Question

我有一个我要过滤的swift数组，这里是数组

Gemfile

我希望过滤数组，使得以搜索字符串开头的项目显示在仅包含搜索字符串的项目之前

所以当我搜索示例p时，结果应该是某种方式

let array = [apple,workshops,shopping,sports,parties,pantry,pen]

我试过这个

let array = [parties,pantry,pen,apple,workshops,shopping,sports]

但是这给了我包含搜索字符串的所有字符串。

那么伙计我怎么能这样做

Answer 1

你可以写

 let result = words
    .filter { $0.contains(keyword) }
    .sorted { ($0.hasPrefix(keyword) ? 0 : 1) < ($1.hasPrefix(keyword) ? 0 : 1) }

实施例

let words = ["apple", "workshops", "shopping", "sports", "parties", "pantry", "pen", "cat", "house"]
let keyword = "p"
let result = words
    .filter { $0.contains(keyword) }
    .sorted { ($0.hasPrefix(keyword) ? 0 : 1) < ($1.hasPrefix(keyword) ? 0 : 1) }

// ["pen", "pantry", "parties", "apple", "workshops", "shopping", "sports"]

Answer 2

试试这个。首先，它过滤数组以删除那些不包含搜索字符串的元素，然后使用自定义排序来优先选择以搜索字符串开头的项目。一般来说，使用笛卡尔方法将问题分解为较小的子问题，而不是试图一步解决问题。

let searchString = "p"
let array = ["apple", "workshops", "shopping", "sports", "parties", "pantry", "pen", "xyzzy"]

let filteredArray = array.filter({ $0.contains(searchString) })
filteredArray // drops xyzzy

let sortedArray = filteredArray.sorted(isOrderedBefore:  {
    switch ($0.hasPrefix(searchString), $1.hasPrefix(searchString)) {
    case (true, true):
        return $0 < $1
    case (true, false):
        return true
    case (false, true):
        return false
    case (false, false):
        return $0 < $1
    }

})
sortedArray // "pantry", "parties", "pen", "apple", "shopping", "sports", "workshops"] as required

Answer 3

另一种方式：

let searchString = "p"
let array = ["apple", "workshops", "shopping", "sports", "parties", "pantry", "pen", "xyzzy"]
let result = array.filter{$0.containsString(searchString)}
    .map{($0.hasPrefix(searchString) ? 0 : 1, $0)}
    .sort{$0 < $1}
    .map{$1}
print(result) //->["pantry", "parties", "pen", "apple", "shopping", "sports", "workshops"]

（我不知道为什么，但我的Xcode花了很多时间来编译这些行。相信这个编译并最终按预期运行。）

过滤和排序swift数组

3 个答案:

实施例