合併幾個列表到字典

Question

我有3個表：

a = ["John", "Archie", "Levi"] 
b = ["13", "20"] 
c = ["m", "m", "m", "m"] 

我想用字典合併成一個列表如下：

result = [ 
{"name": "John", "age": "13", "gender": "m"}, 
{"name": "Archie", "age": "20", "gender": "m"}, 
{"name": "Levi", "age": "", "gender": "m"}, 
{"name": "", "age": "", "gender": "m"}, 
    ] 

Answer 1

嘗試這樣的：

let a = ["John", "Archie", "Levi"] 
let b = ["13", "20"] 
let c = ["m", "m", "m", "m"] 


var dicArray:[[String:String]] = [] 

for index in 0..<max(a.count,b.count,c.count) { 
    dicArray.append([:]) 
    dicArray[index]["name"] = index < a.count ? a[index] : "" 
    dicArray[index]["age"] = index < b.count ? b[index] : "" 
    dicArray[index]["gender"] = index < c.count ? c[index] : "" 
} 

dicArray // [["gender": "m", "age": "13", "name": "John"], ["gender": "m", "age": "20", "name": "Archie"], ["gender": "m", "age": "", "name": "Levi"], ["gender": "m", "age": "", "name": ""]] 

Answer 2

好吧，這是非常正常的計算機科學問題。這裏是如何去解決它的輪廓：

1. 首先，確定輸入和所需的輸出。你已經做到了。
2. 請注意您需要處理的任何邊緣情況
3. 接下來，使用僞代碼映射出邏輯流。
4. 從僞代碼轉換爲真實代碼。
5. 測試和調試。

對於第2步，您的數據表明您想要處理每個源陣列中元素數量不同的情況。它看起來像是要爲每個數組中的所有0個元素創建一個字典，然後爲1個元素創建一個字典等。當您用完給定數組的元素時，它看起來像要跳過您的鍵產生字典輸入。

我們的僞代碼：

Find the array with the maximum number of elements. Make this your output_count (the number of items in your output array.) 
Create an output array large enough for output_count entries. 
Loop from 0 to output_count - 1. 
    create a dictionary variable 
    for each input array, if there are enough elements, add a key/value pair 
    for that array's key and the value at the current index. Otherwise skip 
    that key. 
    add the new dictionary to the end of your output array. 

這應該足以讓你開始。現在看看您是否可以將該僞代碼轉換爲實際代碼，進行測試並進行調試。如果您的代碼無法正常工作，請返回此處並附上您的實際代碼，並隨時尋求幫助。

Answer 3

這是我服用。我首先創建陣列，然後用enumerate()和forEach來處理每個陣列並將其添加到詞典的數組：

let a = ["John", "Archie", "Levi"] 
let b = ["13", "20"] 
let c = ["m", "m", "m", "m"] 

let count = max(a.count, b.count, c.count) 

var result = Array(count: count, repeatedValue: ["name":"", "age":"", "gender":""]) 

a.enumerate().forEach { idx, val in result[idx]["name"] = val } 
b.enumerate().forEach { idx, val in result[idx]["age"] = val } 
c.enumerate().forEach { idx, val in result[idx]["gender"] = val } 

print(result) 

[[「性別」：「M」，「年齡」：「13 「，」性別「：」m「，」年齡「： 」20「，」姓名「：」Archie「]，[」性別「：」m「 ： 「」， 「名」： 「李維斯」]， [ 「性別」： 「M」， 「年齡」： 「」， 「名」： 「」]]

Answer 4

這應該做的工作

let maxLength = max(a.count, b.count, c.count) 
let paddedA = a + [String](count: maxLength-a.count, repeatedValue: "") 
let paddedB = b + [String](count: maxLength-b.count, repeatedValue: "") 
let paddedC = c + [String](count: maxLength-c.count, repeatedValue: "") 

let res = zip(paddedA, zip(paddedB, paddedC)).map { 
    ["name": $0.0, "age": $0.1.0, "gender": $0.1.1] 
} 

輸出

[["gender": "m", "age": "13", "name": "John"], ["gender": "m", "age": "20", "name": "Archie"], ["gender": "m", "age": "", "name": "Levi"], ["gender": "m", "age": "", "name": ""]]