可以在目标 c 中使用 swift 中的数组减少概念吗？

Question

我在 Swift 中有这行代码：

let graphPoints:[Int] = [4, 2, 6, 4, 5, 8, 3]

let average = graphPoints.reduce(0, combine: +) / graphPoints.count

是否可以将这行代码“翻译”成目标 c 代码？

reduce combine 概念的工作原理对我来说不是很清楚。我读过它，但仍然不清楚。

我从本教程中获取了代码：http://www.raywenderlich.com/90693/modern-core-graphics-with-swift-part-2

请帮忙。谢谢。

Answer 1

假设您有一些 NSNumbers 存储在 NSArray 中，您可以使用这个 KVC 集合运算符：

NSArray *someNumbers = @[@0, @1.1, @2, @3.4, @5, @6.7];
NSNumber *average = [someNumbers valueForKeyPath:@"@avg.self"];

Answer 2

对于 Objective-C，我会将高阶函数添加到此答案列表中：https://github.com/fanpyi/Higher-Order-Functions

#import <Foundation/Foundation.h>
typedef id (^ReduceBlock)(id accumulator,id item);
@interface NSArray (HigherOrderFunctions)
-(id)reduce:(id)initial combine:(ReduceBlock)combine;
@end

#import "NSArray+HigherOrderFunctions.h"
@implementation NSArray (HigherOrderFunctions)
-(id)reduce:(id)initial combine:(ReduceBlock)combine{
    id accumulator = initial;
    for (id item in self) {
        accumulator = combine(accumulator, item);
    }
    return accumulator;
}
@end

示例：

   NSArray *numbers = @[@5,@7,@3,@8];
    NSNumber *sum = [numbers reduce:@0 combine:^id(id accumulator, id item) {
        return @([item intValue] + [accumulator intValue]);
    }];
    NSNumber *multiplier = [numbers reduce:@1 combine:^id(id accumulator, id item) {
        return @([item intValue] * [accumulator intValue]);
    }];
    NSLog(@"sum=%@,multiplier=%@",sum,multiplier);

Answer 3

reduce 函数在 Objective-C 中不是标准的。不过，您可以将其实现为 NSArray 的扩展。

在您的情况下，您在 Swift 中有一个 Int 数组。你不能在 Objective-C 中拥有它，你需要一个 NSNumber 的数组。

这是reduce 的实现，应该适用于您的情况：

@implementation NSArray (Helpers)

- (NSInteger)reduceInt:(NSInteger)initial combine:(NSInteger (^)(NSInteger acum, NSInteger element))block {
    if (!self) {
        return initial;
    }

    NSInteger acum = initial;
    for (id element in self) {
        if ([element isKindOfClass:[NSNumber class]]) {
            acum = block(acum, [(NSNumber *)element integerValue]);
        }
    }

    return acum;
}

@end

您可以将它与您的数组一起使用，如下所示：

NSArray *a = @[@1, @2, @3];
NSInteger result = [a reduceInt:0 combine:^NSInteger(NSInteger acum, NSInteger element) {
    return acum + element;
}];

Answer 4

如何将 reduce 转换为 ObjC（或者更好地说是如何解决 Objective C 中的“普通问题”）由 André Slotta 完美回答。快速减少远不止于此。我会试着回答你问题的第二部分，这个概念是如何在 swift 中起作用的

func reduce<T>(initial: T, @noescape combine: (T, Self.Generator.Element) throws -> T) rethrows -> T  

返回结果 重复调用组合与初始化为的累积值 初始和 self 的每个元素，依次，即返回 结合（结合（...结合（结合（初始，自我[0]）， self[1]),...self[count-2]), self[count-1]).

let arr: Array<Int> = [1,2,3,4,5]
let sum = arr.reduce(0) { (sum, i) -> Int in
    return sum + i
}
print(sum) // 15

// this is an quasi equivalent of
var sum1 = 0 // ..... reduce(0)....
arr.forEach { (elementValue) -> Void in
    sum1 = sum1 + elementValue   // ...{ return sum + i }
}
print(sum1) // 15 reduce function will return accumulated inital value

// reduce is part of SequenceType protocol, that is why

let arr1 = ["H","e","l","l","o"," ","w","o","r","l","d"]
let str = arr1.reduce("") { (str, s) -> String in
    str + s
}
// works the same way
print(str) // "Hello world"

// let have a litle bit more complex example, to see how powerful, useful and easy to use reduce can be
let dict = arr1.reduce([:]) { (var dict, s) -> Dictionary<Int,String> in
    let i = dict.count
    dict.updateValue(s, forKey: i+1)
    return dict
}
print(dict) // [11: "d", 10: "l", 2: "e", 4: "l", 9: "r", 5: "o", 6: " ", 7: "w", 3: "l", 1: "H", 8: "o"]

Answer 5

编写一个 NSArray 扩展

- (NSInteger)reduceStart:(NSInteger)start combine:(NSInteger(^)(NSInteger x, NSInteger y))combine {
    for (NSNumber* n in self) {
        if ([n isKindOfClass:[NSNumber class]]) {
            start = combine (start, n.integerValue);
        }
    }
    return start;
}

纠正我犯的所有错误，就是这样。只是不如 Swift 灵活。