Question

这是我的两个数组

 Arrayone = [{keyphraseId:"tcpid1234",name:"shakti"},{keyphraseId:"tcpid456",name:"shakti"},{keyphraseId:"tcpid897",name:"srichandan "},{keyphraseId:"tcpid779",name:"prakash"}]

Arraytwo  = [{idstring:"tcpid1234",name:"shakti"},{idstring:"tcpid456",name:"shakti"}]

我想在数组2的基础上过滤数组一，即（我想得到数组1的所有元素，其中没有数组二中的idstring

Answer 1

如果你不介意O（n ^ 2），那么直接一步测试。

let arrayOne = [["keyphraseId":"tcpid1234", "name":"shakti"], ["keyphraseId":"tcpid456", "name":"shakti"], ["keyphraseId":"tcpid897", "name":"srichandan "], ["keyphraseId":"tcpid779", "name":"prakash"]]
let arrayTwo = [["idstring":"tcpid1234", "name":"shakti"],["idstring":"tcpid456", "name":"shakti"]]

arrayOne.filter { one in
    !arrayTwo.contains { two in
        one["keyphraseId"] == two["idstring"]
    }
}

如果您想要更好的性能O（n），请将stringid放入集合

let arrayOne = [["keyphraseId":"tcpid1234", "name":"shakti"], ["keyphraseId":"tcpid456", "name":"shakti"], ["keyphraseId":"tcpid897", "name":"srichandan "], ["keyphraseId":"tcpid779", "name":"prakash"]]
let arrayTwo  = [["idstring":"tcpid1234", "name":"shakti"],["idstring":"tcpid456", "name":"shakti"]]

let idstrings = Set(arrayTwo.flatMap { $0["idstring"] })

arrayOne.filter {
    guard let keyphraseId = $0["keyphraseId"] else { return false }
    return !idstrings.contains(keyphraseId)
}

Answer 2

您有2个数组，如下所示：

idStrings

要过滤数组，您需要执行以下两个步骤：



从arrayTwo

列出arrayOne

过滤idString并在步骤1中抓取keyphraseId

从arrayOne获取元素

其idstring位于arrayTwo中let arrayTwoIds = arrayTwo.map { $0["idstring"] } let filteredResults = arrayOne.filter { arrayTwoIds.contains($0["keyphraseId"]) } print(filteredResults)：

keyphraseId

其idstring不在于arrayTwo !：

只需将let filteredResults = arrayOne.filter { !arrayTwoIds.contains($0["keyphraseId"]) }置于过滤条件：

{{1}}

希望这会有所帮助。

如何在swift 4中基于另一个对象数组过滤对象数组

2 个答案: