NSDateComponents 时间配置文件 [英] NSDateComponents Time Profile

问题描述

我的问题与下面代码片段中给出的 NSDateComponents 操作的时间配置文件有关.

My problem is to do with the Time Profile of NSDateComponents operations given in the code snippet below.

• 我有一个方法可以遍历大量 NSDate 对象
• 部分方法要求我丢弃每个日期的小时、分钟和秒.

• 为此，我有 3 个步骤:

• I have a method that iterates over a large array of NSDate objects
• Part of the method requires that I discard the Hour, Minute and seconds of each date.

• To do this I have 3 steps:

1. 我从原始日期创建了一个 NSDateComponents 对象.
2. 我将 H:M:S 元素设置为 0
3. 我根据需要将 H:M:S 设置为 0 的新日期对象

问题:

步骤 1 &以上 3 个在我的整体方法执行时间中占据了很大比例，分别为 22.4% 和 56.8%.

Steps 1 & 3 above are taking a huge percentage of my overall method execution time, 22.4% and 56.8% respectively.

我正在寻找有关如何优化这部分代码的建议或将 H:M:S 归零的替代方法，这些方法可能会表现得更好.

I'm looking for suggestions as to how I might optimize this part of my code or alternative ways of zeroing the H:M:S that might perform better.

NSDate* date = [[NSDate alloc] init];
// Next line takes 22.4% of the overall method execution time
NSDateComponents* components =  [calendar components:(NSYearCalendarUnit | NSMonthCalendarUnit 
                                | NSDayCalendarUnit | NSHourCalendarUnit  
                                | NSSecondCalendarUnit) fromDate:date]; 

[components setHour:0];
[components setMinute:0];
[components setSecond:0];

// Next line takes 56.8% of the overall method execution time
date = [calendar dateFromComponents:components];

推荐答案

更新

时钟:

时间 1:≈ 0.000139 - 选项 1 - 原始

Time 1: ≈ 0.000139 - Option 1 - Original

时间 2:≈ 0.000108 - 选项 2

Time 2: ≈ 0.000108 - Option 2

时间 3:≈ 0.000013 - 选项 3

Time 3: ≈ 0.000013 - Option 3

时间 4:≈ 0.000004 - 选项 4

Time 4: ≈ 0.000004 - Option 4

NSDate* originDate = [[NSDate alloc] init];
NSDate* date = [[NSDate alloc] init];
NSCalendar *calendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents* components =  [calendar components:(NSYearCalendarUnit | NSMonthCalendarUnit
                                                      | NSDayCalendarUnit | NSHourCalendarUnit
                                                      | NSSecondCalendarUnit) fromDate:date];

[components setHour:0];
[components setMinute:0];
[components setSecond:0];
date = [calendar dateFromComponents:components];
NSLog(@"Time 1: %f", -[originDate timeIntervalSinceNow]);

选项 2

你不需要

[components setHour:0];
[components setMinute:0];
[components setSecond:0];

甚至不要将这些添加为日期组件的选项

Just don't even add these as options to date components

NSDate* originDate2 = [[NSDate alloc] init];
NSDate* date2 = [[NSDate alloc] init];
NSCalendar *calendar2 = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];

NSDateComponents* components2 =  [calendar2 components:(NSYearCalendarUnit | NSMonthCalendarUnit
                                                        | NSDayCalendarUnit
                                                        ) fromDate:date2];
date2 = [calendar2 dateFromComponents:components2];
NSLog(@"Time 2: %f", -[originDate2 timeIntervalSinceNow]);

来源:类似问题 + David 的评论

Sources: Similar Question + David's Comment

基本上，我发现如果我使用已经创建的 NSCalendar 运行该方法，我的速度始终会提高 85% 到 90%.

Basically, I found that if I run the method with an already created NSCalendar, I get 85% to 90% faster speed consistently.

声明 calendarProperty 并在运行该方法之前对其进行初始化

Declare calendarProperty and initialize it before you run the method

NSDate* originDate3 = [[NSDate alloc] init];
NSDate* date3 = [[NSDate alloc] init];
NSDateComponents* components3 =  [calendarProperty components:(NSYearCalendarUnit | NSMonthCalendarUnit
                                                               | NSDayCalendarUnit
                                                               ) fromDate:date3];
date3 = [calendarProperty dateFromComponents:components3];
NSLog(@"Time 3: %f", -[originDate3 timeIntervalSinceNow]);

选项 4

* 基于@HotLicks 的评论 *

Option 4

* based on @HotLicks's comment *

同时避免所有那些烦人的 dateComponents - 大约快 95%

Avoid all those annoying dateComponents all together - about 95% faster

我不确定选项 4 在实践中的一致性如何，但它的计时速度最快，到目前为止对我来说是可靠的.

I'm not sure about how consistent Option 4 is in practice, but it's clocking in the fastest and it has been reliable for me so far.

NSDate* originDate4 = [[NSDate alloc] init];
NSDate* date4 = [[NSDate alloc] init];
float date4Interval = [date4 timeIntervalSince1970];
int totalDays = date4Interval / (60 * 60 * 24);
date4 = [NSDate dateWithTimeIntervalSince1970:totalDays * 60 * 60 * 24];
NSLog(@"Time 4: %f", -[originDate4 timeIntervalSinceNow]);

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