map一个数组是大部分高级语言都有的, OC 没有, 有几个方案让它实现, 我优选出三个:
原生实现
其实就是valueForKeyPath的活用:
NSArray *names = [allEmployees valueForKeyPath: @"[collect].{daysOff<10}.name"];
NSArray *albumCovers = [records valueForKeyPath:@"[collect].{artist like 'Bon Iver'}.<NSUnarchiveFromDataTransformerName>.albumCoverImageData"];
category
这个大家肯定早想到过了, 你没有, 我给你扩展出来一个:
定义:
@interface NSArray (Map)
- (NSArray *)mapObjectsUsingBlock:(id (^)(id obj, NSUInteger idx))block;
@end
@implementation NSArray (Map)
- (NSArray *)mapObjectsUsingBlock:(id (^)(id obj, NSUInteger idx))block {
NSMutableArray *result = [NSMutableArray arrayWithCapacity:[self count]];
[self enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
[result addObject:block(obj, idx)];
}];
return result;o
}
@end
使用:
NSArray *people = @[
@{ @"name": @"Bob", @"city": @"Boston" },
@{ @"name": @"Rob", @"city": @"Cambridge" },
@{ @"name": @"Robert", @"city": @"Somerville" }
];
// per the original question
NSArray *names = [people mapObjectsUsingBlock:^(id obj, NSUInteger idx) {
return obj[@"name"];
}];
// (Bob, Rob, Robert)
// you can do just about anything in a block
NSArray *fancyNames = [people mapObjectsUsingBlock:^(id obj, NSUInteger idx) {
return [NSString stringWithFormat:@"%@ of %@", obj[@"name"], obj[@"city"]];
}];
// (Bob of Boston, Rob of Cambridge, Robert of Somerville)
三方库
是的, 一般简单功能能自己实现就自己实现, xcode 项目还是不能像 nodejs项目一样, 哪怕有的包里也只有一句话, 我也要从用第三方的...ddddd
js 的世界里, underscore用来处理数组可算神器, 自然, 我也挑中了同样名字的 OC 库: Underscore
NSArray *tweets = Underscore.array(results)
// Let's make sure that we only operate on NSDictionaries, you never
// know with these APIs ;-)
.filter(Underscore.isDictionary)
// Remove all tweets that are in English
.reject(^BOOL (NSDictionary *tweet) {
return [tweet[@"iso_language_code"] isEqualToString:@"en"];
})
// Create a simple string representation for every tweet
.map(^NSString *(NSDictionary *tweet) {
NSString *name = tweet[@"from_user_name"];
NSString *text = tweet[@"text"];
return [NSString stringWithFormat:@"%@: %@", name, text];
})
.unwrap;
当然, 所有方案来源于StackOverflow 上的答案, 一些其它方案, 其它库(如BlocksKit), 都可以试试, 也挺简洁的
