根據條件將數組合併到字典中？

Question

我有兩個陣列：

First array = ["lines","lines","lines","lines"] 
Second array = ["one","two","","three","four","","five","six","","seven","eight"] 

我想將它們合併到具有嵌套陣列的字典：

Dict = ["lines":["one","two"],"lines":["three","four"],"lines":["five","six"],"lines":["seven","eight"]] 

基本上值從所述第二陣列應拆分由於空白空間標準（」 「）並在字典中分配爲獨立的數組對象。

Answer 1

嘗試這種解決方案：

let array2 = ["one","two","","three","four","","five","six","","seven","eight"] 
var splittedArray = array2.split { $0 == "" } 

結果從遊樂場：

重要提示：我使用Xcode中的iOS 7 9

Answer 2

這會做到這一點：

for value in firstArray { 
    if value == "" { 
     // Add a blank subarray whenever we find an empty string. 
     combinedArray.append([String]()) 
    } else if combinedArray.count == 0 { 
     combinedArray.append([value]) 
    } else { 
     // Append the value to the last subarray. 
     combinedArray[combinedArray.count - 1].append(value) 
    } 
} 

Answer 3

這麼多高雅，但這個工作就像你想要的：

var First = ["lines","lines","lines","lines"] 
var Second = ["one","two","","three","four","","five","six","","seven","eight"] 

var counter = 0 
var counters = 0 
var counterss = 0 
var response = NSMutableArray(capacity: First.count) 

for obj in First { 
    var re = NSMutableArray() 
    while counterss != Second.count-1 && Second[counterss] != "" { 
     if (counterss < (Second.count)) { 
      re[counters] = Second[counterss] 
      counterss = counterss + 1 
      counters = counters + 1 
     } 
    } 
    if Second[counterss] == "" { 
     counterss = counterss + 1 
    } 
    if counterss == (Second.count-1) { 
     re[counters] = Second.last! 
    } 
    counters = 0 
    response[counter] = [First[counter],re] 
    counter = counter + 1 
} 
println("heres the final array \(response)") 

截圖：

他水庫是與作爲主陣列內所需的字典對象的解決方案：

var First = ["lines","lines","lines","lines"] 
var Second = ["one","two","","three","four","","five","six","","seven","eight"] 

var counter = 0 
var counters = 0 
var counterss = 0 
var response = NSMutableArray(capacity: First.count) 

for obj in First { 
    var re = NSMutableArray() 
    while counterss != Second.count-1 && Second[counterss] != "" { 
     if (counterss < (Second.count)) { 
      re[counters] = Second[counterss] 
      counterss = counterss + 1 
      counters = counters + 1 
     } 
    } 
    if Second[counterss] == "" { 
     counterss = counterss + 1 
    } 
    if counterss == (Second.count-1) { 
     re[counters] = Second.last! 
    } 
    counters = 0 
    var ss = [First[counter] : re] as Dictionary 
    response[counter] = ss 
    counter = counter + 1 
} 
println("heres the final array \(response)") 

和這裏的決賽中，詞典的詞典：

var First = ["lines","lines","lines","lines"] 
var Second = ["one","two","","three","four","","five","six","","seven","eight"] 

var counter = 0 
var counters = 0 
var counterss = 0 
var response = NSMutableDictionary(capacity: First.count) 

for obj in First { 
    var re = NSMutableArray() 
    while counterss != Second.count-1 && Second[counterss] != "" { 
     if (counterss < (Second.count)) { 
      re[counters] = Second[counterss] 
      counterss = counterss + 1 
      counters = counters + 1 
     } 
    } 
    if Second[counterss] == "" { 
     counterss = counterss + 1 
    } 
    if counterss == (Second.count-1) { 
     re[counters] = Second.last! 
    } 
    counters = 0 
    var ss = [First[counter] : re] as Dictionary 
    response[counter] = ss 
    counter = counter + 1 
} 
println("heres the final array \(response)") 

這不是一個有序字典的字典索引，但是這些指標與您想要的信息相匹配：

是的，這是一本字典詞典，它包含每本詞典中的「名稱」和「數組」，就像要求的那樣。