objective-c - 加速循环以创建长字符串

Question

我通过在四个不同的数组中格式化 NSNumber 来创建一个非常长的字符串:

NSString *datos = @"";

for (NSInteger counter = 0; counter < [latOut count]; counter++) {
    datos = [datos stringByAppendingString:[NSString stringWithFormat:@"%.0005f,", [[latOut objectAtIndex:counter] floatValue]]];
    datos = [datos stringByAppendingString:[NSString stringWithFormat:@"%.0005f,", [[lonOut objectAtIndex:counter] floatValue]]];
    datos = [datos stringByAppendingString:[NSString stringWithFormat:@"%ld,", [[tipoOut objectAtIndex:counter] integerValue]]];
    datos = [datos stringByAppendingString:[NSString stringWithFormat:@"%ld\n", [[velocidadOut objectAtIndex:counter] integerValue]]];
}

NSString *curDir = [[NSFileManager defaultManager] currentDirectoryPath];

NSString *path = @"";
path = [path stringByAppendingPathComponent:curDir];
path = [path stringByAppendingPathComponent:@"data.csv"];

// Y luego lo grabamos
[datos writeToFile:path atomically:YES encoding:NSASCIIStringEncoding error:&error];

计数为 18,000 个条目，此循环需要近 2 分钟才能完成。

我怎样才能让它更快？

Answer 1

由于您经常使用字符串，因此我在这里看到的主要建议是使用 NSMutableString ，它应该更高效:

// give a good estimate of final capacity
NSMutableString *datos = [[NSMutableString alloc] initWithCapacity:100000];

for (NSInteger counter = 0; counter < [latOut count]; counter++) {
  [datos appendFormat:@"%.0005f", [[latOut objectAtIndex:counter] floatValue]];
  ...
}

这将避免大量不必要的临时不可变字符串分配。