两个NSArrays组合成具有交替值的NSDictionary [英] Two NSArrays combine to a NSDictionary with alternating values

问题描述

您好，我是Objective-c的新手，请原谅我的无知。所以基本上这就是我想发生的事情。我有两个数字数组

  Array1：（1,3,5）; 
 Array2：（2,4,6）; 
  

我希望它们在词典中组合后像这样

 字典：{ 1：2 ,: 3：4， 5：6} 
  

任何反馈将不胜感激！

解决方案

数组

我有两个数字数组， Array1：（1、3、5）& Array2：（2,4,6）

我假设您有它们放在 NSArray 中，并且您知道 NSNumber & Objective-C文字。换句话说，您将：

  NSArray * keys = @ [@@@@ 3， @ 5]; // NSNumber数组
 NSArray * objects = @ [@ 2，@ 4，@ 6]; // NSNumber数组
  

dictionary：{ 1： 2 ,: 3：4， 5：6}

我认为这意味着：

  @ {
 @ dictionary：@ {
 @ 1：@ 2， 
 @ 3：@ 4，
 @ 5：@ 6 
} 
} 
  

步骤1-字符串化字符串

  NSArray *键= @ [@@，@ 3，@ 5]; 
 NSArray *对象= @ [@ 2，@ 4，@ 6]; 
 
 NSMutableArray * stringifiedKeys = [NSMutableArray arrayWithCapacity：keys.count]; 
（NSNumber *键中的键）{
 [stringifiedKeys addObject：key.stringValue]; 
} 
 
  

步骤2-创建字典

dictionaryWithObjects：forKeys： ：

  +（ instancetype）dictionaryWithObjects：（NSArray< ObjectType> *）对象
 forKeys：（NSArray< id< NSCopying> * *）keys; 
  

您可以通过以下方式使用它：

< preclass = lang-c prettyprint-override> NSArray * keys = @ [@ 1，@ 3，@ 5];
NSArray *对象= @ [@ 2，@ 4，@ 6];

NSMutableArray * stringifiedKeys = [NSMutableArray arrayWithCapacity：keys.count];
（NSNumber *键中的键）{
[stringifiedKeys addObject：key.stringValue];
}

NSDictionary * dictionary = [NSDictionary dictionaryWithObjects：objects
forKeys：stringifiedKeys];


步骤3-将其包装在字典中

< preclass = lang-c prettyprint-override> NSArray * keys = @ [@ 1，@ 3，@ 5];
NSArray *对象= @ [@ 2，@ 4，@ 6];

NSMutableArray * stringifiedKeys = [NSMutableArray arrayWithCapacity：keys.count];
（NSNumber *键中的键）{
[stringifiedKeys addObject：key.stringValue];
}

NSDictionary * dictionary = [NSDictionary dictionaryWithObjects：objects
forKeys：stringifiedKeys];
NSDictionary * result = @ {@ dictionary：字典};

NSLog（@％@，结果）；


结果：

  {
字典= {
 1 = 2; 
 3 = 4; 
 5 = 6; 
}; 
} 
  

手动

<课前= lang-c prettyprint-override> NSArray * keys = @ [@ 1，@ 3，@ 5];
NSArray *对象= @ [@ 2，@ 4，@ 6];

// Mimick dictionaryWithObjects：forKeys：行为
if（objects.count！= keys.count）{
NSString * reason = [NSString stringWithFormat：@对象数（ ％lu）与键计数（％lu），（无符号长）objects.count，（无符号长）keys.count]不同；

@throw [NSException exceptionWithName：NSInvalidArgumentException
原因：原因
userInfo：nil];
}

NSMutableDictionary * inner = [NSMutableDictionary dictionaryWithCapacity：keys.count]；

for（NSUInteger index = 0; index< keys.count; index ++）{
NSString * key = [keys [index] stringValue];
NSString * object = objects [index];
inner [key] =对象；
}

NSDictionary * result = @ {@ dictionary：内部};


脚注

由于我对Objective-C陌生，我确实有意避免：

• Blocks&比较安全的枚举方式


• 轻量级仿制药


• 可空性

Hi I am very new to Objective-c please excuse my ignorance. So basically this is what I want to happened. I have two arrays of numbers

Array1: (1,3,5);
Array2: (2,4,6);

and I want them to be like this after combining them in a dictionary

"dictionary":{"1":2,: "3":4, "5":6}

Any feedback will be appreciated!

解决方案

Arrays

I have two arrays of numbers Array1: (1,3,5) & Array2: (2,4,6)

I assume that you have them in NSArray and that you're aware of NSNumber & Objective-C Literals. In other words, you have:

NSArray *keys    = @[@1, @3, @5]; // Array of NSNumber
NSArray *objects = @[@2, @4, @6]; // Array of NSNumber

"dictionary":{"1":2,: "3":4, "5":6}

I assume that it means:

@{
    @"dictionary": @{
        @"1": @2,
        @"3": @4,
        @"5": @6
    }
}

Step 1 - Stringify keys

NSArray *keys = @[@1, @3, @5];
NSArray *objects = @[@2, @4, @6];

NSMutableArray *stringifiedKeys = [NSMutableArray arrayWithCapacity:keys.count];
for (NSNumber *key in keys) {
    [stringifiedKeys addObject:key.stringValue];
}

Step 2 - Create a dictionary

dictionaryWithObjects:forKeys::

+ (instancetype)dictionaryWithObjects:(NSArray<ObjectType> *)objects 
                              forKeys:(NSArray<id<NSCopying>> *)keys;

You can use it in this way:

NSArray *keys = @[@1, @3, @5];
NSArray *objects = @[@2, @4, @6];

NSMutableArray *stringifiedKeys = [NSMutableArray arrayWithCapacity:keys.count];
for (NSNumber *key in keys) {
    [stringifiedKeys addObject:key.stringValue];
}

NSDictionary *dictionary = [NSDictionary dictionaryWithObjects:objects
                                                       forKeys:stringifiedKeys];

Step 3 - Wrap it in a dictionary

NSArray *keys = @[@1, @3, @5];
NSArray *objects = @[@2, @4, @6];

NSMutableArray *stringifiedKeys = [NSMutableArray arrayWithCapacity:keys.count];
for (NSNumber *key in keys) {
    [stringifiedKeys addObject:key.stringValue];
}

NSDictionary *dictionary = [NSDictionary dictionaryWithObjects:objects
                                                       forKeys:stringifiedKeys];
NSDictionary *result = @{ @"dictionary": dictionary };

NSLog(@"%@", result);

Result:

{
    dictionary = {
        1 = 2;
        3 = 4;
        5 = 6;
    };
}

Manually

NSArray *keys = @[@1, @3, @5];
NSArray *objects = @[@2, @4, @6];

// Mimick dictionaryWithObjects:forKeys: behavior
if (objects.count != keys.count) {
    NSString *reason = [NSString stringWithFormat:@"count of objects (%lu) differs from count of keys (%lu)", (unsigned long)objects.count, (unsigned long)keys.count];

    @throw [NSException exceptionWithName:NSInvalidArgumentException
                                   reason:reason
                                 userInfo:nil];
}

NSMutableDictionary *inner = [NSMutableDictionary dictionaryWithCapacity:keys.count];

for (NSUInteger index = 0 ; index < keys.count ; index++) {
    NSString *key = [keys[index] stringValue];
    NSString *object = objects[index];
    inner[key] = object;
}

NSDictionary *result = @{ @"dictionary": inner };

Footnotes

Because of I am very new to Objective-C, I did intentionally avoid:

• Blocks & safer ways to enumerate
• Lightweight generics
• Nullability stuff

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