1

最近想根据indexPaths删除单元格，所以该函数的输入参数是[IndexPath]type，我需要将[IndexPath]根据 拆分成几个数组indexPath.section，有什么简单的方法吗？例如

indexPaths = 
[IndexPath(row: 0, section: 1),
 IndexPath(row: 1, section: 1), 
 IndexPath(row: 2, section: 1), 
 IndexPath(row: 2, section: 0)]

想将其转换为

indexPath1 = 
[IndexPath(row: 0, section: 1),
 IndexPath(row: 1, section: 1), 
 IndexPath(row: 2, section: 1)]

indexPath0 = 
[IndexPath(row: 2, section: 0)]

// maybe get a [Array]
[indexPath0, indexPath1]
4

3 回答 3

3

一种可能的解决方案是首先构建一个字典，其中键是节号，值是该IndexPath节中的数组。

let indexPaths = [
    IndexPath(row: 0, section: 1),
    IndexPath(row: 1, section: 1),
    IndexPath(row: 2, section: 1),
    IndexPath(row: 2, section: 0),
]

let pathDict = Dictionary(grouping: indexPaths) { (path) in
    return path.section
}

然后您可以将此字典映射到路径数组的数组中。但首先按部分对这些数组进行排序。

let sectionPaths = pathDict.sorted { (arg0, arg1) -> Bool in
    return arg0.key < arg1.key // sort by section
}.map { $0.value } // get just the arrays of IndexPath

print(sectionPaths)

输出：

[[[0, 2]], [[1, 0], [1, 1], [1, 2]]]

于 2019-04-27T04:45:31.773 回答
0

使用过滤器：

let indexPath0 = indexPaths.filter { $0.section == 0 }
let indexPath1 = indexPaths.filter { $0.section == 1 }
于 2019-04-27T05:42:13.360 回答
0
  • 我们需要一个 HashMap，映射到键上。
  • 我们需要对字典进行排序
  • 我们需要提取字典的值并将它们附加到数组
  • 我们需要返回那个数组
var indexPaths = [IndexPath(row: 0, section: 1),
     IndexPath(row: 1, section: 1),
     IndexPath(row: 2, section: 1),
     IndexPath(row: 2, section: 0)
]

extension Array where Element == IndexPath {
    func splitArray() ->  Array<[IndexPath]> {
        var tempDict = [String : [IndexPath]]()
        for element in self {
            let section = element.section
            if tempDict[String(section)] != nil {
                // some element append
                if var array = tempDict[String(section)] {
                    array.append(element)
                    tempDict[String(section)] = array
                }

            } else {
                tempDict[String(section)] = [element]
            }
        }

        // dictionary mayn't have sorted, sort the dictionary 

        tempDict.sorted{ $0.key > $1.key }

        var returnedArray = Array<[IndexPath]>()
        for (key, value) in tempDict {
           returnedArray.append(value)
        }

        return returnedArray
    }
}

print(indexPaths.splitArray())
于 2019-04-27T04:57:41.997 回答