"name":"James Cromwell","region":"Australia"},{"name":"Janet April","region":"Australia"},{"name":"David Smith","region":"USA"},{"name":"Tracey Partridge","region":"USA"}]
const result = fn(people)
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
const people = [{
name: "James Cromwell",
region: "Australia"
}, {
name: "Janet April",
region: "Australia"
}, {
name: "David Smith",
region: "USA"
}, {
name: "Tracey Partridge",
region: "USA"
}];
let output = people
.filter((p1, i1) =>
people.some((p2, i2) =>
i1 !== i2 && p1.region === p2.region && p1.name[0] === p2.name[0]));
console.log(output);