Question

我有一组类型为<Int,String>的字典数组：

[14: "2", 17: "5", 6: "5", 12: "Ali", 11: "0", 2: "4", 5: "It it it", 15: "5", 18: "2", 16: "5", 8: "2", 13: "4", 19: "4", 1: "2", 4: "12-09-2017 - 9:52"]

我需要单独获取密钥并将它们保存在字符串中，并且仅将值保存在另一个字符串中。

结果应如下：

string key = "12,17,6,12,11,2,5,15,18,16,8,13,19,1,4"
string values = "2,5,5,Ali,0,4,It it ti,5,2,5,2,4,4,2,12-09-2017 - 9:52"

Answer 1

字典有一个keys和一个values属性返回键/值作为（懒惰）集合。对于你刚才拥有的价值观加入他们：

let dict = [14: "2", 17: "5", 6: "5", 12: "Ali", 11: "0", 2: "4", 5: "It it it", 15: "5", 18: "2", 16: "5", 8: "2", 13: "4", 19: "4", 1: "2", 4: "12-09-2017 - 9:52"]

let values = dict.values.joined(separator: ",")
// Ali,5,2,It it it,5,0,4,5,4,4,2,12-09-2017 - 9:52,5,2,2

键是整数，必须先转换为字符串：

let keys = dict.keys.map(String.init).joined(separator: ",")
// 12,17,14,5,15,11,13,16,19,2,18,4,6,8,1

订单未指定，但键和值相同。

Answer 2

试试这个，这可以重构我相信，这是第一个打击我的逻辑。

从字典数组中获取元素并将键保存为字符串，将值保存为字符串

2 个答案: